If $G/M\simeq H/N$ then is it true that both $G\simeq H$ or $M\simeq N$?

Question

Let $G$ and $H$ be two groups. Suppose that $M$ and $N$ be two normal subgroups of $G$ and $H$ respectively such that we have the following,

• $G$ and $H$ are isomorphic.

• $M$ and $N$ are isomorphic.

Then we know that $G/M$ and $H/N$ are isomorphic if for an isomorphism $\varphi:G\to H$ we have $\varphi(M)=N$.

My Question is,

If $G$ and $H$ be two groups and $M$ and $N$ are two normal subgroups of $G$ and $H$ respectively such that $G/M$ and $H/N$ are isomorphic then is it true that both $G$ and $H$ are isomorphic or $M$ and $N$ are isomorphic?

I don't know how to approach this problem. Can anyone help me?

Answer 1

Not at all.

Just let $G$ and $H$ be arbitary group and $M=G$, $N=H$.

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Apart from this, $G\cong H$ and $M\cong N$ does not imply $G/M\cong H/N$. Just pick $G=H=\Bbb Z$, $M=2\Bbb Z$, $N=H$.

Answer 2

No, this is not true. Consider $G=GL_n(K)$ and $M=SL_n(K)$, and $H=K^{\times}$, $N=1$. Then $G/M\cong H/N\cong K^{\times}$, but neither $G\cong H$ nor $M\cong N$.

Answer 3

This is false.

Let $$G = \mathbb{Z}_2 \times \mathbb{Z}_2 = \textrm{ The Klein Four Group }$$ and $$M = \textrm{ trivial subgroup}.$$ Now let $$H = D_4 = \textrm{Dihedral Group of Order 8}$$ and $$N = Z(D_4) = \textrm{ The Center of } D_4.$$

You can show that $G/M \cong H/N$.