# How to recover the integral group ring?

I would like to solve the following exercise:

Suppose $R$ is a commutative semisimple ring of characteristic $p^t, t\geq1$, and we have two finite groups $G_1=H_1 \times A_1$ and $G_2=H_2 \times A_2$. Now $H_1$ and $H_2$ are finite $p$-groups, where $p \nmid \vert A_i \vert, i=1,2$. Prove that $RG_1 \cong RG_2$ iff $\mathbb Z A_{1} \cong \mathbb Z A_2$ and $\mathbb{Z} H_1 \cong \mathbb{Z} H_2$.

Having no idea how to start, I first tried solving a different exercise, also related to direct products:

If $G_1=H_1 \times A_1$ is a finite group, where $(\vert H_1 \vert,\vert A_1 \vert)=1$, and $G_2$ is any other group, then: $\mathbb Z G_1 \cong \mathbb Z G_2$ iff $G_2=H_2 \times A_2$, $\mathbb{Z}A_1 \cong \mathbb Z A_2$ and $\mathbb Z H_1 \cong \mathbb Z H_2$.

Proof: One direction is clear, since $\mathbb Z (G \times H) \cong \mathbb Z G \otimes \mathbb Z H$. For the other direction, since $H_1 \trianglelefteq G_1$, by the normal subgroup correspondence, there exists a unique normal subgroup $H_2 \trianglelefteq G_2$ such that $\mathbb Z (G_2 / H_2) \cong \mathbb Z (G_1/H_1) \cong \mathbb Z A_1$. Doing the same for $A_1 \trianglelefteq G_1$ gives $\mathbb{Z}(G_2/A_2) \cong \mathbb Z H_1$. The correspondence respects intersections, so $A_2 \cap H_2= \emptyset$. Since $\vert A_2 \times H_2 \vert= \vert A_2 \vert \cdot \vert H_2 \vert=\vert A_1 \vert \cdot \vert H_1 \vert =\vert G_1 \vert = \vert G_2 \vert$ ($\mathbb Z G_1$ and $\mathbb Z G_2$ have the same rank over $\mathbb Z$), I think we can conclude that $G_2=A_2 \times H_2$. $\square$

EDIT: Does anyone know why the orders are supposed to be coprime?

This doesn't really seem to help for the first exercise though. Since $R=k_1 \times k_2 \times \ldots \times k_n$, where the $k_i$ are fields of characteristic $p$, it is probably helpful to first consider $R=k$. Any help would be appreciated.

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## 1 Answer

Let $k$ be a field of characteristic $p$ and $G = H \times A$ where $H$ is a Sylow $p$-subgroup of $G$.

In this case we can recover $kG$ from $kH$ and $kA$; and vice versa.

$kG$ from $kH$ and $kA$: This is easy (you made a similar observation in the question too) since $kG = k[H \times A] \cong kH \otimes_k kA$. In particular we can recover $kG$ from $\mathbb{Z}H$ and $\mathbb{Z}A$ because $kH \cong k \otimes_{\mathbb{Z}} \mathbb{Z}H$ and $kA \cong k \otimes_{\mathbb{Z}} \mathbb{Z}A$.

$kH$ and $kA$ from $kG$: This can also be done because of the following facts:

• The group algebra $kA$ is isomorphic to the radical quotient $kG/ \operatorname{Rad} kG$. To see this, consider the surjective $k$-algebra map $$\alpha: kG \rightarrow k[G/H] \cong kA$$ given by the projection $G \rightarrow G/H$. Since $kA$ is semisimple, $\operatorname{Rad} kG \subseteq \ker \alpha$.

For the other inclusion, first note that $\sum_{g \in G}c_g g \in \ker \alpha$ means that if we sum $c_g$'s over a coset of $H$, we get $0$. From here we can see that $\ker \alpha = kG \cdot IH$ where $IH$ is the augmentation ideal of $kH$. Since $H$ is a $p$-group, actually $IH = \operatorname{Rad} kH$. On the other hand by Clifford's theorem, semisimple $kG$-modules remain semisimple when regarded as $kH$-modules by restriction; hence $\operatorname{Rad} kH \subseteq \operatorname{Rad}kG$. Therefore $\ker \alpha \subseteq \operatorname{Rad} kG$.

• $kH$ is isomorphic to the endomorphism algebra of the projective cover of the trivial $kG$-module. To see this, consider the "other" surjection $$\beta : kG \rightarrow k[G/A] \cong kH \, .$$ Consider the left $kH$-module $P = {_{kH}kH}$. Since $H$ is a $p$-group, $P$ is indecomposable. Via $\beta$ we can regard $P$ as a $kG$-module and since $\beta$ is surjective, the $kH$-submodules and $kG$-submodules of $P$ are the same. Therefore $P$ is indecomposable as a $kG$-module too. Moreover, $P$ is clearly a projective $kH$-module and since $|G:H|$ is invertible in $k$, we can deduce that (Maschke type argument / transfer theory) $P$ is also a projective $kG$-module.

As a permutation module $P$ certainly surjects on the trivial module $k$. Thus by the correspondence between the indecomposable projective modules and the simple modules, $P$ must be the projective cover of $k$. Finally, we have $$kH \cong \operatorname{End}_{kH}(P) = \operatorname{End}_{kG}(P) \, .$$ (The equality above stems from $\beta$ being surjective)

So the original question (when $R = k$ with $\operatorname{char} k = p$) reduces to the following two questions:

1. If $G$ is a $p$-group, can we recover $\mathbb{Z}G$ from $kG$?
2. If $G$ is a $p$'-group ($p \nmid |G|$), can we recover $\mathbb{Z}G$ from $kG$?

Edit: After reading this question, I think I'm making a mistake in the second bullet point by treating the trivial module $k$ as an invariant of a group algebra $kG$ up to an algebra isomorphism. I am not sure if I can justify that the trivial $kG$-module $k$ regarded as a $kG'$-module via an abstract isomorphism $kG \cong kG'$ is actually the trivial $kG'$-module.

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