$H,H'$ normal in $G,G'$ respectively and $G \cong G' $ and $H \cong H' $ $\implies$ $G/H \cong G'/H'$ ?

Question

Let $H$ be a normal subgroup of $G$ and $H'$ be a normal subgroup of $G'$ such that $G \cong G' $ and $H \cong H' $ , then is it true that $G/H \cong G'/H'$ ? Denoting by $f$ the isomorphism between $G , G'$ I tried the map $g(xH)=f(x)H'$ , but this didn't work , I know I somehow also have to incorporate the isomorphism between $H,H'$ . Please help

Answer 1

$G=G'=\mathbb{Z}$, $H=\mathbb{Z}$, $H'=2\mathbb{Z}$.

Answer 2

No. There is an abelian counter-example of order $8$. Can you find it?