Recently, I've been studying a little group cohomology and in a set of notes, I saw the following fact assumed: $H_0(G,I)=I/I^2$. Here $G$ is a group and $H_0(G,I)$ is the zeroth homology group with respect to the augmentation ideal $I$ which is the kernel of the natural map from $\mathbb Z[G]\rightarrow G$. Why is this true? Can anyone explain in simple terms. I'm a beginner in these concepts. Than you in advance.
How to show that $H_0(G,I)=I/I^2$?
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group-theory
group-cohomology
1 Answers
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The trivial $\mathbb ZG$ module is (sort of by definition) the quotient of $\mathbb ZG$ by the augmentation ideal $I$ which is the kernel of $\mathbb ZG\longrightarrow \mathbb Z$, and because $H_0(G,-)$ is $\mathbb Z\otimes_{\mathbb ZG} -$, this gives
$$H_0(G,I)= \frac{\mathbb ZG}{I}\otimes I \simeq \frac{I}{I^2}$$
the last being a general isomorphism $\dfrac{A}{I}\otimes_A M\simeq \dfrac{M}{IM}$.