$|G:H|=2$ and $H$ abelian then $H \subset Z(G)$

Question

I have to show whether this is true or false:

$|G:H|=2$ and $H$ abelian then $H \subset Z(G)$

I have proved that $H \triangleleft G$, but with this I can show that if $h \in H, g\in G$ then exists a $h_1 \in H$ such that $ghg^{-1}=h_1$. I would need to show that $h = h_1$ or find a counter example. I've been thinking of one, since I can't find why should be that $h = h_1$ with no luck

Thanks

Answer 1

Hint: Consider $G=S_3$ and $H=\langle (123) \rangle$.

Answer 2

$[S_3:A_3]=2$ and $A_3$ is abelian, but $A_3 \not \subseteq Z(S_3)$.