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

$$\DeclareMathOperator{\Hom}{Hom}$$ $$\DeclareMathOperator{\im}{im}$$ $$\DeclareMathOperator{\id}{id}$$ $$\DeclareMathOperator{\ext}{Ext}$$ $$\newcommand{\Z}{\mathbb{Z}}$$

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 $$A$$.

• Would it count to give a proof that uses a projective resolution, but does not specify what those projectives are? Apr 30, 2019 at 22:17
• Yes, I'm expecting the proof to use an arbitrary projective resolution (or an injective resolution), seeing as that's how group cohomology is defined. Apr 30, 2019 at 22:24
• $H^2(G,A) = \mathrm{Ext}^2_{\mathbb{Z}[G]}(\mathbb{Z}, A)$; so all you have to do is relate extensions (as $\mathbb{Z}[G]$-modules) $0\to A \to X_2\to X_1 \to \mathbb{Z}\to 0$ to extensions $0\to A\to E\to G\to 1$. I don't know if it's easier but it may help. May 5, 2019 at 20:22
• As for the transgression, there is a construction of the Lyndon-Hochschild-Serre spectral sequence that, as far as I recall, doesn't use a specific projective resolution; and on its second page (with coefficients in the conjugation $E$-module $A$) has $H^p(G,H^q(A,A))$ (and it converges to $H^{p+q}(E,A)$) so the differential $d_2^{0,1} : H^0(G,H^1(A,A))\to H^2(G,H^0(A,A))$ seems to be it, as it is actually $H^1(A,A)^G\to H^2(G,A)$. I think this interpretation can be used : a map between extensions will induce a map between the spectral sequences that is the identity on all of these guys May 5, 2019 at 20:30
• @j.p. It is true that Robinson proves that equivalence classes of extensions are in bijection with elements of $Z^2(G,A)/B^2(G,A)$. The problem is that his definitions of $Z^2$ and $B^2$ are in terms of 2-cocycles and 2-cobundaries. The only way that I know of to convert between Robinson's definition of $Z^2/B^2$ and my definition of $H^2$ is to use the bar resolution. May 10, 2019 at 17:26