What can be said about $G$ non-abelian if $\mathbb{Z}_2\rightarrow G\rightarrow\mathbb{Z}_2^3$ is a short exact sequence and $Z(G)\cong\mathbb{Z}_2$?
So $G$ clearly has order 16, and if there exists such a non-abelian group then maybe someone could just let me know which it is, it's got to be either the group generated by the Pauli matrices or $\mathbb{Z}_2^2\rtimes \mathbb{Z}_4$, since I've ruled out all the others. If there doesn't exist such a group then I would prefer not to use the classification of groups of order 16 to simply rule it out.
I have knowledge of group theory up through proofs of the Sylow theorems. I know the center is contained in every normal subgroup of $G$. $\mathbb{Z}_2^3$ has a seven subgroups of order 2 so I've been trying to use the correspondence theorem to get some idea of what this implies for the structure of $G$, but no luck so far. I've found several paths to the fact that $G$ has no element of order 8, but that still leaves a lot of possibilities for its subgroup of order 8. Anyways I've been banging my head against this one for a while now, can anyone help me out with it? Thanks.