## An isomorphism between different Ext’s coming from group cohomology

Let $G$ be an abelian group and $M$ a $G$-module with trivial action. It is well-known that $H^2(G,M)$ classifies extensions of $G$ by $M$, which is $\mathrm{Ext}^1_{Ab}(G,M)$.

On the other hand $H^2$ is $\mathrm{Ext}^2_{G-mod}(\mathbb Z,M)$ (by the definition of group cohomology), where $\mathbb Z$ is given the trivial $G$-action.

Is there a nice way to see this isomorphism $$\mathrm{Ext}^1_{Ab}(G,M) \cong \mathrm{Ext}^2_{G-mod}(\mathbb Z,M)?$$

Or is it just an accident that both happen to classify the same type of object?

Is there a high-brow reason? Is there a generalization? Why should $\mathrm{Ext}^1$ and $\mathrm{Ext}^2$ be related in such a way? Might this result from a spectral sequence? I wonder if this might be a special case of something well-known.

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You mean $\mathrm{Ext}^1_{Ab}(G,M)$, not $\mathrm{Ext}^1_{Ab}(M,G)$ – Eric Wofsey Jan 4 at 12:00
Right...I was thinking of the order they appear in the exact sequence. – David Corwin Jan 4 at 12:02
Your $Ext^1$ classifies abelian extensions, while $H^2$ classifies all extensions. – Angelo Jan 4 at 12:19

The group $H^2(G,M)$ classifies different types of extensions than $\operatorname{Ext}^1(G,M)$. On the one hand, $H^2(G,M)$ classifies extensions

$$M\hookrightarrow H\twoheadrightarrow G$$

where $H$ may be non-abelian, and the action of $H$ on $M$ by conjugation is encoded in the $G$-module structure of $M$. On the other hand, $\operatorname{Ext}^1(G,M)$ classifies extensions

$$M\hookrightarrow A\twoheadrightarrow G$$

where $A$ is an abelian group, in particular $A$ acts trivially on $M$ by conjugation, i.e. $M$ can only be regarded as a trivial $G$-module here.

Even if you regard $M$ as a trivial $G$-module in both cases, $H^2(G,M)$ and $\operatorname{Ext}^1(G,M)$ may be different due to the existence of non-abelian but central extensions. In general, there is a universal coefficient split short exact sequence

$$\operatorname{Ext}^1(G,M)\hookrightarrow H^2(G,M)\twoheadrightarrow \operatorname{Hom}(H_2G,M).$$

You can find this in most books on group cohomology. The first morphism represents the inclusion of abelian extensions into central (but possibly non-abelian) extensions (recall that $M$ carries here the trivial $G$-module structure). The group $\operatorname{Hom}(H_2G,M)$ measures the amount of really non-abelian central extensions of $G$ by $M$.

Fortunately, $H_2G$ is very easy to compute, it is the exterior square $H_2G=\wedge^2G$, i.e. the quotient of $G\otimes G$ by the relations $g\otimes g=0$, $g\in G$. This functor is quadratic, $$\wedge^2(G_1\oplus G_2)=\wedge^2(G_1)\oplus (G_1\otimes G_2) \oplus \wedge^2(G_2)$$ and vanishes on (finite or infinite) ciclyc groups $\wedge^2(\mathbb{Z}/n)=0$, $n\in\mathbb Z$. This gives a recipe to compute $H_2G$ for any finitely generated abelian group. In particular, if you take $G=(\mathbb{Z}/2)^2$ and $M=\mathbb{Z}/2$ you get

$$\operatorname{Ext}^1(G,M)=(\mathbb{Z}/2)^2,\qquad \operatorname{Hom}(H_2G,M)=\mathbb Z/2.$$ Hence $$H^2(G,M)=(\mathbb{Z}/2)^3.$$

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 In case it isn't clear, this "universal coefficient sequence" literally is the usual universal coefficient theorem for the cohomology of the space $BG$ with coefficients in $M$, noting that $G=H_1(BG)$. – Eric Wofsey Jan 4 at 12:41 For an explicit example, you can take the Heisenberg group $H_3$ of strictly upper-triangular matrices in $\text{SL}_3(\mathbb{Z})$. This is generated by the elementary matrices $x=E_{12}$, $y=E_{23}$ and $z=E_{13}$, subject only to the relations $[x,y]=z$ and $[x,z]=[y,z]=1$. The center is $\langle z\rangle \simeq \mathbb{Z}$, and the quotient is $\langle \overline{x},\overline{y} \vert [\overline{x},\overline{y}]=1\rangle\simeq \mathbb{Z}^2$, so you have a central extension $1\to \mathbb{Z}\to H_3\to \mathbb{Z}^2\to 1$. But $H_3$ is not abelian. – Tom Church Jan 6 at 16:18