Suppose $A$ is an abelian group and $$0\to A\to H\to G\to 0$$ is exact. Does it follow that that this SES is isomorphic to one of the form $$ 0\to A\to H'\to G\to 0$$ such that $A$ is contained in the center $Z(H')$ of $H'$?

I ask for the following reason:

  • According to this article, there's a bijection between the second cohomology group $H^2(G, A)$ and the set of equivalence classes of extensions of $G$ by $A$.

  • On the other hand, according to this Wikipedia article, there's a bijection between $H^2(G,A)$ and the set of equivalence classes of central extensions of $G$ by $A$.

Therefore, the above two facts seem to be contradictory if not every extension of $G$ by $A$ is equivalent to some central extension of $G$ by $A$. However, this fact doesn't seem true to me.

If this isn't true, then how can one reconcile the above two facts? For example, $$H^2(\mathbb{Z};\mathbb{Z}) = H^2(K(\mathbb{Z},1);\mathbb{Z}) = H^2(S^1;\mathbb{Z}) = 0$$ so the second fact then tell us the only central extension of $\mathbb{Z}$ by $\mathbb{Z}$ is the trivial one $$0\to\mathbb{Z}\to\mathbb{Z}^{\oplus 2}\to\mathbb{Z}\to 0.$$ However the first fact then seems to be wrong, as for example $$0\to\mathbb{Z}\to\mathbb{Z}\rtimes\mathbb{Z}\to \mathbb{Z}\to 0$$ seems like an extension which I suspect isn't isomorphic to the trivial one, so the first fact would imply that $H^2(\mathbb{Z};\mathbb{Z})$ has at least two elements (the equivalence classes corresponding to both the extensions I just gave), a contradiction to the fact that $H^2(\mathbb{Z};\mathbb{Z}) = 0$.


1 Answer 1


This is indeed false, an example is $S_3$ with it's normal cyclic subgroup of order $3$ and quotient of order $2$. One needs to be quite careful with the equivalence between extensions and ext groups, as there are some subtleties, that should be understood through concrete examples.

To answer where this line of thinking goes wrong, to define the ext group $H^i(G,A)$, there is implicitly an action of $G$ on $A$, which is sometimes (depending on context) taken to be trivial. This is the case for which $H^2$ classifies central extensions, but in general group cohomology is a functor defined on $G$ modules, so the action is crucially important.


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