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

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 $A$.

Thomas Browning
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  • Would it count to give a proof that uses a projective resolution, but does not specify what those projectives are? –  Apr 30 '19 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. – Thomas Browning Apr 30 '19 at 22:24
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    $H^2(G,A) = \mathrm{Ext}^2_{\mathbb{Z}[G]}(\mathbb{Z}, A)$; 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. – Maxime Ramzi May 05 '19 at 20:22
  • I thought about that formulation but the fact that $E$ is not a $G$-module kept making things tricky. – Thomas Browning May 05 '19 at 20:24
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    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 – Maxime Ramzi May 05 '19 at 20:30
  • Therefore equivalent extensions yield equal elements of $H^2(G,A)$, so the map is well-defined. I wouldn't know how to prove either injectivity or surjectivity from that, but someone who knows more about this spectral sequence might – Maxime Ramzi May 05 '19 at 20:32
  • This spectral sequence approach looks promising. One worry I have is whether or not the transgression map $H^1(A,A)^G\to H^2(G,A)$ "remembers" the group $E$. The domain and codomain of the map are defined independently of $E$ but hopefully the transgression map depends on the choice of $E$. I have no idea whether this is the case or not. – Thomas Browning May 05 '19 at 20:36
  • Do you know Derek Robinson's "A Course in the Theory of Groups"? Section 11.1 "Group Extensions and Covering Groups" might be interesting for you. – j.p. May 06 '19 at 06:14
  • That section looks interesting, but it seems to take the usual cocycle/factor set approach in proving that equivalence classes of extensions are counted by $H^2(G,A)$. In particular, the book uses a specific projective resolution to go between elements of $H^2(G,A)$ and factor sets. – Thomas Browning May 07 '19 at 18:58
  • @ThomasBrowning: I never understood this part of group theory, but I have the impression that Robinson proves the equivalence in 11.1.4 before ever speaking about projective resolutions. – j.p. May 10 '19 at 05:52
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    @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. – Thomas Browning May 10 '19 at 17:26

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