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Let $G$ be a finite group and $A_1$ and $A_2$ be $\mathbb ZG$ modules. Is it always true that for any $n\in \mathbb N$, the cohomology group of $G$ with coefficients in $A_1\oplus A_2$ is the same as the following: $$H^n(G; A_1\oplus A_2)=H^n(G; A_1)\oplus H^n(G; A_2) ?$$

If not, then how to calculate $H^n(G; A_1\oplus A_2)$ in terms of $H^n(G; A_1)$ and $H^n(G; A_2)$?

Thanks in advance.

the_fox
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Shivani Sengupta
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    For the question to make sense, you need $A_1$ and $A_2$ to be ${\mathbb Z}G$-modules, not just abelian groups. But then the answer to your question is yes. – Derek Holt Dec 04 '18 at 12:57
  • @DerekHolt This is to clarify myself. If I want to calculate $H^n(Z_3; Z_2 \times Z_2 \times Z_2)$ and suppose the generator of $Z_3$ is acting on the generator $a, b, c$ of $Z_2 \times Z_2 \times Z_2$ by $a$ goes to $b$, $b$ goes to $c$, $c$ goes to $a$. But the above action is very different when we consider an individual factor, i.e., $H^n(Z_3; Z_2)$, here the action of $Z_3$ on $Z_2$ is trivial. Does the above fact is independent of the action? why does it so? – Shivani Sengupta Dec 04 '18 at 13:25
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    For the direct sum isomorphism to hold, $A_1 \oplus A_2$ needs to be a decomposition of ${\mathbb Z}G$-modules, not just of abelian groups. In fact, in your example, there is a ${\mathbb Z}G$-decomposition into $A_1 \oplus A_2$ with $|A_1|=4$ and $|A_2|=2$. – Derek Holt Dec 04 '18 at 14:12

1 Answers1

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The answer to your question is yes. Remind how cohomology is defined: Pick a projective $G$-resolution $P_* \rightarrow \mathbb{Z}$ for the trivial $G$-module $\mathbb{Z}$ and apply $\text{Hom}_\mathbb{Z}(\cdot,A_1 \oplus A_2) = \text{Hom}_\mathbb{Z}(\cdot,A_1) \oplus \text{Hom}_\mathbb{Z}(\cdot,A_2). $ Taking $G$-invariants respects taking direct products, i.e. we have a cochain complex

$$\text{Hom}_G(P_*,A_1 \oplus A_2) = \text{Hom}_\mathbb{Z}(P_*,A_1 \oplus A_2)^G = \text{Hom}_\mathbb{Z}(P_*,A_1)^G \oplus \text{Hom}_\mathbb{Z}(P_*,A_2)^G \\ = \text{Hom}_G(P_*,A_1) \oplus \text{Hom}_G(P_*,A_2).$$

Now $H_*(G; A_1 \oplus A_2)$ is defined to be the cohomology of this complex but since cohomology commutes with finite direct sums of cochain complexes we are done.

Alternatively, you can directly apply $\text{Hom}_G(\cdot, A_1 \oplus A_2) = \text{Hom}_G(\cdot,A_1) \oplus \text{Hom}_G(\cdot,A_2)$ to the complex $P_*$ and then taking cohomology.

Notice that the argument only works for finite direct sums since otherwise in general $\text{Hom}(\cdot, \bigoplus\limits_{i \in I} A_i) \neq \bigoplus\limits_{i \in I} \text{Hom}(\cdot, A_i)$ and also $\text{Hom}(\cdot, \bigoplus\limits_{i \in I} A_i) \neq \prod\limits_{i \in I} \text{Hom}(\cdot, A_i)$. If $G$ is finite one can construct a $G$-resolution of finitely generated free $G$-modules for $\mathbb{Z}$ hence one gets $H^*\left(G;\bigoplus\limits_{i \in I} A_i \right) = \prod\limits_{i \in I} H^*(G;A_i)$ with a little help from Hom and direct sums and since in general cohomology takes arbitrary direct sums to direct products.

Lukas
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