Given two groups $G$ and $H$, an extension of $G$ by $H$ is a group $K$ with a normal subgroup $N$ isomorphic to $H$ such that the quotient group $K/N$ is isomorphic to $G$ (equivalently, $1 \to H \to K \to G \to 1$ is an exact sequence).

So, up to isomorphism, how many (not necessarily abelian) extensions of $\mathbb{Z}$ by $\mathbb{Z}$ are there?

One obvious extension is the direct product $\mathbb{Z} \times \mathbb{Z}$. Another extension is the group with the same underlying set as $\mathbb{Z} \times \mathbb{Z}$ and $(a,b) \ast (c,d)$ still defined to be $(a+c,b+d)$ if $b$ is even, but it is defined to be $(a-c,b+d)$ if $b$ is odd (or simply, $(a+(-1)^bc,b+d)$). This nontrival extension is the semidirect product $\mathbb{Z} \rtimes \mathbb{Z}$ corresponding to the homomorphism $\mathbb{Z} \to \operatorname{Aut}(\mathbb{Z})$ where even integers act as the identity and odd integers act as negation.

The answer would be $2$ if the direct product and the even/odd semidirect product were the only extensions up to isomorphism. Is this in fact the case?


2 Answers 2


Yes they are the only two. Since the infinite cyclic group is free, any such extension must split, and $|{\rm Aut}({\mathbb Z})| = 2$, so there are only two possible actions in the semidirect product: the trivial action and inversion.


Conveniently, there is a standard line of attack for such questions which works when the subgroup in the extension is abelian: classify actions $\rho$ of the quotient $G$ on the subgroup $A$, and for each such $G$-module structure on $A$, compute $H^2(G, A_\rho)$ (group cohomology).

  1. How many group actions are there of $\mathbb{Z}$ on $\mathbb{Z}$ (compatible with $+$ on the right): answer: 2, since there are two maps from $\mathbb{Z}$ to $\text{Aut}(\mathbb{Z}) = \pm 1$.

  2. For each of these $\mathbb{Z}$-modules, compute the group cohomology $H^2(\mathbb{Z}, M)$. Since $\mathbb{Z}$ has homological dimension $1$, these are both trivial, and so the only two extensions are the ones you wrote down.


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