Isomorphism of two simple normal subgroups of a group $G$, given that $G$ has composition series Let $G$ be a group having composition series. Suppose that $H$ and $K$ are normal simple subgroups of $G$ and that $G/H\cong G/K$. Then $H\cong K$.
My attempt (edited with hints from @DerekHolt):
Since $G$ has composition series and $H$ and $K$ are normal in $G$, then $G/H$ and $G/K$ and $H$ and $K$ have composition series. Choose composition series for $G/H$ and $G/K$:
\begin{align*}
\{1_{G/H}\}\subseteq H_{1}/H\subseteq H_{2}/H\subseteq \dots\subseteq H_{n-1}/H\subseteq H_{n}/H&=G/H
\\
\{1_{G/K}\}\subseteq K_{1}/K\subseteq K_{2}/K\subseteq \dots\subseteq K_{m-1}/K\subseteq K_{m}/K&=G/K
\end{align*}
and composition series for $H$ and $K$
\begin{align*}
\{1_{G}\}\subseteq L_{1}\subseteq L_{2}\subseteq\dots\subseteq L_{s}=H
\\
\{1_{G}\}\subseteq M_{1}\subseteq M_{2}\subseteq\dots\subseteq M_{t}=K
\end{align*}
Combining the composition series for $H$, $G/H$ and $K$, $G/K$, we get
\begin{align*}
\{1_{G}\}\subseteq L_{1}\subseteq L_{2}\subseteq\dots\subseteq L_{s}=H\subseteq H_{1}\subseteq\dots\subseteq H_{n}&=G
\\
\{1_{G}\}\subseteq M_{1}\subseteq M_{2}\subseteq\dots\subseteq M_{t}=K\subseteq K_{1}\subseteq\dots\subseteq K_{m}&=G.
\end{align*}
By the Jordan-Holder Theorem, $m=n$. (This is where I am stuck)

I tried to consider a smaller case where the length is $4$ (things work out well for length $2$ and $3$.
If we have (noting that $H$ and $K$ are simple, so there are no nontrivial normal subgroups of $H$ and $K$)
\begin{align*}
\{1_{G}\}\subseteq H=L_{1}\subseteq L_{2}\subseteq L_{3}\subseteq L_{4}&=G
\\
\{1_{G}\}\subseteq K=K_{1}\subseteq K_{2}\subseteq K_{3}\subseteq K_{4}&=G
\end{align*}
then by the third isomorphism theorem
\begin{align*}
G/K_{\sigma(3)}\cong(G/K)/(K_{\sigma(3)}/K)\cong(G/H)/(L_{3}/H)\cong G/L_{3}
\end{align*}
for some $\sigma\in S_{n}$. (I don't think I'm supposed to use the isomorphism in this manner, but I cannot see how else to use it)

A similar question has been posted to How are composition series for two isomorphic groups related? but the answer states that the composition factors are isomorphic (which is precisely the Jordan-H\"older theorem), but nothing much is said about the individual subgroups $G_{i}$ $H_{j}$ (e.g. is $G_{i}\cong H_{j}$ for some $i,j$?) in the composition series of $G$ and $H$.
I would appreciate any comments, hints on this problem.
 A: I'm answering my question based on @DerekHolt 's hints.
Since $G$ has composition series and $H$ and $K$ are normal and simple in $G$, then $G/H$ and $G/K$ and $H$ and $K$ have composition series. Choose composition series for $G/H$ and $G/K$:
\begin{align*}
\{1_{G/H}\}\subset H_{2}/H\subset H_{3}/H\subset \dots\subset H_{n-1}/H\subseteq H_{n}/H&=G/H
\\
\{1_{G/K}\}\subseteq K_{2}/K\subseteq K_{3}/K\subseteq \dots\subseteq K_{m-1}/K\subseteq K_{m}/K&=G/K
\end{align*}
and composition series for $H$ and $K$ (which are simple and normal in $G$)
\begin{align*}
\{1_{G}\}\subset H
\\
\{1_{G}\}\subset K
\end{align*}
Combining the composition series for $H$, $G/H$ and $K$, $G/K$, we get
\begin{equation*}
\{1_{G}\}\subset H\subset H_{2}\subset\dots\subset H_{n-1}\subset H_{n}=G
\\
\{1_{G}\}\subset K\subset K_{2}\subset\dots\subset K_{m-1}\subset K_{m}=G
\end{equation*}
By the Jordan-H\"older theorem, $m=n$ and we have
\begin{equation*}
\{1_{G}\}\subset H\subset H_{2}\subset\dots\subset H_{n-1}\subset H_{n}=G
\\
\{1_{G}\}\subset K\subset K_{2}\subset\dots\subset K_{n-1}\subset K_{n}=G
\end{equation*}
Now, the composition series for $G/H$ and $G/K$ are
\begin{equation*}
\{1_{G/H}\}\subset H_{2}/H\subset H_{3}/H\subset \dots\subset H_{n-1}/H\subset H_{n}/H=G/H
\\
\{1_{G/K}\}\subset K_{2}/K\subset K_{3}/K\subset \dots\subset K_{n-1}/K\subset K_{n}/K=G/K
\end{equation*}
Since $G/H\cong G/K$, then there exists an isomorphism
\begin{equation*}
\psi:G/H\tilde{\longrightarrow} G/K
\end{equation*}
and if $H_{i}/H\leq G/H$, by the fourth isomorphism theorem for groups, $\psi(H_{i}/H)\leq G/K$ and moreover, if $H_{i}/H\triangleleft H_{i+1}/H$ in $G/H$, then $\psi(H_{i}/H)\triangleleft \psi(H_{i+1}/H)$ as normal subgroups in $G/K$. This gives us
\begin{equation*}
\psi(H_{i}/H)=K_{i}/K
\end{equation*}
for each $i=2,3,\dots,n$ and so $H_{i}/H\cong K_{i}/K$ for each $i=2,3,\dots,n$. In particular, $H_{2}/H\cong K_{2}/K$ as simple quotient groups. Then we have composition series for $H_{2}$ and $K_{2}$ as
\begin{equation*}
1_{G}\subset H\subset H_{2}
\\
1_{G}\subset K\subset K_{2}
\end{equation*}
so that $H\cong K$.
