Why is first cohomology of a finite group torsion? I am trying to solve Exercise B.2 in Silverman's Arithmetic of Elliptic Curves, which asks:

Let $G$ be a finite group and $M$ be a $G$-module.
a) If $G$ has order $n$, prove every element of $H^1 (G,M)$ is killed by $n$.
b) If $M$ is finitely generated as a $G$-module, prove that $H^1 (G, M)$ is finite.

I've been trying to solve part a) by looking at $f(g^k)$ for a crossed homomorphism $f: G \to M$, but the best this gives is $(g^{n-1} + g^{n-2} + \cdots + g + 1) \cdot f(g) = 0$ for all $g\in G$. Then part b) should follow from a) if we can prove $H^1$ must be a finitely generated abelian group, but I'm not sure how to show this.
I'd really appreciate any help!
 A: I'll assume $M$ is a left $G$-module, since this coincides with your notation above.
For part $(a)$, let $f: G \to M$ be a crossed homomorphism, which, by definition, must satisfy $$f(gh) = g\cdot f(h) + f(g)$$ for all $g$, $h \in G$. For fixed $g$, summing this equation over all $h \in G$ gives $$\sum_{h \in G} f(h) = g\cdot \sum_{h\in G} f(h) + n f(g) \implies n f(g) = m - g\cdot m$$ where $m = \sum_{h\in G} f(h)$. Thus, $nf$ is a principal crossed homomorphism, and it follows $n$ kills $f$ in $H^1(G, M)$.
For part $(b)$, let $x_1, \cdots, x_k$ be generators of $M$ as a $G$-module. Then $\{g_i x_j\}$ are generators of $M$ as a $\mathbb{Z}$-module. It follows that $\text{Maps}(G, M)$ is finitely generated as a $\mathbb{Z}$-module (by the maps $f_{ijk}$ satisfying $f_{ijk}(g_i) = g_j x_k$ and $f_{ijk} (g_r) = 0$ for $r\neq i$). Since $Z^1 (G, M)$ (the 1-cocycles) are a subgroup of $\text{Maps}(G, M)$, it follows $Z^1 (G, M)$ is a finitely generated $\mathbb{Z}$-module, hence so is $H^1 (G, M)$. Finally, since we know from part $(a)$ that $H^1 (G, M)$ is torsion, it has to be finite.
