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

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.

• Use the bar resolution. – Mariano Suárez-Álvarez Nov 5 '15 at 17:02
• @MarianoSuárez-Alvarez I know this is definitely a way to do it. However, as I mentioned in my question, the author didn't introduce the bar resolution but stated this question anyway. So, I guess I am supposed to calculate it using previous knowledge. Otherwise, I've known there are many ways to do this. Thank you, anyway – YYF Nov 5 '15 at 17:06

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 $\hom_G(IG,A)\to\hom_{\mathrm{groups}}(G,A)$.
• Thank you so much. I tried this method, but failed somehow. Maybe, I just didn't keep my head straight. – YYF Nov 5 '15 at 17:35
• Well, that is why when you ask a question you should tell us what you tried... – Mariano Suárez-Álvarez Nov 5 '15 at 17:35

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*}

• This is slick only because it pushes all the details to the behind-the-scenes action involved in proving all the equalities and the connection between group cohomology and E-M spaces. – Mariano Suárez-Álvarez Nov 5 '15 at 17:15
• That's easy, though: It just comes from the fact that group cohomology is independent of the particular choice of injective resolution (which is part of the definition of group cohomology), and $K(G, 1)$ has a particularly nice one. It's certainly much nicer than doing a direct computation. – anomaly Nov 5 '15 at 17:54
• Well, the proof is in the pudding. I suggest you add the details (keeping in mind thata this is to answer a question which appears on page 3 of the chapter introducing group cohomology, of course) – Mariano Suárez-Álvarez Nov 5 '15 at 17:57
• I don't have Weibel's book; how would you define group cohomology (in general, not just $H^1$ or $H^2$) without having some proof of the independence of choice of resolution? The OP mentioned constructing $H^*$ as a right-derived functor, and every treatment of derived functors I've seen includes a proof of that fact. – anomaly Nov 5 '15 at 18:04
• He defines cohomology using Exts, $H^\bullet(G,-)=Ext_G^\bullet(\mathbb Z,-)$; independence of resolutions is not the problem. What is much less trivial in your approach is that you use the conneection with $K(G,1)$, its existence, the universal coefficient theorem, and the Hurewicz isomorphism. You replaced a half page computation with half a course in algebraic topology :-) – Mariano Suárez-Álvarez Nov 5 '15 at 18:06