0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.