Unit groups of integral finite group rings with no noncyclic abelian finite subgroups

TL;DR

This paper investigates the structure of units in integral group rings, establishing conditions under which certain noncyclic subgroups exist, linking properties of units to the original group structure.

Contribution

It proves that noncyclic subgroups of order p^2 in units correspond exactly to those in the original group, extending known results for p=2.

Findings

Noncyclic subgroups of order p^2 in units only exist if they exist in G for odd p.

For p=2, the Brauer--Suzuki theorem applies, confirming the correspondence.

The results connect the subgroup structure of units with that of the original finite group.

Abstract

It is shown that in the units of augmentation one of an integral group ring $\mathbb{Z} G$ of a finite group $G$, a noncyclic subgroup of order $p^{2}$, for some odd prime $p$, exists only if such a subgroup exists in $G$. The corresponding statement for $p=2$ holds by the Brauer--Suzuki theorem, as recently observed by W. Kimmerle.

Equations

$$\sum _{i=1}^k{\epsilon }_{x_i}(u^s)\chi (x_i)=\chi (u^s)=\chi (u{)}^{\sigma }=\sum _{i=1}^k{\epsilon }_{x_i}(u)\chi (x_i{)}^{\sigma }$$

$$=\sum _{i=1}^k{\epsilon }_{x_i}(u)\chi (x_i^s)=\sum _{i=1}^k{\epsilon }_{x_i^t}(u)\chi (x_i).$$

$$\sum_{s\in(\mbox{$\scriptstyle\mathbb{Z}$}/n\mbox{$\scriptstyle\mathbb{Z}$})^{\times}}\varepsilon_{x}(u^{s})=\sum_{s\in(\mbox{$\scriptstyle\mathbb{Z}$}/n\mbox{$\scriptstyle\mathbb{Z}$})^{\times}}\varepsilon_{x^{s}}(u).$$

$$\varepsilon_{x}\Bigg{(}\sum_{i=1}^{p-1}u^{i}\Bigg{)}=\sum_{i=1}^{p-1}\varepsilon_{x^{i}}(u)=\frac{p-1}{k}\sum_{y^{G}\colon y\in\langle{x}\rangle}\varepsilon_{y}(u)=\frac{p-1}{k}=\varepsilon_{x}\Bigg{(}\sum_{i=1}^{p-1}x^{i}\Bigg{)}.$$

$$\chi\Bigg{(}\frac{1}{p^{2}}\sum_{u\in U}u\Bigg{)}=\frac{1}{p^{2}}\Bigg{(}\chi(1)+(p+1)\sum_{i=1}^{p-1}\chi(x^{i})\Bigg{)}.$$

$$\frac{1}{p^2}(|G:P|+|N:P|(p^2-1)).$$

$$\frac{1}{p^2}(|G:P|-|N:P|(p+1)).$$

Full text

1. Introduction

Is a finite subgroup $H$ of units in the integral group ring \mbox{\mathbb{Z}}G of a finite group $G$ necessarily isomorphic to a subgroup of $G$? Of course, torsion coming from the coefficient ring should be excluded, that is, only finite subgroups $H$ in \mbox{\rm V}(\mbox{\mathbb{Z}}G), the group of units of augmentation one in \mbox{\mathbb{Z}}G, will be considered. The question was raised by Higman in his thesis (1940), where he gave an affirmative answer when $G$ is metabelian nilpotent or the affine group over a prime field; cf. Sandling (1981). In the survey of Sandling (1984) it is included as Problem 5.4, and noted that an affirmative answer for metabelian $G$ was finally given by Roggenkamp (1981); but see also Cliff, Sehgal and Weiss (1981), and Marciniak and Sehgal (2003) for a more recent result, giving a generalization based on a theorem of Weiss (1988). These results are really about certain ‘large’ torsion-free normal subgroups of \mbox{\rm V}(\mbox{\mathbb{Z}}G). For a more complete discussion, see Chapter 4 in Sehgal’s book (1993).

As a sort of converse, one may fix a finite group $H$ and look for groups $G$ for which $H$ embeds into \mbox{\rm V}(\mbox{\mathbb{Z}}G), again hoping for the best, but little is known in this respect. What is known is that if a cyclic group $H$ of prime power order embeds into some unit group \mbox{\rm V}(\mbox{\mathbb{Z}}G), then $H$ also embeds into $G$ (due to an observation of Cohn and Livingstone (1965); see also Zassenhaus (1974)), and only recently in Hertweck (2007b) it was shown that the restriction on the order can be removed if in addition $G$ is assumed to be solvable. In this spirit, Marciniak, at a satellite conference of the ICM 2006, asked whether a group $G$ necessarily has a subgroup isomorphic to Klein’s four group provided this is the case for \mbox{\rm V}(\mbox{\mathbb{Z}}G). Kimmerle immediately observed that this is implied by the Brauer–Suzuki theorem (rendered in Kimmerle (2006)), see Section 2. Our complementary result is as follows.

Theorem A**.**

Let $G$ be a finite group. Suppose that \mbox{\rm V}(\mbox{\mathbb{Z}}G) has a noncyclic abelian subgroup of order $p^{2}$, for some odd prime $p$. Then the same is true for $G$ (i.e., Sylow $p$-subgroups of $G$ are not cyclic).

It is easy to verify that a finite $p$-group with no noncyclic abelian subgroup is either cyclic or a (generalized) quaternion group, see Theorem 4.10 in Gorenstein (1968). It comes to mind that the theory of cyclic blocks might be used in the proof, but it is pretty simple and makes only use of a fact about vanishing of partial augmentations of torsion units, established in Hertweck (2006, 2007a).

We remark that both results (whether $p$ is even or odd) for a solvable group $G$ are covered by Theorem 5.1 in Dokuchaev and Juriaans (1996).

Note that a group $G$ whose Sylow $2$-subgroups are cyclic has a normal $2$-complement, by Burnside’s well known criterion, see Theorem 4.3 in Gorenstein (1968). We obtain the following corollary.

Corollary 1**.**

Let $G$ be a finite group having cyclic Sylow $p$-subgroups for some prime $p$. Then any finite $p$-subgroup of \mbox{\rm V}(\mbox{\mathbb{Z}}G) is isomorphic to a subgroup of $G$.

Finally, we remark that, as with other results in this field, the theorem can be formulated for more general coefficient rings than $\mathbb{Z}$, notably for the semilocalization of $\mathbb{Z}$ at the prime divisors of the order of $G$. Unfortunately, it is definitely wrong for $p$-adic coefficient rings.

2. Kimmerle’s observation

Coming back to the initial question, we mention that in the hope for further positive results, it is natural to impose restrictions on the prime divisors of the finite subgroup $H$, i.e., to consider only $\pi$-groups $H$ for some set $\pi$ of primes (a singleton $\{p\}$, to begin with), as has been done before in work on the stronger Zassenhaus conjecture (ZC3), cf. Dokuchaev and Juriaans (1996). It is well known that then, one can assume that $\text{\rm O}_{\pi^{\prime}}(G)$, the largest normal $\pi^{\prime}$-subgroup of $G$, is trivial, for $H$ has an isomorphic image under the natural map \mbox{\mathbb{Z}}G\rightarrow\mbox{\mathbb{Z}}G/\text{\rm O}_{\pi^{\prime}}(G), see the remark after Theorem 2.2 in Dokuchaev and Juriaans (1996).

This derives from the vanishing of certain partial augmentations of the elements of $H$. Recall that for a group ring element $u=\sum_{g\in G}a_{g}g$ (all $a_{g}$ in $\mathbb{Z}$), its partial augmentation with respect to an element $x$ of $G$, or rather its conjugacy class $x^{G}$ in $G$, is the sum $\sum_{g\in x^{G}}a_{g}$; we will denote it by $\varepsilon_{x}(u)$. The result of Cohn and Livingstone mentioned in the introduction really says that if an element $h$ of $H$ is of prime power order, then there exists an element $x$ in $G$ of the same order such that $\varepsilon_{x}(h)\neq 0$. Note that $\varepsilon_{z}(u)=a_{z}$ for an element $z$ in the center of $G$. An old yet fundamental result from Berman (1955) and Higman (1940) asserts that if $\varepsilon_{z}(h)\neq 0$ for an element $h$ in $H$ and some $z$ in the center of $G$, then $h=z$.

Coming to Marciniak’s question, suppose that $G$ has no subgroups isomorphic to Klein’s four group. For our purpose, we can assume that $\text{\rm O}_{2^{\prime}}(G)=1$ and that Sylow $2$-subgroups of $G$ are not cyclic. Thus Sylow $2$-subgroups of $G$ are (generalized) quaternion, and by the Brauer–Suzuki theorem, from Brauer and Suzuki (1959), $G$ contains a unique involution $z$. For an involution $u$ in \mbox{\rm V}(\mbox{\mathbb{Z}}G), the Cohn–Livingstone result gives $\varepsilon_{z}(u)\neq 0$, and therefore $u=z$ by the Berman–Higman result, answering Marciniak’s question in the affirmative.

Theorem B** (Kimmerle).**

Let $G$ be a finite group. Suppose that \mbox{\rm V}(\mbox{\mathbb{Z}}G) has a subgroup isomorphic to Klein’s four group. Then the same is true for $G$.

We do not know of a proof avoiding the use of the Brauer–Suzuki theorem.

Suppose that Sylow $2$-subgroups of $G$ are quaternion groups. Then the theorem implies that finite $2$-subgroups of \mbox{\rm V}(\mbox{\mathbb{Z}}G) are cyclic or quaternion groups. Taking into account the structure of the quaternion groups, and the Cohn–Livingstone result, one obtains the following corollary.

Corollary 2**.**

Let $G$ be a finite group whose Sylow $2$-subgroups are quaternion groups (ordinary or generalized). Then any finite $2$-subgroup of \mbox{\rm V}(\mbox{\mathbb{Z}}G) is isomorphic to a subgroup of $G$.

3. Proof of Theorem A

The partial augmentations of a torsion unit in \mbox{\rm V}(\mbox{\mathbb{Z}}G) encode its character values in a way establishing a connection to group elements which respects a divisibility relation between orders. We will make use of a lemma which is an easy consequence of this fact.

Lemma 3**.**

Let $u$ be a torsion unit in \mbox{\rm V}(\mbox{\mathbb{Z}}G) of, say, order $n$. Let $s$ be a natural integer coprime to $n$, so that $st\equiv 1\mod{n}$ for another natural integer $t$. Then for all $x$ in $G$ whose order divide $n$, we have $\varepsilon_{x}(u^{s})=\varepsilon_{x^{t}}(u)$.

Proof.

Let $\zeta$ be a primitive $n$-th complex root of unity, and let $\sigma$ be the Galois automorphism of \mbox{\mathbb{Q}}(\zeta) sending $\zeta$ to $\zeta^{s}$. Let $x_{1},\dotsc,x_{k}$ be representatives of the conjugacy classes of $G$ whose elements have order dividing $n$. Note that then $x_{1}^{t},\dotsc,x_{k}^{t}$ is another system of representatives. By Theorem 2.3 in Hertweck (2007a), $\varepsilon_{x}(u)\neq 0$ is possible only for elements $x$ whose order divide $n$. Thus for any ordinary irreducible character $\chi$ of $G$, we have

[TABLE]

Since the character table of $G$, stripped off from any additional information, is an invertible matrix, it follows that $\varepsilon_{x_{i}}(u^{s})=\varepsilon_{x_{i}^{t}}(u)$ for all indices $i$, which proves the lemma. ∎

Corollary 4**.**

Let $u$ be a torsion unit in \mbox{\rm V}(\mbox{\mathbb{Z}}G) of, say, order $n$. Then for any $x$ in $G$ whose order divides $n$,

[TABLE]

Corollary 5**.**

Suppose that for a prime divisor $p$ of the order of $G$, all elements of order $p$ in $G$ are conjugate to a power of some fixed element $x$. Let $u$ be a torsion unit in \mbox{\rm V}(\mbox{\mathbb{Z}}G) of order $p$. Then $\sum_{i=1}^{p-1}u^{i}$ and $\sum_{i=1}^{p-1}x^{i}$ have the same partial augmentations.

Proof.

Let $k$ be the number of conjugacy classes of elements of order $p$ in $G$. By Corollary 4 and Theorem 2.3 in Hertweck (2007a),

[TABLE]

Applying again Theorem 2.3 from Hertweck (2007a), the corollary follows. ∎

We will apply this by means of the following formula relating ranks of an idempotent to arithmetical properties of the group.

Corollary 6**.**

Suppose that for a prime divisor $p$ of the order of $G$, all elements of order $p$ in $G$ are conjugate to a power of some fixed element $x$. Suppose further that \mbox{\rm V}(\mbox{\mathbb{Z}}G) contains an elementary abelian subgroup $U$ of order $p^{2}$. Then for any ordinary character $\chi$ of $G$,

[TABLE]

We now turn to the proof of Theorem A. Suppose that $G$ has a cyclic Sylow $p$-subgroup $P$ ($p=2$ is allowed). Let $x$ be an element of order $p$ in $P$, and set $N=\text{N}_{G}(\langle{x}\rangle)$. Suppose further that \mbox{\rm V}(\mbox{\mathbb{Z}}G) contains an elementary abelian subgroup $U$ of order $p^{2}$. Let $\chi$ be the character of $G$ which is induced from the principal irreducible character of $P$. Then the rank in (1) is

[TABLE]

If $\chi$ is a character of $G$ which is induced from a faithful irreducible character of $P$, the rank in (1) is

[TABLE]

The difference of these ranks is $|N:P|(p^{2}+p)/p^{2}$, which is not an integer. This contradiction proves the theorem.

In view of Corollaries 1 and 2, one may be tempted to investigate the analogous problem for groups with dihedral Sylow $2$-subgroups. These groups were classified by Gorenstein and Walter, and listed, for example, on p. 462 in Gorenstein (1968). To indicate what can be done by now, we end with an example.

Note that the order of a finite subgroup of \mbox{\rm V}(\mbox{\mathbb{Z}}G) divides the order of $G$, see Lemma 37.3 in Sehgal (1993); a fact which, surprisingly enough from today’s point of view, is in this generality not recorded in Higman’s thesis.

Example 7**.**

For the alternating group $A_{7}$, any finite $2$-subgroup of \mbox{\rm V}(\mbox{\mathbb{Z}}A_{7}) is isomorphic to a subgroup of $A_{7}$.

Proof.

Sylow $2$-subgroups of $A_{7}$ are dihedral of order $8$. Let $x$ be an element of order $4$ in $A_{7}$. Then $x^{G}$ and $(x^{2})^{G}$ are the only conjugacy classes of elements of order $4$ and $2$, respectively. There is an (irreducible) character $\chi$ of $A_{7}$ of degree $6$ which is afforded by a deleted permutation representation. We have $\chi(x)=0$ and $\chi(x^{2})=2$.

Let $U$ be a finite $2$-subgroup of \mbox{\rm V}(\mbox{\mathbb{Z}}A_{7}). If $U$ is of order $2$, then $U$ is rationally conjugate to a subgroup of $A_{7}$ by Corollary 3.5 in Hertweck (2006). If $U$ is of order $4$, the Luthar–Passi method as described in Hertweck (2007a) is not sufficient to guarantee rational conjugacy to a subgroup of $A_{7}$: for a unit $u$ of order $4$ in \mbox{\rm V}(\mbox{\mathbb{Z}}A_{7}) one cannot exclude the possibility of having $(\varepsilon_{x^{2}}(u),\varepsilon_{x}(u))=(2,-1)$ when $\chi(u)=4$. In this case, also $\chi(u^{-1})=4$. Anyway, $U$ is isomorphic to a subgroup of $A_{7}$, and the same is true if $U$ is a Klein’s four group.

Suppose that $U$ is abelian of order $8$. By the Cohn–Livingstone result, $U$ is not cyclic. Set $e=\frac{1}{8}\sum_{u\in U}u$. Since $e$ is an idempotent, $\chi(u)$ is a rational integer. If $U$ is elementary abelian, then $\chi(e)=\frac{1}{8}(\chi(1)+7\chi(x^{2}))=\frac{20}{8}$, which is impossible. Thus $U$ contains $3$ elements of order $2$ and $4$ elements of order $4$. Trying out all possibilities shows that again $\chi(e)$ is not a rational integer.

It remains to consider the case when $U$ is the quaternion group. Let $u$ be an element of order $4$ in $U$. Since $\chi(u^{2})=\chi(x^{2})$, the restriction of the character $\chi$ to $U$ is the sum of four linear characters and the one of degree two. But this is impossible since $\chi$ is afforded by a rational representation, while the character of degree two of the quaternion group comes from the block of the rational quaternion algebra (whence the name of the group). ∎