If $G/K\cong H/K$ must $G\cong H$? Let K be a normal subgroup of $H$ and $G$ so that $G/K$ is isomorphic to $H/K$, must we have $G\cong H$?
I can't really tell you what I have tries since I haven't really done anything worth telling.
Thanks in advance.
 A: This is false. Let $G=C_4, H=V_4, K=C_2$.
A: Counterexample:
Let $A = {\mathbb Z}_4 \times {\mathbb Z}_2$, and let $G$ and $H$ be the subgroups of $A$ given by ${\mathbb Z}_4 \times \{0\}$ and $\langle 2\rangle \times {\mathbb Z}_2$ respectively. Now let $K = \langle 2\rangle \times  \{0\}$ which is a normal subgroup of both $G$ and $H$.
A: A neat example is $S_3$. This group is the semidirect-product of the cyclic group of order two with the cyclic group of order three.
$$S_3\cong C_3\rtimes C_2$$
However, it is not isomorphic to the cross product $C_3\times C_2\cong C_6$. (Note that I write $C_n$ for the cyclic group of order $n$, so $C_n\cong\mathbb{Z}_n$.)
More generally, if $G$ is a semidirect product $G=N\rtimes_{\phi}H$ where the automorphism $\phi$ of $N$ is not inner, then (in general), $G/N\cong H$ and we have the following.
$$
N\rtimes_{\phi}H\not\cong N\times H
$$
Combining this with the fact that $\frac{N\times H}{N}\cong H$ yields a large class of examples.
Other concrete examples are then easily found:


*

*$D_{2n}\cong C_n\rtimes C_2\not\cong C_2\times C_n$

*The fundamental group of the Klein bottle, which has presentation $\langle a, b; a^{-1}ba=b^{-1}\rangle\cong \mathbb{Z}\rtimes\mathbb{Z}$ is not isomorphic to $\mathbb{Z}\times\mathbb{Z}$.

*$\operatorname{O}(n)\cong \operatorname{SO}(n)\rtimes C_2\not\cong \operatorname{SO}(n)\times C_2$.
Generalising this idea, if you have a short exact sequence
$$
1\rightarrow N\rightarrow G\rightarrow H\rightarrow 1
$$
which does not split then, in general, $G\not\cong N\times H$ so you obtain a counter-example. This is another massive class of counter)examples. For example,
$$
1\rightarrow A_n\rightarrow S_n\rightarrow C_2\rightarrow 1$$
but $S_n\not\cong A_n\times C_2$.
A: The simplest possible counter example is also the first one you would check. The smallest groups of equal order which are non-isomorphic are of order $4$, and both are abelian. Each group of order $4$ must have an element of order $2$, and thus a subgroup of order $2$. The quotient in the above two groups would be a group of order $2$, thus isomorphic to $\mathbb Z_2$.
