Division algebras graded by a finite group

TL;DR

This paper characterizes and constructs division algebras with finite group gradings over fields of characteristic zero, linking algebraic data to gradings and classifying certain G-simple algebras.

Contribution

It provides a converse construction linking group gradings on division algebras to algebraic data, and classifies G-simple algebras admitting division algebra forms.

Findings

Associates gradings with abelian subgroups, integers, and homomorphisms from Schur multipliers.

Main theorem constructs division algebras from given algebraic data.

Classifies G-simple algebras with division algebra forms over algebraically closed fields.

Abstract

Let $k$ be a field containing an algebraically closed field of characteristic zero. If $G$ is a finite group and $D$ is a division algebra over $k$, finite dimensional over its center, we can associate to a faithful $G$-grading on $D$ a normal abelian subgroup $H$, a positive integer $d$ and an element of $Hom(M(H), k^\times)^G$, where $M(H)$ is the Schur multiplier of $H$. Our main theorem is the converse: Given an extension $1\rightarrow H\rightarrow G\rightarrow G/H\rightarrow 1$, where $H$ is abelian, a positive integer $d$, and an element of $Hom(M(H), k^\times)^G$, there is a division algebra with center containing $k$ that realizes these data. We apply this result to classify the $G$-simple algebras over an algebraically closed field of characteristic zero that admit a division algebra form over a field containing an algebraically closed field.

Equations

$$l{u}_{\sigma }y{u}_{\tau }=l\sigma (y)u_{\sigma \tau }.$$

$$l{u}_{\sigma }y{u}_{\tau }=l\sigma (y)\alpha (\sigma ,\tau )u_{\sigma \tau }.$$

$$1\to H\to G\to G/H\to 1,$$

$$\beta :1\to H\to G\to Q\to 1$$

$$1\to H\to Q_8\to Q_8/H\to 1$$

$$1\to H\to D_4\to D_4/H\to 1$$

$$A_g=\text{mbox}{span}_F\left\{u_h\otimes e_{i,j}:g=g_i^{-1}h{g}_j\right\}.$$

$$y_g^{-1}{x}_h{y}_g{y}_g^{-1}{x}_{h^{\prime }}{y}_g=y_g^{-1}\alpha (h,h^{\prime })y_g{y}_g^{-1}{x}_{h{h}^{\prime }}{y}_g.$$

$${\gamma }_h{x}_{g^{-1}hg}{\gamma }_{h^{\prime }}{x}_{g^{-1}{h}^{\prime }g}=y_g^{-1}\alpha (h,h^{\prime })y_g{\gamma }_{h{h}^{\prime }}{x}_{g^{-1}h{h}^{\prime }g}$$

$${\gamma }_h{\gamma }_{h^{\prime }}\alpha (g^{-1}hg,g^{-1}{h}^{\prime }g)x_{g^{-1}h{h}^{\prime }g}=y_g^{-1}\alpha (h,h^{\prime })y_g{\gamma }_{h{h}^{\prime }}{x}_{g^{-1}h{h}^{\prime }g}.$$

$${\gamma }_h{\gamma }_{h^{\prime }}\alpha (g^{-1}hg,g^{-1}{h}^{\prime }g)=y_g^{-1}\alpha (h,h^{\prime })y_g{\gamma }_{h{h}^{\prime }},$$

$$y{z}^{p^{s-r_1}}{y}^{-1}=(z^{\alpha +1}{)}^{p^{s-r_1}}=z^{p^{s-r_1}(\alpha +1)}.$$

$$x=y^{p^r}x{y}^{-p^r}=z^{{(\alpha +1)}^{p^r-1}}{z}^{{(\alpha +1)}^{p^r-2}}\cdots z^{(\alpha +1)}z\cdot x$$

$$M_1=((\alpha +1{)}^{p^r}-1)/\alpha =p^r+(p^r(p^r-1)/2)\alpha +a_3{\alpha }^2+\cdots +p^r{\alpha }^{p^r-2}+1{\alpha }^{p^r-1}$$

$$M_1\equiv p^r+(p^r(p^r-1)/2)\alpha \mathrm{mod}(p^{r+1}).$$

$$z={\alpha }_e{u}_e+{\alpha }_{g_2}{u}_{g_2}+\cdots +{\alpha }_{g_n}{u}_{g_n}.$$

$$g_1{g}_2(x_{s_0})=g_1(g_2(x_{s_0}))$$

$$\beta (g_1{g}_2)x_{s_0}$$

$$g_1(\beta (g_2)x_{s_0})=g_1(\beta (g_2))\beta (g_1)x_{s_0}$$

$$z(t^{-1}{x}_{s_0})=\sum _{g\in G}{y}_g{t}^{-1}{x}_{s_0}{y}_g^{-1}=\sum _{g\in C_G(s_0)}{y}_g{t}^{-1}{x}_{s_0}{y}_g^{-1}+\sum _{g\notin C_G(s_0)}{y}_g{t}^{-1}{x}_{s_0}{y}_g^{-1}=$$

$$\sum _{g\in C_G(s_0)}g(t^{-1}{x}_{s_0})+\sum _{g\notin C_G(s_0)}{y}_g{t}^{-1}{x}_{s_0}{y}_g^{-1}=\sum _{g\in C_G(s_0)}{t}^{-1}{x}_{s_0}+\sum _{g\notin C_G(s_0)}{y}_g{t}^{-1}{x}_{s_0}{y}_g^{-1}=$$

$$t^{-1}|C_G(s_0)|x_{s_0}+\sum _{g\notin C_G(s_0)}{y}_g{t}^{-1}{x}_{s_0}{y}_g^{-1},$$

$$\beta :1\to H\to G\to Q\to 1$$

$$1\to R\to F\to G\to 1$$

$$\beta :1\to N=R/[R,R]\to \Gamma =F/[R,R]\to G\to 1.$$

$$1\to R\to F_H\to H\to 1$$

$$1\to R/[R,F_H]\to F_H/[R,F_H]\to H\to 1.$$

$$(R/[R,F_H])/((R\cap [F_H,F_H])/[R,F_H])\cong R/(R\cap [F_H,F_H])$$

$$\beta :1\to U\to {\Gamma }_H\to {\Gamma }_H/U\to 1$$

Full text

Our main theorem is the converse: Given an extension $1\rightarrow H\rightarrow G\rightarrow Q\rightarrow 1$, where $H$ is abelian, a positive integer $d$, and an element of $Hom(M(H),\mu_{n_{H}})^{Q}$, there is a division algebra as above that realizes these data. We apply this result to classify the $G$-graded simple algebras whose $e$-center is an algebraically closed field of characteristic zero that admit a division algebra form whose $e$-center contains $\mu_{n_{G}}$.

The first author was partially supported by the ISRAEL SCIENCE FOUNDATION (grant No. 1516/16).

Much of this work was done while the second author was a visitor at Guangtong-Technion Israel Institute of Technology. He would like to thank the institute for its support

1. introduction

If $A$ is an algebra over a field $k$ and $G$ is a finite group, we say $A$ is $G$-graded if there is a decomposition $A=\oplus_{g\in G}A_{g}$ of $A$ into a direct sum of $k$-vector spaces such that $A_{g_{1}}A_{g_{2}}\subset A_{g_{1}g_{2}}$ for all $g_{1},g_{2}\in G$. We say the grading is faithful if $A_{g}\not=0$ for all $g\in G$. In this paper we are interested in the case where $A=D$ is a division algebra over $k$, finite dimensional over its center $K$ and faithfully graded by a finite group $G$. We assume throughout that the field $k$ is of characteristic zero and that, unless otherwise stated, $k$ contains the group $\mu_{n_{G}}$ of $n_{G}$-th roots of unity, where $n=exp(G)$.

Remark 1.1*.*

We emphasize that we are not assuming there is a field $F$ containing $k$ such that $D$ is both $G$-graded and finite dimensional over $F$. In fact the existence of such a field will be a consequence of our results (see Theorem 1.5).

We begin with a brief description of our results followed by a more detailed discussion: Writing $D=\oplus_{g\in G}D_{g}$ we see easily that $D_{e}$ is a division algebra of degree $d$ (say) over its center $L$, and $k=k\cdot 1\subseteq D_{e}$ (However it is not necessarily the case that the center $K$ of $D$ is contained in $D_{e}$ or that $K$ is even a graded subalgebra of $D$). The group $G$ acts on the center $L$ of $D_{e}$ and we let $H$ denote the kernel of the action. We obtain in this way an extension $1\rightarrow H\rightarrow G\rightarrow G/H\rightarrow 1$ and a two-cocycle $\alpha$ with class in $H^{2}(H,L^{*})$. It turns out (but is not trivial) that the condition that $k$ contains $\mu_{n_{G}}$ forces $H$ to be abelian. Moreover the class of $\alpha$ is $G/H$-invariant, that is, fixed by the natural action of $G/H$ on $H^{2}(H,L^{*})$ and gives rise to a $G/H$-invariant element in $Hom(M(H),\mu_{n_{H}})$, where $n_{H}=exp(H)$ and $M(H)$ is the Schur multiplier of $H$.

Our main result (Theorem 1.8) is the converse: Given an extension of groups $1\rightarrow H\rightarrow G\rightarrow Q\rightarrow 1$ where $H$ is abelian, a positive integer $d$, and a $Q$-invariant element of $Hom(M(H),\mu_{n_{H}})$, where the action of $Q$ on $\mu_{n_{H}}$ is trivial, there is a field $k$ containing $\mu_{n_{H}}$ and a division algebra $D$ over $k$, faithfully graded by the group $G$ and giving rise to the prescribed data. We apply our result to classify the $G$-graded simple algebras over an algebraically closed field of characteristic zero that admit a division algebra form faithfully $G$-graded over a field $k$ containing $\mu_{n_{G}}$ (see Theorem 1.12).

Remark 1.2*.*

As the reader may have noted two subgroups of the group of roots of unity will play an important role in the paper, namely $\mu_{n_{G}}$ where $n=exp(G)$ and $\mu_{n_{H}}$ where $n=exp(H)$. As mentioned above, a $G$-grading on a division algebra $D$ over $k$, finite dimensional over its center gives rise to a certain normal subgroup $H$ of $G$, and the existence of $\mu_{n_{G}}$ in $k$ forces the group $H$ to be abelian. Also, the $G$-grading on $D$ gives rise to an element in $Hom(M(H),L^{*})$ and it is well known (and easy to prove) that if $H$ is abelian $\exp(M(H))\mid exp(H)$. It follows that because $\mu_{n_{H}}\leq k^{*}\leq L^{*}$, the natural inclusion $Hom(M(H),\mu_{n_{H}})\hookrightarrow Hom(M(H),L^{*})$ is in fact onto.

We proceed to the more detailed description. We begin the paper by determining the general structure of gradings by the group $G$ on $K$-central division algebras (so $K$ contains the base field $k$ which is assumed to contain $\mu_{n_{G}}$) and break the analysis into two cases:

1. (1)

The case in which the center $K$ is contained in the $e$-homogeneous component. 2. (2)

The general case, in which the center $K$ may not be contained in $D_{e}$, and may not be $G$-graded.

There are three types of gradings on division algebras. Our first two theorems (for the two cases mentioned above) will show that any grading by a finite group is a combination of these three types:

1. (1)

The trivial grading: For $D$ any finite dimensional division algebra and $G=\{1\}$. 2. (2)

The case of twisted group algebras: Let $H$ be a finite group and consider a twisted group algebra $L^{\alpha}H$ where $\alpha$ is a $2$-cocycle $H\times H\rightarrow L^{*}$, and where the action of $H$ on $L^{*}$ is trivial. Because $char(L)=0$ the twisted group algebra is semisimple but in general not simple. It seems to be difficult to determine the finite groups $H$ which admit a $2$-cocycle over an aribitrary field $L$ of characteristic zero such that the twisted group algebra $L^{\alpha}H$ is a division algebra. In our case, in which the base field contains $\mu_{n_{H}}$, we will see that the group $H$ must be abelian (Proposition 2.1). It is easy to show that for any abelian group $H$ we can find a cocycle $\alpha$ and a field $L$ such that the twisted group algebra is a field and so in particular a finite dimensional division algebra. The case where $L^{\alpha}H$ is central over $L$ is of special interest. These are obtained precisely when $H$ is abelian of the form $A\times A$ and the cocycle is a suitable nondegenerate cocycle on $H$ (see Remark 1.3 below). 3. (3)

The crossed product grading: If $L/K$ is a Galois extension with $G$ as Galois group, we can construct the skew group algebra $L_{t}G$. As a left $L$-vector space , $L_{t}G$ is isomorphic to the group algebra $LG$. We denote its elements by $\sum_{\sigma\in G}l_{\sigma}u_{\sigma}$, where $l_{\sigma}\in L$ and $u_{\sigma}$ is a symbol in $L_{t}G$ corresponding to the element $\sigma\in G$. The product is defined using distributivity and so as to satisfy the condition

[TABLE]

It is well known that the algebra $L_{t}G$ is isomorphic to $M_{n}(K)$ where $n=ord(G)$. In particular the skew group algebra structure gives a $G$-grading on $M_{n}(K)$. Now we can twist the multiplication by a $2$-cocycle on $G$ with coefficients in $L^{*}$, where the action of $G$ on $L^{*}$ is induced by the Galois action on $L$. If $\alpha:G\times G\rightarrow L^{*}$ is such a $2$-cocycle we construct the crossed product algebra, denoted by $L_{t}^{\alpha}G$, which again is isomorphic to $LG$ and $L_{t}G$ as a left vector space over $L$ with the product determined by the formula

[TABLE]

Cohomologous $2$-cocycles in $H^{2}(G,L^{*})$ yield $G$-graded isomorphic algebras over $K$ and in particular isomorphic algebras over $K$. On the other hand, noncohomologous $2$-cocycles yield nonisomorphic $K$-central simple algebras and in particular nonisomorphic $G$-graded algebras.

The crossed product $L_{t}^{\alpha}G$ is $K$-central simple and is a $G$-graded twisted form of $L_{t}G$. It is well known (we provide a proof below) that for any finite group $G$ one can find a $G$-Galois extension $L/K$ and a cocycle $\alpha$ such that $L_{t}^{\alpha}G$ is a $K$-central division algebra. Moreover one can find such crossed product division algebras for which the center contains an algebraically closed field of characteristic zero and in particular contains $\mu_{n_{G}}$. Note that here the $e$-component is $L$, a maximal subfield in $L_{t}^{\alpha}G$.

Much more generally a finite dimensional algebra $A$ faithfully graded by the group $G$ is called a ring theoretic $G$-crossed product if every homogeneous component contains an invertible element. The standard notation is $A_{e}*G$, where $A_{e}$ denotes the $e$-component of the grading. Of course in this sense every division algebra faithfully graded by $G$ is a $G$-crossed product. The (classical) crossed product $L_{t}^{\alpha}G$ described above is a special case and the source of the name. Moreover if $H$ is a normal subgroup of $G$ then $A$ may also be viewed as a $G/H$-crossed product, where the $e$-component is now $A_{H}=\oplus_{h\in H}A_{h}$. So the crossed product $A_{e}*G$ is also a crossed product $A_{H}*G/H$. This kind of construction, originally due to Teichmuller, can be found in $($[15], [17] and [18]$)$. We will see that in the case of a division algebra faithfully graded by a group $G$ over a field containing $\mu_{n_{G}}$, we can say much more: In particular we can choose the normal subgroup $H$ to be abelian with the division subalgebra $D_{H}$ a twisted group algebra over $D_{e}$ (see case (2) above).

Remark 1.3*.*

Classically, a finite group $G$ is said to be of central type if it has a complex irreducible representation of degree $[G:Z]^{1/2}$ where $Z=Z(G)$ is the center of $G$. In particular the index of $Z$ in $G$ must be a square. Because a linear representation of $G$ gives rise to a $2$-cocycle $\alpha$ on $G/Z$ with values in $\mathbb{C}^{*}$ and to a projective representation of $G/Z$ corresponding to $\alpha$, researchers borrowed that terminology and defined (nonclassically) a group $\Lambda$ to be of central type if $\Lambda$ admits a $2$-cocycle $\alpha\in Z^{2}(\Lambda,\mathbb{C}^{*})$ such that the twisted group algebra $\mathbb{C}^{\alpha}\Lambda\cong M_{n}(\mathbb{C})$, where $n^{2}=ord(\Lambda)$. Such a cocycle $\alpha$ is called nondegenerate. In this article we slightly extend the terminology and say that a $2$-cocycle $Z^{2}(\Lambda,L^{*})$ is nondegenerate if the twisted group algebra $L^{\alpha}\Lambda$ is an $L$-central simple algebra.

We can now state our first structure theorem.

Theorem 1.4**.**

Let $D$ be a division algebra, finite dimensional over its center $K$. Suppose $D$ is faithfully graded by a finite group $G$. Let $D_{e}$ be the identity component of $D$ and denote by $L$ its center. Assume $K$ is contained in $D_{e}$ and that $K$ contains $\mu_{n_{G}}$, $n_{G}=exp(G)$. Then the following conditions hold:

1. (1)

Conjugation on $D_{e}$ by nonzero homogeneous elements $x_{g}\in D_{g}$ determines a $G$-action on $L$, the center of $D_{e}$. Furthermore, $K=L^{G}$ and hence if we denote by $H$ the kernel of the action, $L/K$ is a $G/H$-Galois extension. 2. (2)

The group $H$ is abelian and is isomorphic to $A\times A$, for some abelian group $A$. 3. (3)

There exists a $2$-cocycle $\alpha\in Z^{2}(H,L^{*})$ such that $D_{H}=\sum_{h\in H}D_{h}\cong D_{e}\otimes_{L}L^{\alpha}H$. Furthermore, the cocycle $\alpha$ is nondegenerate, that is $L^{\alpha}H$ is an $L$-central simple algebra. The subalgebras $L^{\alpha}H$ and $D_{H}$ have center $L$. 4. (4)

The cohomology class $[\alpha]\in H^{2}(H,L^{*})$ is $G/H$-invariant. 5. (5)

The algebra $D$ is a $G/H$-crossed product. Its degree is $d\sqrt{|H|}[G:H]$, where $d=\sqrt{dim_{L}(D_{e})}$.

Our second structure theorem deals with the general case; we drop the assumption that $K$, the center of $D$, is contained in the identity component and do not assume $K$ is $G$-graded.

Theorem 1.5**.**

Let $D$ be a division algebra, finite dimensional over its center $K$. Suppose $D$ is faithfully graded by a finite group $G$. Let $D_{e}$ be the identity component of $D$ and denote by $L$ its center. Let $K_{0}=K\cap L$ denote the central $e$-homogeneous elements of $D$. Assume $K_{0}$ contains $\mu_{n_{G}}$, $n_{G}=exp(G)$. With these conditions the following hold:

1. (1)

Conjugation on $D_{e}$ with nonzero homogeneous elements $x_{g}\in D_{g}$ determines a $G$-action on $L$ with $K_{0}=L^{G}$. If we denote by $H$ the kernel of the action, then $L/K_{0}$ is a $G/H$-Galois extension. 2. (2)

The algebra $D$ is finite dimensional over $K_{0}$. 3. (3)

The group $H$ is abelian. 4. (4)

There exists a $2$-cocycle $\alpha\in Z^{2}(H,L^{*})$ such that $D_{H}=\sum_{h\in H}D_{h}\cong D_{e}\otimes_{L}L^{\alpha}H$. Hence, if we denote $L_{1}=Z(L^{\alpha}H)$, then $L_{1}$ is also the center of the division algebra $D_{H}$. 5. (5)

Conjugation on $D_{H}$ by nonzero homogeneous elements induces an action of $G$ on $L_{1}$ with fixed field $K$, the center of $D$. Furthermore, the extension $L_{1}/K$ is $G/H$-Galois. 6. (6)

The cohomology class $[\alpha]\in H^{2}(H,L^{*})$ is $G/H$ invariant. If we let $S=\{s\in H:x_{s}\in L_{1}\}\ \ (=\{s\in H:\alpha(s,h)=\alpha(h,s)\ \ \hbox{for all}\ \ h\in H\})$, then $S$ is a subgroup of $H$ and $L_{1}=L^{\alpha}S$. The group $H/S$ is of central type, hence isomorphic to $A\times A$ for some abelian group $A$. 7. (7)

The division algebra $D$ is a $G/H$-crossed product over the algebra $D_{H}$. Its degree is $d\sqrt{[H:S]}[G:H]$, where $d=\sqrt{dim_{L}(D_{e})}$.

In the next theorem we determine under what circumstances the center $K$ (in Theorem 1.5 above) is $G$-graded. By statement (5) of the theorem, $K$ is a subfield of $D_{H}$ and hence if $K$ is $G$-graded, it is necessarily $H$-graded.

We continue the notation as in Theorem 1.5. For every $h\in H$ we let $x_{h}$ be a representative in $L^{\alpha}H$ (i.e., any nonzero element in $(L^{\alpha}H)_{h}$). Let S=\{h\in H:x_{h}x_{h^{\prime}}=x_{h^{\prime}}x_{h}\ \for all$\ \ h^{\prime}\in H\}$, that is, $S$ is the set of elements in $H$ such that $x_{s}\in L_{1}$, the center of $L^{\alpha}H$. We claim that in fact $L_{1}=L^{\alpha}S$: It is clear that $L^{\alpha}S\subseteq L_{1}$. Conversely if $z=\sum_{h\in H}l_{h}x_{h}$ lies in $L_{1}$, then for all $r\in H$, $x_{r}z=zx_{r}$. But $H$ is abelian and $x_{r}$ commutes with the elements of $L$. It follows that if $l_{h}\not=0$ then $h\in S$.

Note that because $G$ acts on $L_{1}$, it acts also on $S$, so $S$ is a normal subgroup of $G$.

Theorem 1.6**.**

The center $K$ is $G$ graded if and only if $S$ is central in $G$. Moreover if $S$ is central in $G$ then $K$ is a twisted group algebra $K_{0}^{\tilde{\alpha}}S$, where $\tilde{\alpha}:S\times S\rightarrow K_{0}^{*}$ is a two-cocycle cohomologous to $\alpha$ in $H^{2}(S,L^{*})$.

The most substantial part of this paper is devoted to proving a converse to the structure theorems. To this end we adopt the following terminology:

Let $D$ be a finite dimensional division algebra with center $K$ and suppose $D$ is faithfully graded by a finite group $G$. Assume the $e$-center $K_{0}$ of $D$ contains a primitive $n_{G}$-th root of unity where $n_{G}$ is the exponent of $G$. By the structure theorems this gives rise to

1. (1)

a group extension with abelian kernel

[TABLE] 2. (2)

a cohomology class $[\alpha]\in H^{2}(H,L^{*})^{G/H}$, where $L=Z(D_{e})$,

and 3. (3)

a positive integer $d$, the degree of $D_{e}$ over its center.

Let $M(H)$ denote the Schur multiplier of $H$ and let $\phi_{[\alpha]}\in Hom(M(H),L^{*})$ be the map corresponding to $[\alpha]$ by means of the Universal Coefficients Theorem. By naturality, the map $\phi_{[\alpha]}$ is $G$-invariant. Furthermore, because the values of $\phi_{[\alpha]}$ are contained in the group $\mu_{n_{H}}$, and these are assumed to be in $K_{0}$, we have $\phi_{[\alpha]}\in Hom(M(H),\mu_{n_{H}})^{G/H}$, where the action of $G$ on $\mu_{n_{H}}$ is trivial. Because $H$ is abelian the Schur multiplier can be identified with the wedge product $H\wedge H$ and if $h_{1},h_{2}\in H$, then $\phi_{[\alpha]}(h_{1}\wedge h_{2})=\alpha(h_{1},h_{2})\alpha(h_{2},h_{1})^{-1}$. In the twisted group algebra $L^{\alpha}H=\sum_{h\in H}Lx_{h}$ one sees easily that for all $h_{1},h_{2}\in H$, the commutator $x_{h_{1}}x_{h_{2}}x_{h_{1}}^{-1}x_{h_{2}}^{-1}$ is a root of unity and we also have $\phi_{[\alpha]}(h_{1}\wedge h_{2})=x_{h_{1}}x_{h_{2}}x_{h_{1}}^{-1}x_{h_{2}}^{-1}$. Note that if $F$ is an algebraically closed field containing $K_{0}$ then in fact $H^{2}(H,F^{*})^{G/H}\cong Hom(M(H),\mu_{n_{H}})^{G/H}$, and hence it makes sense to speak of the nondegeneracy of $\phi_{[\alpha]}\in Hom(M(H),\mu_{n_{H}})^{G/H}$ (see Remark 1.3). Here, the action of $G$ on $F^{*}$ and on $\mu_{n_{H}}$ is the trivial action.

Letting $[\beta]\in H^{2}(G/H,H)$ denote the cohomology class determined by the group extension given above, we see that any finite dimensional division algebra faithfully graded by the group $G$ determines uniquely an ordered triple $([\beta],\phi,d)\in H^{2}(Q,H)\times Hom(M(H),\mu_{n_{H}})^{Q}\times\mathbb{N}$ where $Q=G/H$ and $\mathbb{N}=\{1,2,3,\dots\}$ denotes the set of natural numbers.

Definition 1.7**.**

Let $Q$ be a finite group and $H$ an abelian group. Let $n_{H}=exp(H)$. We say the triple $([\beta],\phi,d)\in H^{2}(Q,H)\times Hom(M(H),\mu_{n_{H}})^{Q}\times\mathbb{N}$ is realizable if there exists a finite dimensional division algebra, faithfully $G$-graded, over a field of characteristic zero containing $\mu_{n_{H}}$, that yields that triple. Here, the group $G$ is the group determined by the class $[\beta]$.

Here is the main result of the paper. Notation as above.

Theorem 1.8**.**

If $Q$ is a finite group and $H$ is an abelian group, then every ordered triple $([\beta],\phi,d)\in H^{2}(Q,H)\times Hom(M(H),\mu_{n_{H}})^{Q}\times\mathbb{N}$ is realizable. Moreover, it is realizable over a field containing an algebraically closed field of characteristic zero. If $\phi$ is nondegenerate then the center of every realization will lie in the $e$-component of its grading.

We will say that an extension

[TABLE]

with an abelian kernel is realizable if there exists a $Q$-invariant map $\phi:M(H)\rightarrow\mu_{n_{H}}$ such that the triple $([\beta],\phi,1)$ is realizable. We obtain the following corollary (see Cuadra and Etingof result ([10], Theorem $2.1$)).

Corollary 1.9**.**

Every extension

[TABLE]

with an abelian kernel is realizable by a division algebra of degree $[G:H]$.

Proof.

If $\phi$ is trivial then $\phi$ is $Q$-invariant and so by the theorem the triple $([\beta],[\phi],1)$ is realizable by a division algebra $D$. Moreover the $e$-component of $D$ is a field $L$ and, because $\phi$ is trivial, the $H$ component is a field extension $L_{1}$ of $L$. The group $Q$ acts faithfully on $L_{1}$ and the fixed field is the center $K$ of $D$. It follows that $D$ is a (classical) crossed-product algebra of degree $Q$. ∎

Example 1.10**.**

In [10] Cuadra and Etingof give an example of a finite dimensional division algebra graded by $Q_{8}$, the quaternion group of order $8$. In their example the center is not graded. In the following we describe the possible grading structure on division algebras over fields which contain a primitive $4$th root of unity, by the group $Q_{8}$ and by the group $D_{4}$, the dihedral group of order $8$. As above $D$ will denote the division algebra, $K$ its center and $K_{0}$ its $e$-center. We will assume $D_{e}=L$, so $d=1$ $($See the remarks at the beginning of section 3$)$ and $L_{1}$ denotes the center of the $H$-component.

Let

[TABLE]

1. (1)

*If $H=\{1\}$ we obtain a *$Q_{8}-$crossed product division algebra of degree $8$. 2. (2)

If $H\cong\mathbb{Z}_{2}$, the $e$-component $L$ of $D$ is a $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ Galois extension of the $e$-center $K_{0}$. The $H$-component $D_{H}$ is a field, so equal to $L_{1}$ and $K=L_{1}^{G/H}=K_{0}^{\alpha}H$ is graded, a quadratic extension of $K_{0}$. The division algebra $D$ is a $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-crossed product for the extension $L_{1}/K$. 3. (3)

*If $H\cong\mathbb{Z}_{4}$, the group $S$ of Theorem 1.5 equals $H$ $($because $H$ is cyclic$)$. The $e$-component $L$ is an extension of degree 2 over $K_{0}$ and the $H$-component is a field extension *$($so equal to L_{1}$$) of $L$ of degree 4. The center $K=L_{1}^{G/H}$ is ungraded, an extension of degree 4 over $K_{0}$. The division algebra $D$ is a quaternion algebra over $K$. This is the example of Cuadra and Etingof.

Now let

[TABLE]

1. (1)

If $H=\{1\}$ we obtain a $D_{4}$ crossed product division algebra of degree $8$. 2. (2)

If $H\cong\mathbb{Z}_{2}$ the $e$-component $L$ of $D$ is a $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ Galois extension of the $e$-center $K_{0}$. The $H$-component $D_{H}$ is a field, so equal to $L_{1}$ and $K=L_{1}^{G/H}=K_{0}^{\alpha}H$ is graded, a quadratic extension of $K_{0}$. The division algebra $D$ is a $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-crossed product for the extension $L_{1}/K$. 3. (3)

If $H\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and the map $\phi$ is trivial, then the $e$-component $L$ is an extension of degree 2 over $K_{0}$ and the $H$-component is a field (so equal to $L_{1}$), a biquadratic extension of $L$. The division algebra $D$ is a quaternion algebra over the nongraded center $K=L_{1}^{G/H}$ which is of degree 4 over $K_{0}$. Because the sequence is split, $D$ is isomorphic to $D_{0}\otimes_{K_{0}}K$, where $D_{0}$ is a quaternion algebra over $K_{0}$: Indeed, if $D_{4}=\{\sigma,\tau:\sigma^{4}=\tau^{2}=e,\sigma^{-1}\tau=\tau\sigma\}$ and $H=\{e,\sigma^{2},\tau,\sigma^{2}\tau\}$, the element $\sigma\tau$ is of order $2$ and acts nontrivially on $L$. Thus $L$ and $\sigma\tau$ generate a quaternion subalgebra $D_{0}$ over $K_{0}$ and $D\cong D_{0}\otimes_{K_{0}}K$. 4. (4)

If $H\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and the map $\phi$ is nontrivial, then $\phi$ is the only nontrivial element of $Hom(M(H),\mu_{2})$ and so $\phi$ is $G$-invariant. The $e$-component $L$ has degree 2 over $K_{0}$. The $H$-component of $D$ is a quaternion algebra with center $L$. Therefore $L_{1}=L$ and $K=K_{0}$. Because the sequence is split, $D\cong Q_{1}\otimes_{K}Q_{2}$ is a tensor product of two quaternion algebras over the center $K$. Indeed, if $D_{4}=\{\sigma,\tau:\sigma^{4}=\tau^{2}=e,\sigma^{-1}\tau=\tau\sigma\}$ and $H=\{e,\sigma^{2},\tau,\sigma^{2}\tau\}$, the element $\sigma\tau$ is of order $2$ and acts nontrivially on $L$. Thus it generates a quaternion subalgebra $Q_{1}$ over $K$ which can be factored from $D$. 5. (5)

If $H\cong\mathbb{Z}_{4}$, the group $S$ of Theorem 1.5 equals $H$ $($because $H$ is cyclic$)$. The $e$-component $L$ is of degree 2 over $K_{0}$ and the $H$-component is a field extension of $L$ of degree 4. So $L_{1}$ equals the $H$-component. The center $K=L_{1}^{G/H}$ is ungraded, an extension of degree 4 over $K_{0}$. The division algebra $D$ is a quaternion algebra over $K$ and because the extension is necessarily split, as in example (3) there is a quaternion algebra $D_{0}$ over $K_{0}$ such that $D\cong D_{0}\otimes_{K_{0}}K$.

Next we address the following problem: What are the $G$-graded simple algebras over an algebraically closed field of characteristic zero that admit a $G$-graded form division algebra?

Let $D$ be a finite dimensional division algebra faithfully $G$-graded with center $K$. Let $K_{0}=K\cap D_{e}$ as above. Suppose $K_{0}$ contains $\mu_{n_{G}}$, $n_{G}=exp(G)$. Extension of scalars $D_{F}=D\otimes_{K_{0}}F$ where $F$ is any algebraically closed field containing $K_{0}$ yields a finite dimensional $G$-graded simple algebra in which $F$ will be the field of central homogeneous elements of degree $e$. Such algebras have been classified by means of elementary and fine gradings by Bahturin, Sehgal and Zaicev [8]. Here is their result:

Theorem 1.11**.**

Let $A$ be a finite dimensional $G-$simple algebra over an algebraically closed field $F$ of characteristic zero. There is a subgroup $H$ of $G$, a two cocycle $\alpha\in Z^{2}(H,F^{*})$ $($the $H-$action on $F^{*}$ is trivial$)$ and $\mathfrak{g}=(g_{1},\ldots,g_{s})\in G^{(s)}$, such that $A\cong F^{\alpha}H\otimes M_{s}(F)$, where $($by this identification$)$ the $G-$grading on $A$ is given by

[TABLE]

Furthermore, we may replace the elements in the $s$-tuple by right H-cosets representatives and also, by permuting elements of the $s$-tuple we may assume equal representatives are adjacent to each other (see [7], [11], [2]).

Unlike the ungraded case, namely matrix algebras over $F$, not every finite dimensional $G$-graded simple algebra over $F$ admits a division algebra form which is $G$-graded. In fact it may not admit a form which is a $G$-graded division algebra (i.e. nonzero homogeneous elements are invertible) (see [3], Theorems $1.8$ and $1.12$).

We have the following result.

Theorem 1.12**.**

Notation as above. Let $A$ be a finite dimensional $G$-graded simple algebra over an algebraically closed field $F$ of characteristic zero. Then $A$ admits a division algebra $G$-graded $K_{0}$-form, for some field $K_{0}$ which contains $\mu_{n_{G}}$, $n_{G}=exp(G)$, if and only if the following hold:

1. (1)

$H$* is abelian.* 2. (2)

Every coset of $H$ is represented in the $s$-tuple (elementary grading). Moreover the number of representatives is equal for all cosets. 3. (3)

$H$* is normal in $G$.* 4. (4)

The cohomology class $[\alpha]\in H^{2}(H,F^{*})$ is $G$ invariant where the action of $G$ on $F$ is trivial.

Furthermore, with the condition above, $A$ admits a division algebra $G$-graded $K_{0}$-form where $K_{0}$ contains an algebraically closed field.

2. proof of the structure theorems

In this section we prove Theorems 1.4$-$1.6. We start with Theorem 1.4.

Proof.

Let $D$ be a finite dimensional algebra over its center $K$ and suppose $D$ is graded by a finite group $G$. We are assuming the center $K$ is contained in $D_{e}$, the identity component of $D$. Let $L=Z(D_{e})$. Conjugation on $D_{e}$ with nonzero homogeneous elements induces an action of $G$ on $L$ and we denote by $H$ the kernel of this action. We have $K=L^{G}=L^{G/H}$. We claim that $D_{H}\cong D_{e}\otimes_{L}L^{\alpha}H$ for some $\alpha\in Z^{2}(H,L^{*})$: For all $h\in H$, let $u_{h}$ be a nonzero homogeneous element of degree $h$. Because $h$ acts trivially on $L$, and $D_{e}$ is finite dimensional over $L$, conjugation by $u_{h}$ determines an inner automorphism of $D_{e}$ and hence, by the Skolem-Noether Theorem, there is a (nonzero) element $\theta_{h}$ in $D_{e}$ such that $x_{h}=u_{h}\theta_{h}^{-1}$ centralizes $D_{e}$. Let $h$ and $h^{\prime}$ in $H$. Clearly $x_{h}x_{h^{\prime}}=\alpha(h,h^{\prime})x_{hh^{\prime}}$ where $\alpha(h,h^{\prime})\in D_{e}^{*}$, but because $x_{h}$ and $x_{h^{\prime}}$ centralize $D_{e}$, we have $\alpha(h,h^{\prime})\in L^{*}$. It follows that $\alpha:H\times H\rightarrow L^{*}$ is a $2$-cocycle in $Z^{2}(H,L^{*})$. Finally, because the set $\{x_{h}\}_{h\in H}$ is linearly independent over $D_{e}$ (and in particular over $L$) we obtain $D_{H}=D_{e}\otimes_{L}L^{\alpha}H$, as desired.

Next we will show $L$ is the center of the twisted group algebra $L^{\alpha}H$: If $L_{1}=Z(L^{\alpha}H)=Z(D_{H})$, conjugation by nonzero homogeneous elements of $D$ induces an action of $G$ on $L_{1}$ with fixed field $K$. But $H$ acts trivially on $L_{1}$ and so $L_{1}^{G/H}=K$. Because $L\subseteq L_{1}$ and the action of $G/H$ is faithful, we obtain $L=L_{1}$, as desired. It follows that the $2$-cocycle $\alpha$ is nondegenerate and $H$ is of central type.

We claim next that $H$ is abelian: Because this may be of independent interest we state it as a separate proposition. Its proof is presented right after the end of the proof of Theorem 1.4.

Proposition 2.1**.**

Let $L$ be a field of characteristic zero and let $H$ be a finite group of exponent $n_{H}$. Let $L^{\alpha}H$ be a twisted group algebra of $H$ over $L$ where $\alpha\in H^{2}(H,L^{*})$ and the action of $H$ on $L$ is trivial. Suppose the field $L$ contains the group $\mu_{n_{H}}$ of $n_{H}$th roots of unity. If $L^{\alpha}H$ is a division algebra then $H$ is abelian.

Note that the commutativity of the group $H$ in Theorem 1.4 follows because $\mu_{n_{H}}\leq\mu_{n_{G}}\leq k^{*}\leq L^{*}$.

Next we will show that the class $[\alpha]\in H^{2}(H,L^{*})$ is $G$ invariant, where the action of $G$ on the cohomology is induced by the given action on $H$ and $L$. Let $y_{g}$ be a nonzero homogeneous element of degree $g\in G$. If $h\in H$, then $y_{g}x_{h}y_{g}^{-1}$ lies in $D_{H}$ and commutes with the elements of $D_{e}$. It follows that conjugation by $y_{g}$ stabilizes $L^{\alpha}H$. We need to show the cocycle $g(\alpha)$ determined by $g(\alpha)(h,h^{\prime})=y_{g}^{-1}\alpha(ghg^{-1},gh^{\prime}g^{-1})y_{g}$ is cohomologous to $\alpha$. Write $x_{h}x_{h^{\prime}}=\alpha(h,h^{\prime})x_{hh^{\prime}}$. Conjugating both sides by $y_{g}^{-1}$ we obtain

[TABLE]

Because $y_{g}^{-1}x_{h}y_{g}\in D_{g^{-1}hg}$ and centralizes $D_{e}$, we have $y_{g}^{-1}x_{h}y_{g}=\gamma_{h}x_{g^{-1}hg}$ where $\gamma=\gamma(g):H\rightarrow L^{*}$. Note that the elements of $D_{e}$ centralize $x_{h}$, so the value $\gamma_{h}$ does not depend on the choice of $y_{g}\in D_{g}=D_{e}y_{g}$.

Thus the equation above yields

[TABLE]

and so

[TABLE]

We conclude that

[TABLE]

showing the cocycles are cohomologous.

The algebra $D_{H}$ is an $L$-central division of degree $\sqrt{\dim_{L}(D_{e})|H|}$ and $D$ is a $G/H$-crossed product. Moreover $K\subseteq L\subseteq D_{H}\subseteq D$ and $[D:D_{H}]=[L:K]=|G/H|$, so $deg(D)=\sqrt{\dim_{K}(D)}=deg(D_{H})|G/H|=\sqrt{\dim_{L}(D_{e})|H|}[G:H]$. ∎

We proceed with the proof of Proposition 2.1.

Proof.

Our objective in the first part of the proof is to show the group $H$ must be nilpotent. This will reduce the problem to $p$-groups, a case which is treated in Lemma 2.3.

Lemma 2.2**.**

Let $H$ be a finite group. Suppose

1. (1)

The commutator subgroup $H^{\prime}$ is nilpotent. 2. (2)

If $p$ and $q$ are different primes, every $p$-element $z\in H$ commutes with every $q$-element $w\in H^{\prime}$.

Then $H$ is nilpotent.

Proof.

We show every Sylow $p$-subgroup of $H$ is normal. Let $y\in H$ and $x\in P$, a Sylow $p$-subgroup of $H$. Write $yxy^{-1}=cx$ with $c\in H^{\prime}$ and by the nilpotency of $H^{\prime}$ let $c=c_{1}c_{2}$ where $c_{1}$ and $c_{2}$ are elements in $H^{\prime}$ of orders $p^{\prime}$ (prime to $p$) and $p$ respectively. Because $yxy^{-1}$ commutes with $c_{1}$ it commutes with $c_{2}x$. We claim $c_{2}x$ is a $p$-element: Indeed, because $(H^{\prime})_{p}$, the $p$-Sylow subgroup of $H^{\prime}$, is characteristic in $H^{\prime}$, it is normal in $H$ and hence it is contained in every $Sylow$ $p$-subgroup of $H$ proving the claim. But $yxy^{-1}$ and $c_{2}x$ commute and hence $yxy^{-1}(c_{2}x)^{-1}=c_{1}$ is both a $p$ and a $p^{\prime}$ element. It follows that $yxy^{-1}=c_{2}x$ and because $c_{2}\in(H^{\prime})_{p}\leq P$, we have $c_{2}x\in P$ and the lemma is proved. ∎

We now resume the proof of the proposition. Let $\Gamma$ be the group of trivial units in $L^{\alpha}H$, that is the group of elements of the form $l_{h}u_{h}$, where $h\in H$ and $l_{h}\in L^{*}$. Because $\Gamma$ is center by finite, it follows by a theorem of Schur that the commutator subgroup $\Gamma^{\prime}$ of $\Gamma$ is finite. Note that $\Gamma/L^{*}\cong H$ and $\Gamma^{\prime}/(L^{*}\cap\Gamma)\cong H^{\prime}$. Invoking Amitsur’s classification of finite subgroups of $D^{*}$ where $D$ is an arbitrary division algebra ([5]), we have that $\Gamma^{\prime}$ is one of the following: ($1$) metacyclic (including cyclic) ($2$) $\hat{A}_{4}$ the binary tetrahedral group of order $24$ ($3$) $\hat{S}_{4}$ the binary octahedral group of order $48$ ($4$) $\hat{A}_{5}$ the binary icosahedral group of order $120$.

We first show that under our assumptions cases $2-4$ are not possible. Then in case ($1$) we will show that the group $H$ satisfies the conditions of Lemma 2.2 and hence is nilpotent.

For cases $2-4$ we apply ([4] Section $2$, Main Lemma). In that lemma it is proved that $\Gamma^{\prime}$ contains a subgroup $U$, normal in $\Gamma$, with the property that $U$ is isomorphic to $Q_{8}$, the quaternion group of order $8$. We claim that $U\cap L^{*}\cong\mathbb{Z}_{2}$: If not $U\cap L^{*}=\{e\}$ and so $U$ embeds in $H$. By the Universal Coefficient Theorem together with the fact that the Schur multiplier $M(U)$ is trivial, we have the group algebra $LU^{\prime}\subseteq L^{\alpha}H$, showing $L^{\alpha}H$ cannot be a division algebra. This proves the claim. It follows that because $L^{\alpha}H$ is a division algebra, the group $U$ spans $(-1,-1)_{2,L}$, the Hamilton quaternions over $L$. Next we claim the algebra $(-1,-1)_{2,L}$ is split contradicting $L^{\alpha}H$ is a division algebra. To see this, note that in cases $2-4$ the $3$-Sylow subgroup of $\Gamma^{\prime}$ is not central, and so $3\mid exp(H)$. It follows by our assumption that $\zeta_{3}\in L^{*}$ and so $(-1,-1)_{2,L}$ is split.

We turn now to case ($1$) when $\Gamma^{\prime}$ is metacyclic. The treatment in this case is divided into two cases, denoted $(1.1)$ and $(1.2)$ in the notation of ([4] Section $2$, Main Lemma). In case $(1.1)$ it is shown in that lemma that $\Gamma^{\prime}$ must be cyclic. In case $(1.2)$ it is shown that $\Gamma^{\prime}\cong C_{t}\times U$ where $C_{t}$ is cyclic of odd order and $U$ is isomorphic to $Q_{8}$. Furthermore, as above $U\cap L^{*}\cong\mathbb{Z}_{2}$ and $U$ spans $(-1,-1)_{2,L}$, the Hamilton quaternions over $L$. The proof of the nilpotency of $H$ will be completed if we show that in both cases $(1.1)$ and $(1.2)$ the group $H$ satisfies the conditions of Lemma 2.2. The nilpotency of $H^{\prime}\cong L^{*}\Gamma^{\prime}/L^{*}$ is evident. For the second condition we assume first $\Gamma^{\prime}$ is cyclic: Because a $q$-Sylow subgroup $(\Gamma^{\prime})_{q}$ of $\Gamma^{\prime}$ is characteristic in $\Gamma^{\prime}$, it is normal in $\Gamma$. But the presence of $\zeta_{q}\in L^{*}$ implies the inner automorphisms on $(\Gamma^{\prime})_{q}$ are of $q$-power order which means that for $p\neq q$, any $p$-element of $H$ centralizes $(\Gamma^{\prime})_{q}$. Suppose now $\Gamma^{\prime}\cong C_{t}\times U$ where $t$ is odd. As above every $q$-Sylow subgroup $(\Gamma^{\prime})_{q}$ is characteristic in $\Gamma^{\prime}$ and hence normal in $\Gamma$. If $q$ is odd the argument is as above. Suppose $q=2$. The automorphism group of $U$ is $S_{4}$ and the only way the second condition of the lemma can fail is if there exists an element $y\in H$ whose order is divisible by $3$ that acts nontrivially on $L^{*}U/L^{*}$ by conjugation. But this is not possible because in that case $\zeta_{3}\in L^{*}$ and the algebra $(-1,-1)_{2,L}$ is split. ∎

As promised we complete the proof of Proposition  2.1 with the following lemma.

Lemma 2.3**.**

Let $H$ be a finite $p$-group, say of exponent $p^{r},r\geq 1$. Let $L$ be a field of characteristic zero containing $\zeta_{p^{r}}$, a primitive $p^{r}th$ root of unity. If a twisted group algebra $L^{\alpha}H$ is a division algebra, the group $H$ must be abelian.

Proof.

Notation as above. Let $p^{s}$ the order of $\Gamma^{\prime}$. We claim $\Gamma^{\prime}$ is cyclic. Indeed, by the analysis given in the first part of the proof of Proposition 2.1 the only way this can be false is if $\Gamma^{\prime}\cong Q_{8}$ which spans $(-1,-1)_{2,L}$ over $L$. The field $L$ cannot contain $\mu_{4}$ because otherwise $(-1,-1)_{2,L}$ is split. Therefore the exponent of $H$ must be $2$ and hence $H$ is abelian in this case. This proves the claim. So, let $\Gamma^{\prime}=\langle z\rangle$ where $z$ has order $p^{s}$. Because $\Gamma^{\prime}$ is a cyclic $p$-group the element $z$ is a single commutator and so there are elements $x$ and $y$ such that $yxy^{-1}=zx$. Following the assumption that $L$ contains $\zeta_{p^{r}}$, we let $\zeta_{p^{r_{1}}}$, $r_{1}\geq r$ be the maximal $p$-power root of unity in $L^{*}$. Note that we may assume such a maximum exists, for otherwise the finite cyclic group $\Gamma^{\prime}$ would be contained in $L^{*}$ showing $H$ is abelian. For the proof of the lemma we may assume the integers ${r,r_{1},s}$ satisfy the following relations: $s\geq r_{1}\geq r\geq 1$ and $s-r_{1}\leq r$. We will show that $s\leq r$. In that case $\Gamma^{\prime}\subset L^{*}$, so $H$ is abelian.

Let $\alpha$ be a nonnegative integer such that $yzy^{-1}=z^{\alpha+1}$. Because $z^{p^{s-r_{1}}}$ is central, we have on the one hand $yz^{p^{s-r_{1}}}y^{-1}=z^{p^{s-r_{1}}}$ and on the other hand

[TABLE]

This implies $p^{s}$ divides $p^{s-r_{1}}\alpha$, or equivalently that $p^{r_{1}}$ divides $\alpha$. Hence we can write $\alpha=k_{0}p^{t}$ where $(k_{0},p)=1$ and $t\geq r_{1}.$ Now, applying conjugation by $y$, $p^{r}$ times on $x$, we obtain

[TABLE]

and so $z^{({(\alpha+1)^{p^{r}}-1})/\alpha}=1$ in $\Gamma$. This shows $p^{s}$ divides $M_{1}={({(\alpha+1)^{p^{r}}-1})/\alpha}.$ Claim: $s=r_{1}$. Write

[TABLE]

where $a_{3}$ is a binomial coefficient. Because $\alpha$ is divisible by $p^{t}$ where $t\geq r_{1}\geq r\geq 1$, we have

[TABLE]

Now, if $p$ is odd or $p=2$ and $r\geq 2$, $M_{1}\equiv p^{r}$ $\mod(p^{r+1})$ and so the only way $M_{1}$ can be divisible by $p^{s}$ is if $s\leq r$. We are left with the case where $exp(H)=2$, which implies $H$ is abelian. This completes the proof of the lemma and also of the proposition. ∎

In the next example we show the presence of $\mu_{n_{H}}$ in $L$ is necessary to ensure the group $H$ is abelian (see [1]).

Example 2.4**.**

Let $\mathbb{Z}_{p^{2}}\rtimes\mathbb{Z}_{p^{2}}=\langle b,\sigma:\sigma b\sigma^{-1}=b^{p+1}\rangle$ and $L=\mathbb{Q}(x,y)(\zeta_{p})$ where $p$ is an odd prime. There exists a cohomology class $\alpha\in H^{2}(H,L^{*})$ such that the corresponding twisted group algebra $L^{\alpha}H$ is a division algebra.

We proceed to prove Theorem 1.5, that is, we drop the assumption that the center is contained in the $e$-component.

Proof.

We have $D=D_{e}\oplus D_{g_{2}}\oplus\cdots\oplus D_{g_{n}}$ and $L=Z(D_{e})$. As above, the group $G$ acts on $L$ and if $H$ denotes the kernel of the action, then $K_{0}=L^{G/H}$. We want to show, as above, that we can find representatives $x_{h},h\in H$, which centralize $D_{e}$ so that we have $D_{H}\cong D_{e}\otimes_{L}L^{\alpha}H$, for some $\alpha\in Z^{2}(H,L^{*})$. In order to apply Skolem-Noether we need to know $D_{e}$ is finite dimensional over its center $L$. We are given that $D$ is finite dimensional over $K$.

Claim: the center $K$ is contained in $D_{\tilde{H}}$, where $\tilde{H}$ is the subgroup of $H$ consisting of all elements $g\in G$ for which there exists $u_{g}\in D_{g}$ that induces an inner automorphism on $D_{e}$ by conjugation. Note that in that case there exists (as above) an element $x_{g}\in D_{g}$ that centralizes $D_{e}$. To prove the claim let $z\in K$. Then there are elements $\alpha_{e},\alpha_{g_{2}},\dots,\alpha_{g_{n}}\in D_{e}$ such that

[TABLE]

We show first $\alpha_{g}=0$ if $g\not\in H$. Conjugation by elements of $L$ centralizes $z$ and preserves all homogeneous components. But if $g\not\in H$, we can find $l_{g}\in L^{*}$ which does not commute with $u_{g}$, so $\alpha_{g}=0$, as desired. Suppose now that there is an element $h\in H$ such that conjugation by $u_{h}$ does not give an inner automorphism on $D_{e}$. Need to show $\alpha_{h}=0$. Conjugation by any nonzero element of $D_{e}$ fixes $z$ and so must centralize $\alpha_{h}u_{h}$. This shows conjugation of $D_{e}$ by $\alpha^{-1}_{h}$ and $u_{h}$ determine the same automorphism and hence the action by $u_{h}$ is inner if $\alpha$ is nonzero. This proves that $K$ is contained in $D_{\tilde{H}}$, as desired. It follows that $K$ is contained in the center of $D_{\tilde{H}}$ and that $D_{\tilde{H}}$ is finite dimensional over its center.

For every element $h\in\tilde{H}$ there is a nonzero element $x_{h}\in D_{h}$ that centralizes $D_{e}$. If $h_{1},h_{2}\in\tilde{H}$ then the product $x_{h_{1}}x_{h_{2}}$ also centralizes $D_{e}$. It follows that $x_{h_{1}}x_{h_{2}}\in L^{*}x_{h_{1}h_{2}}$. Therefore the algebra generated over $L$ by $\{x_{h}:h\in\tilde{H}\}$ is isomorphic to a twisted group algebra $L^{\alpha}\tilde{H}$ where $\alpha:\tilde{H}\times\tilde{H}\rightarrow L^{*}$ is a $2$-cocycle. Now, if we let $L_{1}=Z(L^{\alpha}\tilde{H})$, we see that $L_{1}$, which is a finite extension of $L$, is the center of $D_{\tilde{H}}\cong D_{e}\otimes_{L}L^{\alpha}\tilde{H}$. But we have seen that $D_{\tilde{H}}$ is finite dimensional over its center and so we have that $D_{\tilde{H}}$ is finite dimensional over $L$. This shows that $D_{e}$ is finite dimensional over its center $L$ and hence finite dimensional over $K_{0}=K\cap L=L^{G/H}$. Note that in fact $H=\tilde{H}$, because by Skolem-Noether, conjugation by every homogeneous element of $D_{H}$ is inner. We have proved $(2)$ and $(4)$.

The proof that $H$ is abelian and that the cocycle $\alpha$ is $G$-invariant (part $(3)$ and the first statement of part $(6)$) is the same as in Theorem 1.4.

For the proof of $(5)$ note that conjugation of $D_{H}$ by homogeneous elements induces an action of $G$ on $L_{1}$ whose kernel contains $H$. But then the kernel must be equal to $H$ because $L$ is contained $L_{1}$.

Proof of the second part of $(6)$: Let S=\{h\in H:x_{h}x_{h^{\prime}}=x_{h^{\prime}}x_{h}\ \for all$\ \ h^{\prime}\in H\}$, that is, $S$ is the set of elements $s\in H$ such that $x_{s}\in L_{1}$, the center of $L^{\alpha}H$. We claim that in fact $L_{1}=L^{\alpha}S$: It is clear that $L^{\alpha}S\subseteq L_{1}$. Conversely if $z=\sum_{h\in H}l_{h}x_{h}$ lies in $L_{1}$, then for all $r\in H$, $x_{r}z=zx_{r}$. But $H$ is abelian and $x_{r}$ commutes with the elements of $L$. It follows that if $l_{h}\not=0$ then $h\in S$.

Because $L_{1}=L^{\alpha}S$ is the center of $L^{\alpha}H$ the group $H/S$ must be of central type, that is, isomorphic to $A\times A$ for some abelian group $A$.

Proof of $(7)$: The algebra $D_{H}$ is an $L_{1}$-central division of degree $deg(D_{e})\sqrt{|H/S|}$ and $D$ is a $G/H$-crossed product. Moreover $K\subseteq L_{1}\subseteq D_{H}\subseteq D$ and $[D:D_{H}]=[L_{1}:K]=|G/H|$, so $deg(D)=\sqrt{\dim_{K}(D)}=deg(D_{H})|G/H|=deg(D_{e})\sqrt{H/S}|G/H|$, as desired.

∎

We proceed to Theorem 1.6.

Proof.

The set up is as in Theorem 1.5. Recall that $L_{1}=Z(L^{\alpha}H)$ where $H$ is abelian and that $S$ is the subgroup of elements $s\in H$ such that $x_{s}$ commutes with $x_{h}$ for all $h\in H$. We have seen that $L_{1}=L^{\alpha}S$ and that $S$ is normal in $G$.

Let $S_{1}=S\cap Z(G)$. Clearly $L_{2}=L^{\alpha}S_{1}$ is contained in $L_{1}$ and $L_{2}$ is normalized by $G$. Suppose $S_{1}=S$, that is, $S$ is central in $G$. We want to show the center $K$ is $H$-graded and in particular $G$-graded. Indeed, if $z\in K$, $z=\sum_{s\in S}\gamma_{s}x_{s}$, conjugation by nonzero homogeneous elements $y_{g}$ acts on one hand trivially on $z$ and on the other hand by multiplication of each $x_{s}$ by a nonzero scalar. Because the elements $x_{s}$ are linearly independent, the result follows.

Suppose now $S$ is not central in $G$, that is, $S_{1}$ is a proper subgroup of $S$. We want to show the center $K$ is not graded. Take $s_{0}\in S\setminus S_{1}$. Clearly $x_{s_{0}}$ is not in $K$. For every $g\in G$ we choose nonzero $y_{g}\in D_{g}$, where, if $h\in H$, $y_{h}$ is chosen to commute with $D_{e}$. Consider the element $z(x_{s_{0}})=\sum_{g\in G}y_{g}x_{s_{0}}y_{g}^{-1}$. We claim $z(x_{s_{0}})\in K$: Because $x_{s_{0}}$ is in $L_{1}$ and conjugation by $y_{g}$ preserves $L_{1}$, we see that $z(x_{s_{0}})$ lies in $L_{1}$ and in particular commutes with the elements of $D_{e}$. Therefore it suffices to show that for all $g\in G$, $z(x_{s_{0}})$ is fixed under conjugation by $y_{g}$. But this is clear because conjugation by $y_{g}$ permutes the components modulo nonzero elements in $D_{e}$ and the elements of $D_{e}$ commute with $x_{s_{0}}$. So $z(x_{s_{0}})$ lies in $K$. Note that if $r\in L^{*}$ then $rx_{s_{0}}$ lies in $D_{H}$ and commutes with $D_{e}$, so substituting we obtain $z(rx_{s_{0}})=\sum_{g\in G}y_{g}rx_{s_{0}}y_{g}^{-1}$ lies in $K$. We claim we can find such an $r\in L^{*}$ such that $z(rx_{s_{0}})$ is nonzero. Because all of the homogeneous elements in the expression for $z(rx_{s_{0}})$ lie outside of $S_{1}$ it will follow that there are at least two nonzero terms in the sum and so we will have shown $K$ is nongraded, as desired.

To prove the claim, let $C_{G}(s_{0})$ denote the centralizer of $s_{0}$ in $G$. For every $g\in C_{G}(s_{0})$ let $g(x_{s_{0}})$ denote $y_{g}x_{s_{0}}y_{g}^{-1}$. Then $g(x_{s_{0}})=\beta_{g}x_{s_{0}}$ for some $\beta_{g}\in L^{*}$. We let $\beta:C_{G}(s_{0})\rightarrow L^{*}$ be given by $\beta(g)=\beta_{g}$. We want to show $\beta$ is a $1$-cocycle. Indeed, if $g_{1},g_{2}\in C_{G}(s_{0})$, then

[TABLE]

The left hand side yields

[TABLE]

and the right hand side yields

[TABLE]

so the result follows.

Now, because $s_{0}\in S$, we have $\beta(h)=1$ for all $h\in H$ and hence $\beta$ is induced by the inflation map from a $1$-cocycle (which we call $\bar{\beta}$) which represents a class in $H^{1}(C_{G}(s_{0})/H,L^{*})$. The group $C_{G}(s_{0})/H$ acts faithfully on $L$ and so, applying Hilbert’s Theorem $90$ for $\bar{\beta}$, there is an element $t\in L^{*}$ such that $\beta(g)=g(t)t^{-1}$ for every $g\in C_{G}(s_{0})$. We can therefore write $g(x_{s_{0}})=g(t)t^{-1}x_{s_{0}}$ which implies that for all $g\in C_{G}(s_{0})$, $g(t^{-1}x_{s_{0}})=t^{-1}x_{s_{0}}$. We now compute

[TABLE]

[TABLE]

[TABLE]

which is nonzero because $|C_{G}(s_{0})|$ is nonzero because the characteristic of the base field is zero! This finishes the proof of the claim.

We are left with determining the structure of $K$ when $K$ is graded, that is when $S$ is central in $G$. In that case if $s\in S$, then $C_{G}(s)=G$ and the argument in the preceding paragraph shows that there is a representative $x_{s}$ of $s$ such that $x_{s}$ lies in $K$. It follows that $K=K_{0}[x_{s}:s\in S]=K_{0}^{\tilde{\alpha}}S$ where $\tilde{\alpha}:S\times S\rightarrow K^{*}_{0}$ is a two-cocycle and $\tilde{\alpha}$ is cohomologous to $\alpha$ in $H^{2}(S,L^{*})$, as desired. ∎

3. construction of division algebras graded by $G$

In this section we prove that given a group extension

[TABLE]

where $H$ is abelian, a positive integer $d$, and a $Q$-invariant map $\phi:M(H)\rightarrow\mu_{n_{H}}$, $n_{H}=exp(H)$, there is a finite dimensional division algebra over its center, faithfully $G$-graded which realizes the given data. Here, the action of $Q$ on $M(H)$ is the induced action of $Q$ on $H$ and trivial on $\mu_{n_{H}}$.

We construct first division algebras $D$ in which $d$, the degree of $D_{e}$, equals $1$. That is, $L=D_{e}$ is a field. Thereafter we show how to pass from $d=1$ to an arbitrary $d$.

Remark 3.1*.*

In our construction the base field $k$ will contain $\zeta_{n}$, a primitive $n$th root of unity, where $n=l.c.m(d,exp(H))$. There is no “upper bound” on the number of roots of unity in $k$; our construction works even if $k$ contains an algebraically closed field.

Let us start our discussion with two special cases: in case $(1)$ we assume $H=\{1\}$, and in case $(2)$ we assume $G=H$. The case where $H$ is trivial is well known. We recall here a proof which is attributed to S. Rosset and K. Brown (see [9] and [16]). Their idea will appear in the proof of the general case.

Let

[TABLE]

be a presentation of the finite group $G$ where the groups $F$ and $R$ are finitely generated free.

Taking quotient groups modulo the commutator subgroup $[R,R]$ we obtain

[TABLE]

We want the action of $G$ on $N$ in this extension to be faithful, that is, we want the centralizer of $N$ in $\Gamma$ to be $N$ itself. This can be arranged as follows: Let $A$ be the free abelian group on $d_{G}(=|G|)$ generators and let $G$ act on $A$ by left translation on these generators, that is, the regular permutation representation of $G$. We obtain the split exact sequence $1\rightarrow A\rightarrow A\rtimes G\rightarrow G\rightarrow 1$. We take our group $F$ to be free on $2d_{G}$ generators $y_{g},x_{g}$ for all $g\in G$ and map $F$ to $G$ by sending each $y_{g}$ to the identity and each $x_{g}$ to $g$. The group $F$ maps to $A\rtimes G$ by sending the $y_{g}^{\prime}s$ to generators of $A$ and each $x_{g}$ to $g$. The resulting extension denoted above by $\beta$ maps to this split exact sequence and a simple diagram chase then shows that $G$ acts faithfully on $N$.

By a theorem of Higman ([13], Theorem $2$), $\Gamma=F/[R,R]$ is torsion free. Because $\Gamma$ is abelian by finite, it follows from a theorem of Brown that the group algebra $k\Gamma$ is a domain (see [9], Cor. $2$). Let $S$ denote the ring $kN$, a commutative subring of $k\Gamma$ and hence an integral domain. The group $G$ acts faithfully on $S$ and the crossed-product ring $S*G$ is a domain because it is isomorphic to $k\Gamma$. Let $R$ denote the fixed ring of the action of $G$ on $S$. Then $R$ is the center of $S*G$. Because the ring $R$ is the fixed ring of a finite group action, the extension $S/R$ is integral. It follows that if $K$ denotes the field of fractons of $R$ and $L$ the field of fractions of $S$ then $L=SK$. The extenion $L/K$ is a Galois extension with Galois group $G$. The classical crossed-product algebra $L*G$ is a division algebra: $L*G\cong(S*G)\otimes_{R}K$ and every element of $L*G$ can be written as $r^{-1}a$ where $r\in R$ is nonzero and $a\in S*G$. It follows that $L*G$ is a domain and finite dimensional over $K$, so a division algebra. Morevoer $L*G$ realizes the given data for this case.

We next consider case $(2)$ in which $G=H$ is abelian and $\phi$ is any map $\phi:M(H)\rightarrow\mu_{n_{H}}\subset k^{*}$, $n_{H}=exp(H)$. Note that in this case every cohomology class is invariant because the inner action of $H$ acts trivially on $M(H)$.

Let

[TABLE]

be a presentation of $H$. We take quotients modulo $[R,F_{H}]$ and obtain the central extension

[TABLE]

The group $R/[R,F_{H}]$ is finitely generated abelian and so is the direct product of a finite group $U$ and a finitely generated torsion free group $T\cong\mathbb{Z}^{r}$. We claim (and it is well known) that the torsion part is precisely $M(H)$, the Schur multiplier of $H$. Indeed, by the Hopf formula the Schur multiplier is given by $(R\cap[F_{H},F_{H}])/[R,F_{H}]$ so it is necessary and sufficient to show that

[TABLE]

is torsion free. But $R/(R\cap[F_{H},F_{H}])$ is isomorphic to $R[F_{H},F_{H}]/[F_{H},F_{H}]$, a subgroup of the torsion free group $F_{H}/[F_{H},F_{H}]$. This proves the claim.

Let $\Gamma_{H}=F_{H}/[R,F_{H}]$. The group $U$ is characteristic in $R/[R,F_{H}]$ and so normal in $\Gamma_{H}$. We therefore have the extension

[TABLE]

which is easily seen to be the inflation of the extension

[TABLE]

We claim $\Gamma_{H}/U$ is torsion free: Indeed,

[TABLE]

[TABLE]

because $H$ is abelian. The claim follows.

The extension $\beta$ yields a crossed product $(kU)^{\beta}\Gamma_{H}/U$. Composing the map $\phi$, whose image is in $\mu_{n_{H}}$, with the $2$-cocycle $\beta$ we obtain a cocycle which represents a class in $H^{2}(\Gamma_{H}/U,\mu_{n_{H}})$ and which we denote by $\phi(\beta)$. Thus we obtain the twisted group algebra $k^{\phi(\beta)}\Gamma_{H}/U$. Because $\Gamma_{H}/U$ is torsion free and abelian by finite, a theorem of J. Moody (see for instance [15], Lemma, $37.8$) implies that this twisted group algebra is an Ore domain. Because $\beta$ is the inflation of $\bar{\beta}$, the elements of $T$ in $k^{\phi(\beta)}\Gamma_{H}/U$ are central and hence, by inverting the elements of $kT\setminus\{0\}$, we obtain the twisted group division algebra $k(T)^{\phi(\bar{\beta})}H$ over the field $k(T)$. Letting $k(T)^{\phi(\bar{\beta})}H=\sum_{h\in H}k(T)x_{h}$ we compute easily that for all $h_{1},h_{2}\in H$, the commutator $x_{h_{1}}x_{h_{2}}x_{h_{1}}^{-1}x_{h_{2}}^{-1}=\phi(h_{1}\wedge h_{2})$ and so, by the comments preceding Definition 1.7, this division algebra realizes the given data.

We now turn to the proof of the general case. Let

[TABLE]

be a presentation of the finite group $G$ (We will be more precise concerning the choice of $F_{G}$ below). We denote the map $F_{G}\rightarrow G$ by $\pi$.

For the subgroup $H$ we obtain the following induced extension, where $F_{H}=\pi^{-1}(H)$:

[TABLE]

Taking quotients modulo $[R,F_{H}]$ in these two extensions we obtain

[TABLE]

and

[TABLE]

As in case (2), the group $R/[R,F_{H}]$ is the direct product of a finite group $U=M(H)$ and a finitely generated torsion free group $T\cong\mathbb{Z}^{r}$. Moreover $U$ is characteristic in $R/[R,F_{H}]$ and hence normal in both $\Gamma_{H}=F_{H}/[R,F_{H}]$ and $\Gamma_{G}=F_{G}/[R,F_{H}]$. The group $\Gamma_{G}/U$ is torsion free: We have $\Gamma_{G}/U\cong(F_{G}/[R,F_{H}])/((R\cap[F_{H},F_{H}])/[R,F_{H}])\cong F_{G}/[F_{H},F_{H}]$ because $H$ is abelian, and $F_{G}/[F_{H},F_{H}]$ is torsion free by Higman’s theorem.

We have the extension:

[TABLE]

Because $\phi:M(H)\rightarrow\mu_{n_{H}}$ is $G$ invariant, the map $\phi\circ\beta:\Gamma_{G}/U\times\Gamma_{G}/U\rightarrow\mu_{n_{H}}$ is a $2$-cocycle. Therefore we can form the twisted group algebra $k^{\phi\circ\beta}\Gamma_{G}/U$. The group $\Gamma_{G}/U$ is abelian by finite and we have seen that it is torsion free. Again as in case (2) we infer that $k^{\phi\circ\beta}\Gamma_{G}/U$ is an Ore domain.

We claim the ring of quotients of $k^{\phi\circ\beta}\Gamma_{G}/U$ is a division algebra, $G$-graded, satisfying the desired conditions.

What do we have to prove? Because it is an Ore domain the ring of quotients is a division algebra which we denote by $D$. So we need to prove that $D$ is finite dimensional over its center, $G$-graded and the grading gives rise to the given group extension

[TABLE]

where $H$ is abelian, and the given $Q$-invariant map $\phi:M(H)\rightarrow\mu_{n_{H}}$.

We first consider the subalgebra $k^{\phi\circ\beta}\Gamma_{H}/U$. This is the same as the dvision algebra obtained in case (2) and as we saw there the cocycle $\beta$ restricted to $\Gamma_{H}/U$ is inflated from the extension

[TABLE]

Therefore the elements of $T$ in $k^{\phi(\beta)}\Gamma_{H}/U$ are central and hence, by inverting the elements of $kT\setminus\{0\}$, we obtain the twisted group division algebra $k(T)^{\phi(\bar{\beta})}H$ over the field $k(T)$. Also, as in case (2), the cocycle on $H$ maps to $\phi$ as desired.

From the extension $1\rightarrow T\rightarrow\Gamma_{G}/U\rightarrow G\rightarrow 1$ we see that if we set $D_{e}=k(T)$ and choose representatives $x_{g}$ in $\Gamma_{G}/U$ for the elements $g$ in $G$, then $D=\sum_{g\in G}D_{e}x_{g}$ is faithfully $G$-graded. We are then left with showing that $H$ is exactly the kernel of the action of $G$ on $k(T)$. For that we need to be more precise about our choice of $F_{G}$: What is required is that in the sequence $1\rightarrow T\rightarrow F_{G}/U\rightarrow G\rightarrow 1$ the kernel of the action of $G$ on $T$ is exactly $H$ (We already know the kernel contains H). We proceed as in case (1): Let $A$ be the free abelian group on $d_{Q}=|Q|$ generators and let $G$ act on $A$ by left translation on these generators. We obtain the split exact sequence $1\rightarrow A\rightarrow A\rtimes G\rightarrow G\rightarrow 1$. We take our group $F_{G}$ to be free on $d_{Q}+d_{G}$ generators $y_{g_{1}},y_{g_{2}},\dots,y_{g_{d_{Q}}}$, where $g_{1},g_{2},\dots,g_{d_{Q}}$ are distinct coset representatives of $H$ in $G$ together with generators $x_{g}$ for all $g\in G$. We map $F_{G}$ to $G$ by sending each $y_{g_{i}}$ to the identity and each $x_{g}$ to $g$. We map $F_{G}$ to $A\rtimes G$ by sending the ${y_{g_{i}}}^{\prime}s$ to generators of $A$ and each $x_{g}$ to $g$.

With this choice of presentation for $G$ a straightforward calculation shows that we have an induced map of extensions from $1\rightarrow R/[R,F_{H}]\rightarrow F_{G}/[R,F_{H}]\rightarrow G\rightarrow 1$ to $1\rightarrow A\rightarrow A\rtimes G\rightarrow G\rightarrow 1$, where the map on $G$ is the identity. Because $U$ is torsion and $A$ is torsion free we get an induced map of extensions from $1\rightarrow T\rightarrow\Gamma_{G}/U\rightarrow G\rightarrow 1$ to $1\rightarrow A\rightarrow A\rtimes G\rightarrow G\rightarrow 1$. A simple diagram chase now shows that the kernel of the action of $G$ on $T$ is exactly $H$, as desired.

We conclude this section by showing how to pass from $d=1$ to an arbitrary $d$. To this end, we shall assume (as we may by Remark 3.1) the field $k$ in the construction (and hence also $K_{0}$ and $L$) contains $\zeta_{d}$, a primitive $d$th root of unity.

Adjoin new indeterminates $a,b$ to $L$ and let

[TABLE]

be the symbol algebra of degree $d$ over $L(a,b)$ determined by $a$ and $b$. Because $a$ and $b$ are indeterminates the algebra $E$ is a division algebra of degree $d$. Next, we are assuming we have constructed a division algebra $D$ realizing the extension $\beta$, $\phi$ and $d=1$. This means that $D$ is isomorphic to a crossed product of $L^{\alpha}H\ast Q$. We denote by $K_{0}$ its $e$-center. Now, it is well known that a rational extension $E(x)$ of $E$, $E$ any field, induces an injective map $res_{E(x)}^{E}:Br(E)\rightarrow Br(E(x))$ of the Brauer groups (see [12], [6]) and hence extending the $e$-center of $L^{\alpha}H\ast Q$ to $K_{0}(a,b)$ yields a division algebra $L(a,b)^{\alpha}H\ast Q$ which realizes the same data as $D$. Finally, the algebra we are looking for is

[TABLE]

where the outer action of $Q$ on $(a,b)_{L(a,b),d}\otimes L^{\alpha}H$ is determined via the outer action of $Q$ on $L^{\alpha}H$. It is clear the constructed algebra is $G$-graded and realizes the triple with the same $\beta$ and $\phi$ as $D$, but with $d$ rather than $1$. We need to show this is a division algebra. To this end it is convenient to view the algebra $\tilde{D}$ as a Hilbert twist of $D$. Indeed, extend $K_{0}$, the $e$-center of $D$, by the central indeterminate $a^{1/d}$ and denote it by $D_{K_{0}(a^{1/d})}$. This is a division algebra by the fact above and because a transcendental extension of the center yields an injective map

[TABLE]

Consider the Hilbert twist extension $D_{K_{0}(a^{1/d})}(\sigma,y)$. This is the ring of quotients of the twisted polynomial ring $D_{K_{0}(a^{1/d})}[\sigma,y]$ where $\sigma$ is the automorphism of $D_{K_{0}(a^{1/d})}$ mapping $a^{1/d}$ to $\zeta_{d}a^{1/d}$ and $y$ skew-commutes with elements of $D_{K_{0}(a^{1/d})}$ via the rule $zy=y\sigma(z)$. The algebra obtained is $G$-graded isomorphic to $\tilde{D}$ where $b=y^{d}$. We conclude that because $D_{K_{0}(a^{1/d})}$ is a division algebra, the algebra $D_{K_{0}(a^{1/d})}(\sigma,y)$ is a division algebra as desired (see [14], Example $1.8$). This completes the proof of the first part Theorem 1.8. The second part follows easily; if the cocycle is nondegenerate, the center of $L^{\alpha}H$ is $L$ which is contained in $D_{e}$.

4. Forms of finite dimensional $G$-graded simple algebras

In this section we prove Theorem 1.12.

Proof.

Let $A$ be a finite dimensional $G$-graded simple algebra over its $e$-center $F$, where $F$ is algebraically closed of characteristic zero. Let $P_{A}=\{H,(e,g_{2},\ldots,g_{s}),\alpha\}$ be a Bahturin, Sehgal, Zaicev presentation. Suppose first it admits a division algebra $G$-graded form $D$ over $K_{0}$ (the $e$-center of $D$). because $D$ is in particular a $G$-graded division algebra (that is, nonzero homogeneous elements are invertible) the group $H$ which appears in the presentation must be normal, all coset representatives appear and with equal frequency which we denote by $d$. Moreover, the class $[\alpha]$ is $G$-invariant (see [3], Corollary $1.13$). To complete the proof of the necessity it remains to show $H$ is abelian. Now, because the $e$-component of $D$ extends to the $e$-component of $A$, and $D_{N}$ extends to $A_{N}$ for any subgroup $N$ of $G$, the result will follow if we show the elements in $D$ that centralize $Z(D_{e})$ belong to the $N$-homogeneous component where on the one hand $N$ is abelian and on the other hand, these are precisely the elements of $A_{H}$. The first statement follows from the structure theorem. For the proof of the second statement note that the center of $A_{e}$ is spanned over $F$ by the $r$ elements (idempotents)

[TABLE]

where $r=[G:H]$. because $H$ is normal in $G$, $A_{H}$ is spanned by elements of the form $u_{h}\otimes_{F}e_{i,i}$ which centralizes $Z(A_{e})$. Let us show these are all. Let

[TABLE]

be an element in $A$ that centralizes $Z(A_{e})$. Multiplying from left and right by idempotents $u_{e}\otimes(e_{(k-1)d+1,(k-1)d+1}+\ldots+e_{kd,kd})$, $k=1,\ldots,r$, we see $\gamma_{(h,i,j)}=0$ unless $i$ and $j$ belong to the same block, that is $i,j\in\{(k-1)d+1,\ldots,kd\}$, some $k=1,\ldots,r$.

Conversely: We want to show that any such algebra $A$, with presentation

[TABLE]

has a division algebra form. First we say that it is sufficient to consider the case where $d=1$. Indeed, if $P_{A}=\{H,(e,g_{2},\ldots,g_{s}),\alpha\}$, where each coset representative appears $d$ times, that is $dr=s$, we consider the corresponding algebra $\bar{A}$ with $P_{\bar{A}}=\{H,(e,\nu_{2},\ldots,\nu_{r}),\alpha\}$ (i.e. the same presentation but with $d=1$). If $\bar{D}$ is a division algebra form for $\bar{A}$, where its $e$-center contains $\zeta_{d}$, arguing as in last part of the previous section we obtain that the corresponding Hilbert twist is a division algebra $G$-graded form of $A$.

Changing notation we consider a $G$-graded simple with presentation

[TABLE]

where $H$ is abelian and normal, all coset representatives appear exactly once and $[\alpha]$ is $G$-invariant. We need to construct a division algebra form. Consider the triple $(\beta,\pi(\alpha),1)$ induced by $P_{A}$, where $\beta:1\rightarrow H\rightarrow G\rightarrow G/H\rightarrow 1$ and $\pi(\alpha):Hom(M(H)\rightarrow\mu_{n_{H}}$, $n_{H}=exp(H)$, and let $D$ be a division algebra which realizes $(\beta,\pi(\alpha),1)$. In particular we have $D_{e}=L$, $D_{H}\cong L^{\alpha{{}^{\prime}}}H$, $K_{0}=L^{G/H}$ and $H$ is the centralizer of $L$ in $G$. We want to show $D$ is a $G$-graded form of $A$. Suppose extension of scalars of $D$ yields a $G$-graded simple algebra $A^{\prime}$ with presentation $P_{A^{\prime}}=\{H^{\prime},(e,g^{\prime}_{2},\ldots,g^{\prime}_{s}),\alpha^{\prime}\}$. because the $e$-component is commutative and its dimension is on the one hand $[G:H]$ (extension of the form) and on the other hand, from the presentation $P_{A}$, it is $[G:H^{\prime}]$, we have that $H$ and $H^{\prime}$ have the same order and each coset representative in $P_{A^{\prime}}$ must appear once. Furthermore because the homogeneous elements in $D$ which centralize the $e$-component are precisely the $H$-homogeneous elements, this carry out for $A^{\prime}$ and this forces $H=H^{\prime}$. To complete the proof we need to show that extension of scalars yields the cohomology class in $H^{2}(H,F^{*})$ (in the presentation $P_{A^{\prime}}$) which corresponds by means of the Universal Coefficient Theorem to the map $\pi([\alpha]):M(H)\rightarrow\mu_{n_{H}}$. This follows from the fact that the $H$ identities in $D_{H}$ are determined by the map $\pi([\alpha]):M(H)\rightarrow\mu_{n_{H}}$ and the $H$-identities of $A^{\prime}_{H}$ are determined by the map $\pi([\alpha^{\prime}]):M(H)\rightarrow\mu_{n_{H}}$ together with the fact that upon a ($char=0$) field extension of the $e$-center the $H$-graded identities do not change. ∎

Acknowledgment 4.1**.**

We thank the referee for his/her very valuable report.