Normal Subgroups, Isomorphic Quotients, But Not Isomorphic

Group Theory Problems and Solutions in Mathematics

Problem 127

Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$.
Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.

 
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Proof.

We give a counterexample.
Let $G=\Z \times \Zmod{p} \times \Zmod{q}$, where $p$ and $q$ are distinct prime numbers. Then $G$ is an abelian group and thus any subgroups are normal.
We denote $e$ for identity elements of possibly different groups.

Let
\[N_1=\Z \times \Zmod{p} \times \{e\} \subset G\] and
\[N_2=\Z \times \{e\}\times \Zmod{q}\subset G.\] Also define subgroups
\[H_1=q\Z \times \{e\} \times \{e\} \subset N_1\] and
\[H_2=p\Z \times \{e\} \times \{e\} \subset N_2.\] Then both $H_1$ and $H_2$ are isomorphic to $\Z$.

The quotients groups $N_1/H_1$ and $N_2/H_2$ are both isomorphic to $\Zmod{p} \times \Zmod{q}$.

Since $p$ and $q$ are distinct primes, the groups $N_1$ and $N_2$ are not isomorphic.
Therefore, we disprove the claim.


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