This seems pretty trivial, but I used it in an assignment and want to double-, triple check.
If $G \cong G'$ and $H \cong H'$ where $H \trianglelefteq G$ and $H' \trianglelefteq G'$ are normal subgroups, is then also $G/H \cong G'/H'$?
My idea is to map a coset $x+H$ to the coset $f(x)+H'$ where $f$ is the isomorphism between the $G$'s. Does this even require that $H$ and $H'$ be isomorphic?
This is not correct. Try $H=2\Bbb Z\trianglelefteq\Bbb Z=G$ and $H'=\Bbb Z\trianglelefteq G'=\Bbb Z$.
Clearly $H$ and $H$' are isomorphic, and $G$ and $G'$ are equal. Yet $G/H=\Bbb Z/2\Bbb Z$ and $G'/H'=0$ are not isomorphic.
The serious issue is that the isomorphisms $H\cong H'$ and $G\cong G'$ were not related. The trouble with $x+H\mapsto f(x)+H'$ is twofold; you firstly don't know if that's well defined for different representatives $x$, i.e. you don't know if $f(H)\le H'$, and you don't know that it's injective: if $f(x+H)=H'$ then, assuming $f(H)\le H'$, you know $f(x)\in H'$, but you don't know if that means $x\in H$ i.e. if $x+H=H$ is the identity element.
So we see that $f$ must be an isomorphism but also satisfy $f(H)\le H'$ and $f^{-1}(H')\le H$ i.e. we need $f(H)=H'$ together with an isomorphism criterion; so the isomorphism $H\cong H'$ is (i) necessary (ii) necessary to come from the restriction of $G\cong G'$.
$$\require{AMScd}\begin{CD}H@>\cong>>H'\\@VVV@VVV\\G@>>\cong>G'\end{CD}$$