Show no non-abelian group G such that Z(G)=Z2 exists satisfying the mapping

[Poopsilon]

Homework Statement

Show that there is no non-abelian group $G$ such that $Z(G)=\mathbb{Z}_2$, which satisfies the short exact $\mathbb{Z}_2\rightarrow G\rightarrow\mathbb{Z}_2^3$.

The Attempt at a Solution

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.

Note: I want to prove this without resorting to the classification of groups of order 16.

 

[lippyka]

Since \mathbb{Z}_2 is the center of G, it is contained in every normal subgroup of G. This implies that the only nontrivial normal subgroup of G is \mathbb{Z}_2^3. Since G/ \mathbb{Z}_2 \cong \mathbb{Z}_2^3, G must be a semidirect product of \mathbb{Z}_2 and \mathbb{Z}_2^3. However, this is impossible since \mathbb{Z}_2 is abelian and thus can not act nontrivially on \mathbb{Z}_2^3. Therefore, there is no non-abelian group G such that Z(G)=\mathbb{Z}_2, which satisfies the short exact \mathbb{Z}_2\rightarrow G\rightarrow\mathbb{Z}_2^3.