# Do subgroup and quotient group define a group?

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?

• I am shocked that this isn't a duplicate. I don't know of a single beginning student in group theory who hasn't asked this question or made the mistake of assuming it was true at some point. – Patrick Stevens Sep 24 '15 at 11:29
• hey, Patrick Stevens! I see you're a math student going to part 3 in Cambridge! I'm a fresher this year. Churchill College – Asier Calbet Sep 24 '15 at 11:35
• Good luck, although by your question history you're well set for a good start :) – Patrick Stevens Sep 24 '15 at 19:10

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 $C_2$.

• ah! true! how did I miss that easy abelian example! thank you, I should have thought longer before asking the question... – Asier Calbet Sep 24 '15 at 10:28

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")

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.