# What can be said about $G$ non-abelian if $\mathbb{Z}_2\rightarrow G\rightarrow\mathbb{Z}_2^3$ and $Z(G)\cong\mathbb{Z}_2$?

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.

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I believe, of those two, the Pauli matrices is the only one with center of order 2 (generated by $\sigma_3$)... I could be wrong though, I've never been good with the semi-direct product. –  Arthur Nov 23 '12 at 23:54
I looked it up, it's Parseltongue. –  Will Jagy Nov 23 '12 at 23:56
There is no such group. The only groups of order $16$ with a center of order $2$ are $D_{16}$, $SD_{16}$, and $Q_{16}$, none of which have quotients isomorphic to $\mathbb{Z}_2^3$. –  Alexander Gruber Nov 24 '12 at 0:27
Ok thanks Alexander, any thoughts on how to prove it besides just trying every group of order 16? –  Thoth Nov 24 '12 at 0:28
None so far. I'll let you know if I think of something. –  Alexander Gruber Nov 24 '12 at 0:31

You can prove the nonexistence of a group with this property as follows. Let $Z(G) = \langle z \rangle$, and let $a,b,c$ be inverse images of generators of $G/Z(G)$. Since $G/Z(G)$ is abelian, the commutators $[a,b]$, $[a,c]$, $[b,c]$ all lie in $Z(G)$, so they are all equal to 1 or to $z$.
If, say, $[a,b]=[a,c]=1$, then we have $a \in Z(G)$, which we know to be false. Since all commutators are in the centre of the group, the commutator map is bilinear so, if for example $[a,b]=[a,c]=z$, $[b,c]=1$, then $[a,bc] = [b,bc] = [c,bc]=1$, so $bc \in Z(G)$, contradiction. Or if $[a,b]=[a,c]=[b,c]=z$, then $abc \in Z(G)$. So there is no possible assignment of commutators that does not result in an extra element in $Z(G)$.
Equivalently, there is no nondegenerate alternating bilinear map ${\mathbb F}_2^3 \times {\mathbb F}_2^3 \to {\mathbb F}_2$.