# Normal Subgroups, Isomorphic Quotients, But Not Isomorphic ## 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 $N_2$. Add to solve later

## 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. Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Group Theory ##### The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$

Let $p$ be a prime number. Let $G$ be a non-abelian $p$-group. Show that the index of the center of...

Close