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Here is a problem I have been grappling with all day. I started out thinking it might be true but am now inclined to believe it is false, and would like to see a counterexample.

Suppose $G$ is a finite group and $p$ a rational prime, such that $G$ has a unique (normal) sylow $p$-subgroup $P.$ If you want, assume that $G/P$ is cyclic of prime power order.

Let $R$ be either the ring $\mathbb{Z}$ or $\mathbb{Z}_{(p)}.$ By an $R$-representation of $G$ I mean a finitely generated $R[G]$-module which is free as an $R$-module.

Suppose $V$ is such an $R$-representation. Let $\bar{\mathbb{F}_p}$ denote the algebraic closure of the finite field with $p$ elements. By the Krull-Schmidt theorem, the $\bar{\mathbb{F}_p}[G]$-module $V \otimes_R \bar{\mathbb{F}_p}$ can be written as a sum of indecomposables, uniquely so up to reordering. Now in general $\bar{\mathbb{F}_p}[G]$ admits many indecomposable modules, but a particularly nice class is obtained by decomposing $\bar{\mathbb{F}_p}[G/P]$ into one-dimensional constituents (which works because $|G/P|$ is coprime to $p$), and indeed these are precisely the one-dimensional indecomposable $\bar{\mathbb{F}_p}[G]$-modules.

Let me write $\chi(V)$ for the tensor products of all the one-dimensional indecomposables occurring in $V \otimes_R \bar{\mathbb{F}_p}$. I shall say $V$ is good if $\chi(V)^{\otimes 2}$ is the trivial one-dimensional module.

Question: Suppose $V$ is a good $R$-representation and $V = W \oplus W'$ (decomposition as $R$-representations). Does it follow that $W$ is good?

Why one might think this is true: the point is that the representation is integral, so its ordinary character is real. In ordinary representation theory, the number of times a constituent occurs is determined by taking the inner product of characters; it follows that (over $\mathbb{C}$) every one-dimensional constituent with a non-real character occurs precisely as many times as its inverse. Thus multiplying together the one-dimensionals only leaves a real character, which has order two. That is to say, if we do decomposition over $\mathbb{C}$ instead of $\bar{\mathbb{F}_p},$ every $R$-representation is good.

Why one might think this is false: I offer two explanations. For one, the crux of the above argument is the complex conjugation action on a suitable ring of integers, which can be used to identify dual representations. However, this action does not (in general) descend to the finite field. Indeed it may happen that $\mathbb{F}_p$ already contains enough roots of unity to split $\mathbb{F}_p[G/P],$ and so it does not seem possible to use Galois actions like this.

A more direct reason why this proof breaks down is as follows. Instead of trying to decompose over $\bar{\mathbb{F}_p}$ one might try to decompose over the completion $\mathbb{Z}_p$ (this ring of characteristic zero also has Krull-Schmidt). But the problem then is that there can be indecomposable $\mathbb{Z}_p$-modules of dimension $\ge 2$ which have the simple one-dimensional modules as decomposition factors. (And somehow the decomposition factors are the only thing I know how to control via character methods.) Also again note that $\mathbb{Q}_p$ may already contain all the roots of unity we need, so it does not seem possible to use Galois actions.

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  • $\begingroup$ quick clarification: is W' the dual to W, or just another factor? $\endgroup$ – Phil Tosteson Jun 30 '14 at 3:36
  • $\begingroup$ Just some random factor. I want to know if "being good" is stable under summands. $\endgroup$ – Tom Bachmann Jun 30 '14 at 8:47
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For any finite group $G$ and finite-dimensional $\mathbb{F}_p[G]$-module $M$, I'll denote by $\Omega M$ the kernel of an $\mathbb{F}_p[G]$-projective cover $P(M)\to M$.

Let $F(M)$ be a free $\mathbb{Z}[G]$-module with a surjection $F(M)\to M$, and complete to a short exact sequence $$0\to K(M)\to F(M)\to M\to0$$ of $\mathbb{Z}[G]$-modules. Then $K(M)$ is $\mathbb{Z}$-free.

Since $pK(M)\leq pF(M)\leq K(M)$, we get a short exact sequence $$0\to pF(M)/pK(M)\to K(M)/pK(M)\to K(M)/pF(M)\to 0$$ of $\mathbb{F}_p[G]$-modules, where it's easy to check that $pF(M)/pK(M)\cong M$ and $K(M)/pF(M)$, since it's the kernel of a surjection $F(M)/pF(M)\to M$ from a projective $\mathbb{F}_p[G]$-module to $M$, is isomorphic to the direct sum $\Omega M\oplus P$ for some projective $\mathbb{F}_p[G]$-module $P$.

This sequence must certainly split if $\operatorname{Ext}^1_{\mathbb{F}_p[G]}(\Omega M, M)=0$ (or equivalently $\operatorname{Ext}^2_{\mathbb{F}_p[G]}(M,M)=0$), and so in this case $K(M)$ is a $\mathbb{Z}$-free $\mathbb{Z}[G]$-module whose reduction mod $p$ is $M\oplus\Omega M\oplus P$.

Alternatively, it's not hard to see that whether the sequence splits is independent of the choice of $F(M)$, and depends only on the restriction of $M$ to a Sylow $p$-subgroup of $G$. Also, if $M$ lifts to a $\mathbb{Z}$-free $\mathbb{Z}[G]$-module $\tilde{M}$ then one can take $F(M)$ to be a free module that maps onto $\tilde{M}$, and if $\tilde{K}(M)$ is the kernel of $F(M)\to\tilde{M}$ then $K(M)/pK(M)=pF(M)/pK(M)\oplus \tilde{K}(M)/p\tilde{K}(M)$, so there is an explicit splitting. So if the restriction of $M$ to a Sylow $p$-subgroup lifts (which will always be the case if $G$ has a normal Sylow $p$-subgroup and $M$ is simple) then the sequence splits.

Now take, for example, $G=C_7\rtimes C_3$, and $p=7$, and let $M$ be any of the three one-dimensional $\mathbb{F}_p[G]$-modules. Then all of the above applies, and so there is a $\mathbb{Z}$-free $\mathbb{Z}[G]$-module $V$ whose reduction mod $p$ has only one one-dimensional summand, which is isomorphic to $M$ (since $\Omega M$ is indecomposable and $6$-dimensional, and the indecomposable projectives are $7$-dimensional). By taking direct sums of such modules, you can construct a $V$ whose reduction mod $p$ has an arbitrary collection of one-dimensional summands, which shows that the answer to your question is "no".

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