Nonsplit extension of $\mathbb{Z}$ by itself

Question

I was classifying all split extensions on a list of short exact sequences, when I arrived on this one: $$1 \rightarrow \mathbb{Z} \rightarrow G \rightarrow \mathbb{Z} \rightarrow 1$$ That the automorphism group $\text{Aut}(\mathbb{Z})$ is isomorphic to $C_2$ is not hard to work out, and my job to classify split extensions ends pretty much there, once there are only two homomorphisms from $\mathbb{Z}$ to $C_2$.

But when I thought about nonsplit extensions on this sequence, I started wondering, is there one? Over finite groups, some light on this matter can be shed using theorems like Lagrange, Cauchy and Sylow, but my knowledge on infinite groups give me no arguments, neither to claim existence or to deny it.

So, do exist a nonsplit extension of $\mathbb{Z}$ by itself?

Answer 1

Every extension $1\to A\to B\stackrel{p}\to\mathbb{Z}\to 1$ splits, since you can get a splitting $i:\mathbb{Z}\to B$ by choosing any $x\in B$ such that $p(x)=1$ and defining $i(n)=x^n$. More generally, a similar argument applies with $\mathbb{Z}$ replaced by any free group.

Answer 2

There does not exist a non split extension of $Z$ by itself since $H^2(Z,Z)=0$.