# Abstract proof that $\lvert H^2(G,A)\rvert$ counts group extensions

(This question is originally from Math.SE, where it didn't receive any answers.) $$\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\id}{id} \DeclareMathOperator{\ext}{Ext} \newcommand{\Z}{\mathbb{Z}}$$

Let $$G$$ be a group, let $$A$$ be a $$G$$-module, and let $$P_3\to P_2\to P_1\to P_0\to\Z\to0$$ be the start of a projective resolution of the $$G$$-module $$\mathbb{Z}$$. Consider the cohomology group

$$H^2(G,A)=\frac{\ker(\Hom_{\Z G}(P_2,A)\to\Hom_{\Z G}(P_3,A))}{\im(\Hom_{\Z G}(P_1,A)\to\Hom_{\Z G}(P_2,A))}.$$

It can be shown that $$\lvert H^2(G,A)\rvert$$ counts the number of equivalence classes of group extensions $$0\to A\to E\to G\to0$$. The only proof that I know of this result involves choosing a specific projective resolution (namely, the bar resolution).

Is there a proof of this result that does not require choosing a specific projective resolution?

For context, $$\lvert\ext_R^n(M,N)\rvert$$ counts the number of equivalence classes of extensions $$0\to N\to X_n\to\ldots\to X_1\to M\to0$$. The proof of this result is fairly abstract and does not require picking a specific projective resolution of $$M$$ or a specific injective resolution of $$N$$.

Also, I am aware that we actually have isomorphisms in both of these results but I am more interested in the existence of an explicit bijection.

Here is one approach for constructing an element of $$H^2(G,A)$$ from an extension $$0\to A\to E\to G\to0$$: Treat $$A$$ as an $$E$$-module and consider the transgression map $$H^1(A,A)^{E/A}\to H^2(E/A,A^A)$$. Rewriting this gives a homomorphism $$\Hom(A,A)^G\to H^2(G,A)$$. The image of $$\id_A$$ under this map will be an element of $$H^2(G,A)$$.

To make this work, this map would need to be a bijection from equivalence classes of group extensions and elements of $$H^2(G,A)$$.

Another approach that I considered was to work directly with the arbitrary projective resolution (similar to the proof of the Yoneda Ext result). Suppose that we are given a group extension $$0\to A\to E\to G\to0$$. We want to construct an element of $$\ker(\Hom_{\Z G}(P_2,A)\to\Hom_{\Z G}(P_3,A))$$. Equivalently, we want to construct a $$\Z G$$-module homomorphism $$P_2/\im(P_3\to P_2)\to A$$. However, $$\im(P_3\to P_2)=\ker(P_2\to P_1)$$ and $$P_2/\ker(P_2\to P_1)\cong\im(P_2\to P_1)=\ker(P_1\to P_0)$$. Thus, we want to construct a $$\Z G$$-module homomorphism $$f\colon\ker(P_1\to P_0)\to A$$. Furthermore, if we unwind some more definitions, we see that we only need to construct $$f$$ up to the restriction of a $$\Z G$$-module homomorphism $$P_1\to A$$.

Unfortunately, the only information we have about $$A$$ is the short exact sequence $$0\to A\to E\to G\to0$$ which makes it hard to define a $$\Z G$$-module homomorphism to $$A$$.

• I like this question! I wonder why you specify just thinking of $\lvert\operatorname H^2(G, A)\rvert$ as counting extensions, rather than of $\operatorname H^2(G, A)$ as parameterising extensions. – LSpice May 10 '19 at 18:00
• – user2831784 May 10 '19 at 20:02
• Actually your last method works. Since $P_1$ is projective, we get a commutative square $P_1 \to G$ with maps via $E$ and $P_0$. This gives a map $Ker (P_1\to P_0) \to A \cong Ker (E\to G)$ and you are done. – user43326 May 11 '19 at 14:28
• @user43326, could you elaborate a bit more? Are you claiming that we have maps $P_1\to E$ and $P_0\to G$? – Thomas Browning May 11 '19 at 18:55
• Sorry, that comment was a crap. – user43326 May 12 '19 at 6:03

Here is a simple way. The extension $$A \to E \to G$$ induces a map of classifying spaces $$BA\to BE \to BG$$, which is a principal fibration, so classified by (homotopy class of) a map $$BG \to BBA=K(A,2)$$, i.e., an element of $$H^2(BG,A)$$.
• This is interesting, but is there a way to convert between $H^2(BG,M)$ and $H^2(G,M)$ without the choice of a resolution? – Thomas Browning May 10 '19 at 20:21
• @ThomasBrowning Personally, my definition of group cohomology is just cohomology of $BG$... I think it's going to be hard to get off the ground if you don't at least give yourself that. – Kevin Casto May 10 '19 at 23:22
• @ThomasBrowning Any model for $EG$ gives such a projective resolution of $\mathbb{Z}$. – James Cameron May 15 '19 at 16:21
• @ThomasBrowning No you can't get all resolutions that way, for example you'll always get a free resolution from $EG$. Are you trying to avoid using the fact that cohomology doesn't depend on which resolution you use? – James Cameron May 15 '19 at 16:40