How to prove that there either exist $H_2/H_1\cong K_2/K_1$ or $H_2/H_1\cong M_2/M_1$

Question

Suppose I have a group G , and N is a normal subgroup of G. Then, suppose we have subgroups $H_{1}$ and $H_{2}$ of G such that $H_{1}$ is a normal subgroup pf $H_2$. Also, $H_2/H_1$ is a simple sub-quotient group of G(which means $H_2/H_1$ is just a simple group).

Then,prove that either we can find a simple subquotient group $K_2/K_1$ of N($K_2$ and $K_1$ are subgroups of N, and $K_1$ is a normal subgroup pf $K_2$) such that $H_2/H_1\cong K_2/K_1$, or we can find a simple sub-quotient group $M_2/M_1$ of group $G/N$ such that $H_2/H_1\cong M_2/M_1$($M_1$ is a normal subgroup of $M_2$, and $M_1,M_2$ are sungroups of G/N)

I think, if $H_2 \subset N$, then we are done since $H_1,H_2$ are subgroups of N. Similarly, if $H_1\supset N$, then we are also done, since $H_2/H_1\cong (H_2/N)/(H_1/N)$ and ($H_2/N),(H_1/N)$ are subgroups of G/N. But, now I do not know how to contonue to prove this question. Can someone help me solve this question?

Answer 1

First suppose $H_1\cap N\ne H_2\cap N$ then $K=H_2\cap N/H_1\cap N$ is a subquotient group of $N$.

If $K$ is not simple, then there exists a proper subgroup $S$ of $H_2\cap N$ that properly contains $H_1\cap N$. Notice $H_1\subseteq S+H_1\subseteq H_2$. From the definition of $S$, $(S+H_1)\cap N=S\ne H_1\cap N$ so $S+H_1\ne H_1$. Similarly $S+H_1\ne H_2$. But this means $H_2/H_1$ is not simple, so by contradiction $K$ is simple. $H_1\cap N$ is normal in $H_2\cap N$ follows easily from $N$ being normal in $G$ and $H_1$ being normal in $H_2$.

Now suppose $H_1\cap N=H_2\cap N$, then $M=(H_2N/N)/(H_1N/N)$ is a subquotient group of $G/N$ (again, normality follows easily from normality of $N$ in $G$ and $H_1$ in $H_2$).

By the second isomorphism theorem $H_2N/N\cong H_2/(H_2\cap N)$ and $H_1N/N\cong H_1/(H_1\cap N)$ but $H_1\cap N=H_2\cap N$ by assumption, so we get by the third isomorphism theorem $M\cong (H_2/(H_1\cap N))/(H_1/(H_1\cap N))\cong H_2/H_1$ so $M$ is simple.