I need to show that, if I have a non-abelian group G of order 12 with only one element has order 2, then G is soluble and the center Z(G) is such that
$Z(G)\cong \mathbb{Z}_2$ and $\frac{G}{Z(G)}\cong S_3$
I don't know how to solve this problem, and any help would be most welcome.