Compute the first group cohomology with trivial $G$-modules

Question

This is the exercise 6.1.5 in Weibel's book.

Let $G$ be any group and $A$ be a trivial $G$-module. Show that $H^1(G;A)\cong\hom(G,A)\cong\hom(G/[G,G],A)$

In the book, $H^*(G;A)$ is defined to be the right derived functor of the fixed-point functor $(-)^G$, which is naturally isomorphic to $Ext_{\mathbb{Z}G}^*(\mathbb{Z},A)$.

I know there are many ways to calculate this cohomology. For example, one can use the bar resolution of $\mathbb{Z}$ or crossed homomorphism.

However, this exercise is posed right after the calculation of $H_1(G;A)\cong G/[G,G]$ for any trivial $G$-module $A$. Besides, the author hasn't introduced any explicit resolution for $\mathbb{Z}$ or crossed homomorphisms, or any other useful machinary. Therefore, I am wondering if there is some slick calculation taking advantages of the triviality of $A$ and the result of $H_1(G;A)$.

Thank you.

EDIT:

By "slick calculation", i really meant a calculation using just the knowledge introduced before this exercise in the book. I've already known there are many ways to calculate it. However, the author didn't introduce other machinery but stated this exercise anyway. Therefore, I guess I am supposed to be able to calculate it just using what I've learnt so far.

Answer 1

Consider the short exact sequence $0\to IG\to\mathbb ZG\to\mathbb Z\to0$, with $IG$ the augmentation ideal. Since $\mathbb ZG$ is projective, the long exact sequence for $Ext_{\mathbb ZG}(-,A)$ gives us an exact sequence $$0\to\hom_G(\mathbb Z,A)\to\hom_G(\mathbb ZG,A)\to\hom_G(IG,A)\to Ext^1_G(\mathbb Z,A)\to Ext^1_G(\mathbb Z G,A)=0$$ You can compute directly that the map $$\hom_G(\mathbb Z,A)\to\hom_G(\mathbb ZG,A)$$ is an isomorphism because $A$ is trivial, and it follows that $$\hom_G(IG,A)\to Ext^1_G(\mathbb Z,A)$$ is an isomorphism.

Now $IG$ is free over $\mathbb Z$ with basis $\{g-1:g\in G,g\neq1\}$. A $G$-homomorphism $\phi:IG\to A$ is thus uniquel determined by the values $a_g=\phi(g-1)$ for all $g\in G\setminus\{1\}$; set $a_1=0\in A$. $G$-linearlity implies that $$a_g = h\cdot a_g=h\cdot\phi(g-1)=\phi(h\cdot(g-1))=\phi((hg-1)-(h-1))=a_{hg}-a_h,$$ so that $a_{hg}=a_g+a_h$. This means that $\phi$ determines a group homomorphism $\bar\phi:g\in G\mapsto a_g\in A$.

• Show that this induces a bijection $\hom_G(IG,A)\to\hom_{\mathrm{groups}}(G,A)$.

Answer 2

Since you asked for a slick calculation: $H^1(G, A) = H^1(X, A)$ for the space $X = K(G, 1)$ by abstract nonsense, and \begin{align*} H^1(X, A) = \operatorname{Hom}(H_1(X), A) = \operatorname{Hom}(\pi_1(X)^{\operatorname{ab}}, A) = \operatorname{Hom}(G/[G, G], A). \end{align*}