Cohomology of Burnside Rings

TL;DR

This paper studies the cohomology of Burnside rings of finite groups, providing a complete description for square-free group orders and demonstrating unbounded complexity otherwise.

Contribution

It offers a complete characterization of Ext and Tor groups for Burnside rings when the group order is square-free, and shows unbounded growth in non-square-free cases.

Findings

Complete description of Ext and Tor for square-free groups

Unbounded rank of Ext and Tor when group order is not square-free

Provides algebraic invariants related to group structure

Abstract

Let $G$ be a finite group and $A(G)$ its Burnside ring. For $H \subset G$ let $\mathbb{Z}_H$ denote the $A(G)$-module corresponding to the mark homomorphism associated to $H$. When the order of $G$ is square-free we give a complete description of the $A(G)$-modules $\textrm{Ext}^l_{A(G)}(\mathbb{Z}_H, \mathbb{Z}_J)$ and $\textrm{Tor}^{A(G)}_l(\mathbb{Z}_H, \mathbb{Z}_J)$ for any $H, J \subset G$ and $l \geq 0$. We show that if the order of $G$ is not square-free then there exist $H, J \subset G$ such that $\textrm{Ext}^l_{A(G)}(\mathbb{Z}_H, \mathbb{Z}_J)$ and $\textrm{Tor}^{A(G)}_l(\mathbb{Z}_H, \mathbb{Z}_J)$ have unbounded rank as finite groups.

Equations

$$\pi :A(G)\to \prod _{(H)\in \mathrm{ccs}(G)}\mathbb{Z}$$

$$a\mapsto ({\pi }_{(H)}(a){)}_{(H)\in \mathrm{ccs}(G)}.$$

$$N\cdot e_i=N_i{s}_i\in R,$$

$$\mathbb{Q}{\otimes }_{\mathbb{Z}}R\simeq \mathbb{Q}{\otimes }_{\mathbb{Z}}\mathrm{Gh}(I)\simeq \prod _{i\in I}\mathbb{Q},$$

$$\theta :\mathbb{Q}\otimes S\to \prod _{i\in I^{\prime }}\mathbb{Q}$$

$${\mathrm{Ext}}_{R_{\mathbb{Q}}}^l(\mathbb{Q}{\otimes }_{\mathbb{Z}}{\mathbb{Z}}_i,\mathbb{Q}{\otimes }_{\mathbb{Z}}{\mathbb{Z}}_j)=0$$

$$\mathbb{Q}{\otimes }_{\mathbb{Z}}{\mathrm{Ext}}_R^l({\mathbb{Z}}_i,{\mathbb{Z}}_j)\simeq {\mathrm{Ext}}_{R_{\mathbb{Q}}}^l(\mathbb{Q}{\otimes }_{\mathbb{Z}}{\mathbb{Z}}_i,\mathbb{Q}{\otimes }_{\mathbb{Z}}{\mathbb{Z}}_j)$$

$${\mathrm{Ext}}_S^l(M,N)\simeq {\mathrm{Ext}}_{\overline{S}}^l(\overline{M},N)$$

$${\mathrm{Tor}}_l^S(M,N)\simeq {\mathrm{Tor}}_l^{\overline{S}}(\overline{M},N)$$

$${\mathrm{Hom}}_S(X,N)\simeq {\mathrm{Hom}}_{\overline{S}}(\overline{X},N).$$

$$0\to {\mathbb{Z}}_j\stackrel{p}{\to }{\mathbb{Z}}_j\to k_j\to 0,$$

$${\mathrm{Ext}}_R^l({\mathbb{Z}}_i,{\mathbb{Z}}_j)\simeq \underset{q}{⨁}{M}_{l,q}$$

$$a_{l+1}$$

$$={\dim }_k\mathrm{im}\left({\mathrm{Ext}}_{\overline{R}}^l(k_i,k_j)\to {\mathrm{Ext}}_R^{l+1}({\mathbb{Z}}_i,{\mathbb{Z}}_j)\right)$$

$$=b_l-a_l$$

$$z_l=y_{l+1}-z_{l+1}$$

$$\theta :\overline{R}\to \prod _{E\in \mathcal{E}}{k}_E$$

$$R\to \prod _{E\in \mathcal{E}}{k}_E$$

$$r\mapsto (r(i)\mathrm{mod}p{)}_{E\in \mathcal{E}}$$

$$q=p\cdot \sum _{i\in I}{m}_i{s}_i$$

$$a_{l+1}=b_l-a_l$$

$$z_l=y_{l+1}-z_{l+1}$$

$$0\to {\mathrm{Hom}}_R({\mathbb{Z}}_i,{\mathbb{Z}}_i)\stackrel{p}{\to }{\mathrm{Hom}}_R({\mathbb{Z}}_i,{\mathbb{Z}}_i)\to {\mathrm{Hom}}_R({\mathbb{Z}}_i,k_i)\to 0.$$

$${\mathrm{Ext}}_R^l({\mathbb{Z}}_i,{\mathbb{Z}}_j)=0$$

$$a_l=\{\begin{array}{ll}0& \text{ if l odd}\\ 1& \text{ if l even}\end{array}$$

$$a_l^{\prime }=\{\begin{array}{ll}1& \text{ if l odd}\\ 0& \text{ if l even}\end{array}$$

$$\dots \to A\to \dots \to A\to A\to k\to 0$$

$${\mathrm{Ext}}_{A(G)}^l({\mathbb{Z}}_H,{\mathbb{Z}}_J)=0$$

$$M_{H,l}\simeq \{\begin{array}{ll}0& \text{ if l odd}\\ \mathbb{Z}/p\mathbb{Z}& \text{ if l even};\end{array}$$

$$N_{H,J,l}\simeq \{\begin{array}{ll}\mathbb{Z}/p\mathbb{Z}& \text{ if l odd}\\ 0& \text{ if l even}.\end{array}$$

$$d(S)\ge \dim \mathcal{M}/{\mathcal{M}}^2-1,$$

Full text

1. Let $H,J$ be subgroups of $G$.
2. (1)

$\pi_{H}([G/H])=[N_{G}H:H]$. 3. (2)

$\pi_{J}([G/H])\neq 0$ if and only if $J$ is conjugate to a subgroup of $H$. 4. (3)

$\pi_{H}=\pi_{J}$ if and only if $(H)=(J)$.

For each $(H)\in\operatorname{ccs}(G)$ we then have a well-defined homomorphism $\pi_{(H)}=\pi_{H}$. We have a homomorphism of rings

[TABLE]

where for $a\in A(G)$,

[TABLE]

This is an embedding of rings. The ring $\prod_{(H)\in\operatorname{ccs}(G)}\mathbb{Z}$ is called the ghost ring of $G$. Most of our results on the Burnside ring hold more generally for arbitrary subrings of the ghost ring, and in the next section we introduce the notion of a $B$-ring in order to provide the appropriate setting for stating these results.

Let $R$ be a commutative ring. Let Ab be the category of abelian groups and ${R}\textrm{-Mod}$ the category of left $R$-modules. For $R$-modules $M$, $N$, the functors $\operatorname{Hom}_{R}(M,-):{R}\textrm{-Mod}\to\textrm{Ab}$ and $\operatorname{Hom}_{R}(-,N):{R}\textrm{-Mod}\to\textrm{Ab}$ are left exact. For $l$ a non-negative integer, we write $\operatorname{Ext}^{l}_{R}(M,N)$ for the $l$th right derived functor of $\operatorname{Hom}_{R}(M,-)$ applied to the module $N$, or equivalently for the $l$th right derived functor of $\operatorname{Hom}_{R}(-,N)$ applied to the module $M$. The functor $-\otimes_{R}N:{R}\textrm{-Mod}\to\textrm{Ab}$ is right-exact, and we write $\operatorname{Tor}_{l}^{R}(M,N)$ for the $l$th left derived functor of $-\otimes_{R}N$ applied to the module $M$. Since $R$ is a commutative ring, each $\operatorname{Ext}^{l}_{R}(M,N)$ and $\operatorname{Tor}^{R}_{l}(M,N)$ is naturally endowed with the structure of a left $R$-module. In what follows all modules over a commutative ring are left modules.

$B$-rings.

For $I$ a finite set, define $\operatorname{Gh}(I)=\prod_{i\in I}\mathbb{Z}$. Let $R$ be a subring of $\operatorname{Gh}(I)$ and for each $i\in I$ let $\pi_{i}$ be the corresponding projection $R\to\mathbb{Z}$. For $r\in R$, write $r(i)$ for $\pi_{i}(r)$.

If some subring $R\subset\operatorname{Gh}(I)$ is not a $B$-ring, with $i,j\in I$ a pair for which the above condition fails, then it is clear that $r(i)=r(j)$ for all $r\in R$. Then $R$ is isomorphic to the ring $S\subset\operatorname{Gh}(I-\{j\})$ obtained by omitting the factor corresponding to $j$. Repeating this process if necessary we obtain a subset $I^{\prime}\subset I$ and a $B$-ring $R^{\prime}\subset\operatorname{Gh}(I^{\prime})$ with $R$ isomorphic to $R^{\prime}$.

We give an intrinsic definition of these rings as follows.

[TABLE]

and so $N\cdot\operatorname{Gh}(I)\subset R\subset\operatorname{Gh}(I)$. Hence

[TABLE]

i.e. $\mathbb{Q}\otimes R$ is isomorphic to a product of $|I|$ copies of $\mathbb{Q}$.

Suppose $S$ is a commutative ring which is of finite rank and torsion-free as a $\mathbb{Z}$-module, with $\mathbb{Q}\otimes_{\mathbb{Z}}S$ a product of finitely copies of $\mathbb{Q}$. Then we have an isomorphism

[TABLE]

for some finite indexing set $I^{\prime}$.

Since $S$ is torsion-free, $\mathbb{Q}\otimes S$ contains a copy of $S$ as the subring $1\otimes S\subset\mathbb{Q}\otimes S$. Denote the image $\theta(1\otimes S)$ by $S^{\prime}\subset\prod_{i\in I^{\prime}}\mathbb{Q}$, and for each $i\in I^{\prime}$ let $\hat{\pi}_{i}$ denote the projection map $\prod_{i\in I^{\prime}}\mathbb{Q}\to\mathbb{Q}$ onto the $i$th factor. Write $\pi_{i}$ for the restriction of $\hat{\pi}_{i}$ to $S^{\prime}$. Since $S^{\prime}$ is of finite rank as a $\mathbb{Z}$-module we must have that $\pi_{i}(S)\subset\mathbb{Z}\subset\mathbb{Q}$ for each $i\in I^{\prime}$. We can then regard $S^{\prime}$ as sitting inside $\operatorname{Gh}(I^{\prime})=\prod_{i\in I^{\prime}}\mathbb{Z}\subset\prod_{i\in I^{\prime}}\mathbb{Q}$. We claim that this embedding defines a $B$-ring. For $s\in S$, write $s^{\prime}$ for the element $\theta(1\otimes s)$ of $S^{\prime}$. It remains to show that for each distinct pair $i,j\in I^{\prime}$ we can find an element $s\in S$ such that $\pi_{i}(s^{\prime})\neq 0$ and $\pi_{j}(s^{\prime})=0$.

Let $f_{1},\ldots,f_{n}$ be the primitive idempotents of $\operatorname{Gh}(I^{\prime})$, and note that $\hat{\pi}_{j}(f_{i})=1$ if $j=i$ and $\hat{\pi}_{j}(f_{i})=0$ otherwise. For each $i\in I$, we have $f_{i}=\theta(q_{i}\otimes t_{i})$ for some $q_{i}\in\mathbb{Q}$ and $t_{i}\in S$. Then $t^{\prime}_{i}=\theta(1\otimes t_{i})=(1/q_{i})\cdot\theta(q_{i}\otimes t_{i})=(1/q_{i})f_{i}$, and so $\pi_{i}(t^{\prime}_{i})=1/q_{i}\neq 0$ and $\pi_{j}(t^{\prime}_{i})=0$ for each $j\neq i$. Thus $t^{\prime}_{i}$ satisfies the condition of Definition 2 for any $j\neq i$, and the embedding $S^{\prime}\subset\operatorname{Gh}(I^{\prime})$ defines a $B$-ring. ∎

Let $R\subset\operatorname{Gh}(I)$ be a $B$-ring. For each $i\in I$ we have an $R$-module $\mathbb{Z}_{i}$ defined by letting $R$ act on the set $\mathbb{Z}$ by $r\cdot n=r(i)n$ for $r\in R$ and $n\in\mathbb{Z}$.

[TABLE]

for any $l\geq 1$ and $i,j\in I$. But

[TABLE]

(see e.g. [7] Proposition 3.3.10) and so $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ is torsion for any $l\geq 1$ and $i,j\in I$. Since it is also finitely generated, it is finite. The result for $\operatorname{Tor}^{R}_{l}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ follows similarly. ∎

Let $R\subset\operatorname{Gh}(I)$ be a $B$-ring and $M$ a finitely generated $R$-module. Let $\mathcal{F}_{\bullet}$ be a free $R$-module resolution of $M$, where $\mathcal{F}_{l}=R^{\oplus n_{l}}$ for $l\geq 0$. Applying $\operatorname{Hom}_{R}(-,\mathbb{Z}_{j})$ for some $j\in I$ gives a chain complex where each term is of the form $\operatorname{Hom}_{R}(R^{\oplus n_{l}},\mathbb{Z}_{j})\simeq\mathbb{Z}_{j}^{\oplus n_{l}}$. Applying $-\otimes_{R}\mathbb{Z}_{j}$ gives a chain complex where each term is of the form $R^{\oplus n_{l}}\otimes\mathbb{Z}_{j}\simeq\mathbb{Z}_{j}^{\oplus n_{l}}$. $\operatorname{Ext}^{l}_{R}(M,\mathbb{Z}_{j})$ and $\operatorname{Tor}^{R}_{l}(M,\mathbb{Z}_{j})$ are then isomorphic to subquotients of $\mathbb{Z}_{j}^{\oplus n_{l}}$, and it follows that the $R$-module structure of each $\operatorname{Ext}^{l}_{R}(M,\mathbb{Z}_{j})$ and $\operatorname{Tor}^{R}_{l}(M,\mathbb{Z}_{j})$ is given by $r\in R$ acting by $r(j)$. It follows that for $i,j\in I$ and $l\geq 0$, any direct sum decomposition of $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ or $\operatorname{Tor}^{R}_{l}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ as an abelian group is automatically a decomposition as an $R$-module.

By instead considering the functor $\mathbb{Z}_{i}\otimes_{R}-$ for $i\in I$, we note that for any $R$-module $N$, the $R$-module structure of $\operatorname{Tor}^{R}_{l}(\mathbb{Z}_{i},N)$ for each $l\geq 0$ is given by $r$ acting by $r(i)$. Similarly, by considering an injective resolution of $N$, we have that the $R$-module structure of each $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},N)$ is given by $r\in R$ acting by $r(i)$.

Since the $R$-module structure of each $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ and $\operatorname{Tor}^{R}_{l}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ is given by both $r$ acting by $r(i)$ and $r$ acting by $r(j)$, it follows that each indecomposable $R$-module summand of $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ and $\operatorname{Tor}^{R}_{l}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ must be of the form $\mathbb{Z}_{i}/m\mathbb{Z}_{i}\simeq\mathbb{Z}_{j}/m\mathbb{Z}_{j}$ for some $m\mid d(i,j)$.

Fix some rational prime $p$ and put $k=\mathbb{F}_{p}$. For a commutative ring $S$ write $\overline{S}$ for the quotient ring $S/pS\simeq S\otimes_{\mathbb{Z}}k$. For an $S$-module $M$, write $\overline{M}$ for the $\overline{S}$-module $M/pM$. If $M$ is annihilated by $p$, we will also denote the associated $\overline{S}$-module by $M$.

[TABLE]

and

[TABLE]

for each $l\geq 0$.

[TABLE]

Let $(\mathcal{F}_{\bullet},\partial_{\bullet})$ be a free $S$-module resolution of $M$. In particular $\mathcal{F}_{\bullet}$ is a free $\mathbb{Z}$-module resolution of the $\mathbb{Z}$-module $M$, so applying $-\otimes_{\mathbb{Z}}k$ gives a chain complex over $\overline{M}$ with homology groups $\operatorname{Tor}^{\mathbb{Z}}_{l}(M,k)$. But $M$ is torsion-free so the homology groups vanish and the chain complex is exact. Since $\mathcal{F}_{\bullet}$ is a free $S$-module resolution, $\mathcal{F}_{\bullet}\otimes_{\mathbb{Z}}k$ is a free $\overline{S}$-module resolution.

Applying $\operatorname{Hom}_{\overline{S}}(-,N)$ to $\mathcal{F}_{\bullet}\otimes_{\mathbb{Z}}k$ and computing cohomology then computes the groups $\operatorname{Ext}^{l}_{\overline{S}}(\overline{M},N)$. But $\operatorname{Hom}_{\overline{S}}(\overline{S},N)\simeq\operatorname{Hom}_{S}(S,M)$, so this is the same as applying $\operatorname{Hom}_{S}(-,N)$ to $\mathcal{F}_{\bullet}$ and taking cohomology, i.e. computing the groups $\operatorname{Ext}^{l}_{S}(M,N)$. The proof for $\operatorname{Tor}^{S}_{l}(M,N)$ is analogous. ∎

For a $B$-ring $R\subset\operatorname{Gh}(I)$ and $i\in I$, write $k_{i}$ for the $R$-module where $r\in R$ acts on the field $k$ by $r(i)$. Put $\overline{R}=R\otimes_{\mathbb{Z}}k$.

For each $j\in I$ we have a short exact sequence of $R$-modules

[TABLE]

where the map $\mathbb{Z}_{j}\to\mathbb{Z}_{j}$ is given by multiplication by $p$. Applying $\operatorname{Hom}_{R}(\mathbb{Z}_{i},-)$ gives a long exact sequence beginning with

{0}$${\operatorname{Hom}_{{R}}(\mathbb{Z}_{i},\mathbb{Z}_{j})}$${\operatorname{Hom}_{{R}}(\mathbb{Z}_{i},\mathbb{Z}_{j})}$${\operatorname{Hom}_{\overline{{R}}}(k_{i},k_{j})}$${\operatorname{Ext}^{1}_{{R}}(\mathbb{Z}_{i},\mathbb{Z}_{j})}$${\operatorname{Ext}^{1}_{{R}}(\mathbb{Z}_{i},\mathbb{Z}_{j})}$${\operatorname{Ext}^{1}_{\overline{{R}}}(k_{i},k_{j})}$${\operatorname{Ext}^{2}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})}$${\operatorname{Ext}^{2}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})}$${\ldots}$$p$$p$$p

where we make use of the additivity of $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},-)$ for each $l\geq 1$ to identify each map $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})\to\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ as multiplication by $p$, and make use of Lemma 6 above to replace each $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},k_{j})$ with $\operatorname{Ext}^{l}_{\overline{R}}(k_{i},k_{j})$.

For each $l\geq 1$ and each rational prime $q$, let $M_{l,q}$ be the submodule of $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ annihilated by some power of $q$. Since $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ is finite, it follows that we have a decomposition of $R$-modules

[TABLE]

where the sum is over all rational primes $q$.

For $l\geq 1$, let $a_{l}$ be the rank of $M_{l,p}$, and let $b_{l}$ be the $k$-dimension of $\operatorname{Ext}^{l}_{\overline{R}}(k_{i},k_{j})$. Note that $a_{l}$ is equal to the number of summands appearing in a decomposition of $M_{l,p}$ as a sum of indecomposable $R$-modules. For a non-zero indecomposable summand $\mathbb{Z}/p^{\alpha}\mathbb{Z}$ of $M_{l,p}$, the map $\mathbb{Z}/p^{\alpha}\mathbb{Z}\xrightarrow{p}\mathbb{Z}/p^{\alpha}\mathbb{Z}$ has kernel $p^{\alpha-1}\mathbb{Z}/p^{\alpha}\mathbb{Z}\simeq\mathbb{Z}/p\mathbb{Z}$. So for $l\geq 1$, the kernel of the map $\left(\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})\xrightarrow{p}\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})\right)$ is a $k$-vector space with dimension $a_{l}$. Similarly, the cokernel of this map is a $k$-vector space of dimension $a_{l}$, and so the image of the connecting homomorphism $\operatorname{Ext}^{l}_{\overline{R}}(k_{i},k_{j})\to\operatorname{Ext}^{l+1}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ has dimension $b_{l}-a_{l}$. Hence

[TABLE]

for $l\geq 1$. In order to determine the sequence $a_{l}$, it is then sufficient to compute the sequence $b_{l}$.

Applying $\mathbb{Z}_{i}\otimes_{R}-$ to $(\dagger)$ gives a long exact sequence ending with

{\ldots}$${\operatorname{Tor}_{2}^{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})}$${\operatorname{Tor}_{2}^{\overline{{R}}}(k_{i},k_{j})}$${\operatorname{Tor}_{1}^{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})}$${\operatorname{Tor}_{1}^{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})}$${\operatorname{Tor}_{1}^{\overline{{R}}}(k_{i},k_{j})}$${\mathbb{Z}_{i}\otimes\mathbb{Z}_{j}}$${\mathbb{Z}_{i}\otimes\mathbb{Z}_{j}}$${k_{i}\otimes\mathbb{Z}_{j}}$${0.}$$p$$p

Write $z_{l}$ for the $p$-rank of $\operatorname{Tor}^{R}_{l}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ and $y_{l}$ for the dimension $\operatorname{Tor}^{\overline{R}}_{l}(k_{i},k_{j})$. Then repeating the same argument as above gives

[TABLE]

for $l\geq 1$.

$B$-rings modulo a prime.

Since a $B$-ring $R\subset\operatorname{Gh}(I)$ is of finite rank as a $\mathbb{Z}$-module, $\overline{R}$ is a finite-dimensional $k$-algebra, and we have an $\overline{R}$-module decomposition of $\overline{R}$ as a direct sum of finitely many indecomposable projective $\overline{R}$-modules. Since $\overline{R}$ is commutative, this is a decomposition of commutative local $k$-algebras. By the usual block theory considerations (see e.g. [1] Chapter II.5), studying the cohomology of $\overline{R}$ reduces to studying the cohomology of these indecomposable summands.

Define a relation $\sim^{\prime}_{p}$ on $I$ by putting $i\sim^{\prime}_{p}j$ if and only if $p\mid d(i,j)$ for $i\neq j$. Note that by the definition of $d(i,j)$ this relation is symmetric and transitive, and we write $\sim_{p}$ for the equivalence relation defined by taking its reflexive closure. Let $\mathcal{E}$ denote the set of equivalence classes of $I$ with respect to $\sim_{p}$. For an equivalence class $E\in\mathcal{E}$, write $k_{E}$ for the (well-defined) $R$-module which is $k$ as an abelian group and where $r\cdot m=r(i)m$ for $m\in k$ and where $i$ is any element of $E$.

[TABLE]

with kernel the radical of $\overline{R}$.

[TABLE]

given by

[TABLE]

where $i\in E$. By Lemma 7 this homomorphism is surjective. The kernel of this map contains $pR$, and we let $\theta$ be the induced surjective homomorphism of $k$-algebras. Since $\prod_{E\in\mathcal{E}}k_{E}$ is semisimple, the kernel of $\theta$ certainly contains the radical of $\overline{R}$. It remains to show the reverse inclusion.

As in the proof of Proposition 3, choose for each $i\in I$ an element $s_{i}\in R$ such that $s_{i}(j)\neq 0$ if and only if $j=i$. Let $s_{i}(i)=p^{t_{i}}n_{i}$ where $n_{i}$ is coprime to $p$; put $t=\max_{i}t_{i}$ and put $N=\prod_{i\in I}n_{i}$. Let $r\in R$ be such that $r(i)\equiv 0\mod p$ for each $i\in I$. Putting $q=Nr^{t+1}$, we have that $ps_{i}(i)\mid q(i)$ for each $i\in I$, and so we can define integers $m_{i}\in\mathbb{Z}$ by requiring $q(i)=pm_{i}s_{i}(i)$. It follows that

[TABLE]

and so $q\in pR$. Then the image of $q$ in $\overline{R}$ is zero, and hence the image of $r^{t+1}$ in $\overline{R}$ is zero, since $N$ is coprime with $p$.

It follows that the kernel of $\theta$ is nilpotent, and hence equal to the radical of $\overline{R}$. ∎

1. i.

$\overline{R}$ is the direct sum of $|\mathcal{E}|$ indecomposable $k$-algebras; 2. ii.

the set $\{k_{E}\}_{E\in\mathcal{E}}$ is a complete irredundant set of irreducible modules for $\overline{R}$; 3. iii.

the dimension of the indecomposable summand corresponding to $k_{i}$ is $|E|$, the cardinality of the $\sim_{p}$-equivalence class of $i\in I$. The maximal ideal of each indecomposable summand has codimension 1.

It follows that each indecomposable summand of $\overline{R}$ is a commutative local finite-dimensional $k$-algebra $S$ with maximal ideal $\mathcal{M}$ satisfying $S/\mathcal{M}\simeq k$. There is a unique $S$-module structure on $k$ where $\mathcal{M}\cdot k=0$, and we denote this $S$-module by $k$. The following result is standard (see [4] Chapter 2, §3).

[TABLE]

and

[TABLE]

for $l\geq 1$, where $b_{l}$ and $y_{l}$ are the dimensions of the $k$-vector spaces $\operatorname{Ext}^{l}_{\overline{R}}(k_{i},k_{j})$ and $\operatorname{Tor}^{\overline{R}}_{l}(k_{i},k_{j})$ respectively. If $i\not\sim_{p}j$ then $\operatorname{Ext}^{l}_{\overline{R}}(k_{i},k_{j})=\operatorname{Tor}^{\overline{R}}_{l}(k_{i},k_{j})=0$ for each $l$ and the result is trivially true. Otherwise, by Lemma 10 we have $b_{l}=y_{l}$.

Suppose $i=j$. It is clear that $\operatorname{Hom}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{i})\simeq\mathbb{Z}_{i}$ and $\operatorname{Hom}_{R}(\mathbb{Z}_{i},k_{i})\simeq k_{i}$ so the long exact sequence associated to $(\dagger)$ begins with the short exact sequence

[TABLE]

It follows that the map $\operatorname{Ext}^{1}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{i})\xrightarrow{p}\operatorname{Ext}^{1}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{i})$ is injective and hence $\operatorname{Ext}^{1}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{i})$ has zero $p$-part and $a_{1}=0$. Similarly, $z_{1}=b_{1}$, and the recurrence gives $z_{l}=a_{l+1}$ for all $l\geq 1$ as claimed.

Suppose $i\neq j$ with $i\sim_{p}j$. It is clear that $\operatorname{Hom}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})=0$ and $\operatorname{Hom}_{R}(\mathbb{Z}_{i},k_{j})\simeq k_{j}$, so by inspection of the long exact sequence associated to $(\dagger)$ we have $a_{1}=1$. Similarly we have $\mathbb{Z}_{i}\otimes\mathbb{Z}_{j}\simeq\mathbb{Z}_{j}/d(i,j)\mathbb{Z}_{j}$ and $\mathbb{Z}_{i}\otimes k_{j}\simeq k_{j}$ from which we obtain $z_{1}=b_{1}-1$, and once more we have $z_{l}=a_{l+1}$ for each $l\geq 1$. ∎

[TABLE]

for all $l\geq 0$.

Since $p\nmid d(i,j)$, the simples modules $k_{i}$ and $k_{j}$ are non-isomorphic and therefore belong to different blocks of the commutative $k$-algebra $\overline{R}$. Then $\operatorname{Ext}^{l}_{\overline{R}}(k_{i},k_{j})=0$ for each $l\geq 0$, and so the $p$-part of $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ is zero for each $l\geq 0$. Since this is true of any prime $p$, and since $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})$ is finite for $l\geq 1$, it follows that $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{i},\mathbb{Z}_{j})=0$ for all $l\geq 1$. ∎

[TABLE]

[TABLE]

for all $l\geq 1$.

[TABLE]

where each map $A\to A$ is given by $1_{A}\mapsto x$, and so $\operatorname{Ext}^{l}_{A}(k,k)\simeq k$ for all $l\geq 0$, i.e. in the notation of our recurrence relations we have $b_{l}=1$ for all $l\geq 1$. Now $a_{l+1}=b_{l}-a_{l}$ and $a^{\prime}_{l+1}=b_{l}-a^{\prime}_{l}$, where $a_{1}=0$ and $a^{\prime}_{1}=1$, from which the corollary follows immediately. ∎

The Burnside ring as $B$-ring.

For $H\subset G$ and $p$ a prime, let $O^{p}(H)$ denote the smallest normal subgroup $K\triangleleft H$ such that $H/K$ is a $p$-group. We recall the following result due to Dress ([2] Proposition 1).

Since $A(G)$ is a $B$-ring, we have integers $d((H),(J))$ defined for each distinct pair $(H),(J)\in\operatorname{ccs}(G)$. For $H,J\subset G$ non-conjugate subgroups, we write $d(H,J)$ for $d((H),(J))$. It follows immediately from Proposition 16 that $p\mid d(H,J)$ if and only if $O^{p}(J)$ is conjugate to $O^{p}(H)$.

[TABLE]

for all $l\geq 0$.

1. i.

each $E\in\mathcal{E}$ has cardinality $\leq 2$; 2. ii.

$M_{H,l}$ is non-zero for some $l\geq 1$ if and only if $(H)\in E_{i}$ for some $1\leq i\leq q$; 3. iii.

$N_{H,J,l}$ is non-zero for some $l\geq 1$ if and only if $\{(H),(J)\}=E_{i}$ for some $1\leq i\leq q$; 4. iv.

if $(H)\in E_{i}$ for some $1\leq i\leq q$ then

[TABLE] 5. v.

if $\{(H),(J)\}=E_{i}$ for some $1\leq i\leq q$, then

[TABLE]

For parts ii and iii, we know by Corollary 14 that if $(H)\in E_{i}$ for some $i$ then $M_{H,l}$ is non-zero whenever $l$ is even, and if $\{(H),(J)\}=E_{i}$ for some $i$ then $N_{H,J,l}$ is non-zero whenever $l$ is odd. Otherwise $(H)\notin E_{i}$ for each $i$, in which case by part i $\{(H)\}$ is an equivalence class in $\mathcal{E}$ and so the block corresponding to $H$ is 1-dimensional and $M_{H,l}=0$ for all $l\geq 1$. Similarly, if $\{(H),(J)\}\neq E_{i}$ for each $i$, then $(H)$ and $(J)$ belong to distinct equivalence classes and $N_{H,J,l}=0$ for all $l\geq 0$.

Suppose $\{(H),(J)\}$ is an equivalence class in $\mathcal{E}$. By Corollary 14, $M_{H,l}$ has a $p$-power summand only when $l$ is even, and $N_{H,J,l}$ has a $p$-power summand only when $l$ is odd. It remains to show that each $p$-power summand is in fact $\mathbb{Z}/p\mathbb{Z}$. For $N_{H,J,l}$ this is clear since $d(H,J)$ annihilates $N_{H,J,l}$ and $p^{2}\nmid d(H,J)$ by Lemma 18. For $M_{H,l}$ note that by dimension shifting this is the $p$-part of $\operatorname{Ext}^{l-1}_{A(G)}(K_{H},\mathbb{Z}_{H})$ where $K_{H}=\ker\pi_{H}$. Since $p^{2}\nmid d(H,J^{\prime})$ for any $J^{\prime}\subset G$, and since $p\mid d(H,J^{\prime})$ if and only if $(J^{\prime})=(J)$, we can construct an $s_{H}\in A(G)$ such that $\pi_{J^{\prime}}(s_{H})\neq 0$ if and only if $(H)=(J^{\prime})$, and such that $p^{2}\nmid\pi_{H}(s_{H})$. Now $s_{H}$ annihilates $K_{H}$, so it follows that $s_{H}$ annihilates $\operatorname{Ext}^{l}_{A(G)}(\mathbb{Z}_{H},\mathbb{Z}_{H})$, and so a fortiori $s_{H}$ annihilates $M_{H,l}$. So any $p$-power summand of $M_{H,l}$ must be of the form $\mathbb{Z}/p\mathbb{Z}$. ∎

In the remainder we establish the converse, by showing that if $|G|$ is not square-free then there exists $H,J\subset G$ such that the rank of $\operatorname{Ext}^{l}_{R}(\mathbb{Z}_{H},\mathbb{Z}_{J})$ is unbounded as $l\to\infty.$ This is an easy consequence of the following two results. Recall that for a field $F$, a commutative $F$-algebra $S$ is said to be symmetric if there exists an $F$-linear map $\lambda:S\to F$ such that $\ker\lambda$ contains no non-zero ideals of $S$.

[TABLE]

where $d(S)$ denotes the Krull dimension of $S$.

If $\dim\mathcal{M}/\mathcal{M}^{2}=0$ then $S=k$ is clearly symmetric. Suppose $\dim\mathcal{M}/\mathcal{M}^{2}=1$. Let $t\in S$ generate $\mathcal{M}$ and note that $\{1_{S},t,\ldots,t^{q}\}$ is a vector space basis for $S$ for some $q\geq 1$. Define $\lambda:S\to k$ by putting

[TABLE]

Now if $J\subset S$ is a non-zero ideal then we can choose some non-zero element $s=\sum_{i=0}^{q}b_{i}t^{i}$ in $J$, and choose $m$ to be minimal such that $b_{m}\neq 0$. Then $t^{q-m}s=b_{m}t^{q}\in J$ is not in $\ker\lambda$, so $\ker\lambda$ contains no non-zero ideals and $S$ is symmetric, a contradiction. It follows that $\dim\mathcal{M}/\mathcal{M}^{2}>1$ and the sequence $(\dim\operatorname{Tor}^{S}_{l}(k,k))_{l\in\mathbb{N}}=(\dim\operatorname{Ext}^{l}_{S}(k,k))_{l\in\mathbb{N}}$ is unbounded.

By Corollary 9, the summand $S$ corresponds to some equivalence class $E\in\mathcal{E}$, and we have

[TABLE]

Let $H,J$ be subgroups of $G$ with $(H),(J)\in E$, let $\alpha_{l}$ be the $p$-rank of $\operatorname{Ext}^{l}_{A(G)}(\mathbb{Z}_{H},\mathbb{Z}_{J})$, and let $\beta_{l}$ be the dimension of $\operatorname{Ext}^{l}_{\overline{A(G)}}(k_{E},k_{E})$. We have a recurrence

[TABLE]

for $l\geq 1$. Since $\beta_{l}$ is unbounded, it follows immediately that $\alpha_{l}$ is unbounded, and hence the groups $\operatorname{Ext}^{l}_{A(G)}(\mathbb{Z}_{H},\mathbb{Z}_{J})$ have unbounded rank. ∎