Do subgroup and quotient group define a group?

Question

Does a (normal) subgroup along with its corresponding quotient group define a group completely? Or are there groups with isomorphic normal subgroups and isomorphic corresponding quotient groups which are themselves as a whole not isomorphic?

Answer 1

No.

Both $S_3$ and $C_6$ have normal subgroups isomorphic to $C_3$ with quotient isomorphic to $C_2$.

This is not even true of abelian groups: $C_4$ and $C_2 \times C_2$ both have subgroups isomorphic to $C_2$ with quotient isomorphic to $C_2$.

Answer 2

No, there are plenty of examples of groups with isomorphic normal subgroups and quotients. Probably, the smallest example: $C_2^2 / C_2 \cong C_2$ and $C_4 / C_2 \cong C_2$.

In general, determining all groups $G$ such that $G/N \cong H$ is quite difficult; this is known as the group extension problem.

Answer 3

No, your question is concerned with "Extension theory for groups". If you have $N \unlhd G$ a normal subgroup then let $\phi$ be the natural inclusion (injective) $\phi \colon N \hookrightarrow G$ and $N$ is the kernel of the natural map $\pi \colon G \twoheadrightarrow G / N$. Note that $im(\phi)=\phi(N)=N=ker(\pi)$. That is to say that $G$ fits into the sequence $$ 0 \rightarrow N \rightarrow G \rightarrow G /N \rightarrow 0$$ which is exact, i.e. the second arrow is injective (this is $\phi$), the third is surjective (this is $\pi$) and $im(\phi)=ker(\pi)$. One now asks which groups $G$ fit into such an exact sequence, i.e for given groups $\Gamma$ and $Q$:

$$ 0 \rightarrow \Gamma \hookrightarrow G \twoheadrightarrow Q \rightarrow 0.$$

This problem is concerned with cohomology theory of groups (see the corresponding wikipedia article and probably many books with subject "groups")