% jordan-holder-b
% https://docs.google.com/document/d/1NnyjOIWqDY3UuRrg-o8ALYLMwB4f_8J4Ok9tD_My_Lw/edit
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\begin{document}% \tiny \scriptsize \footnotesize \small \normalsize \large \Large \LARGE \huge \Huge 
\begin{center}{\Huge The Jordan-Hölder Theorem}\bigskip\\ Pierre-Yves Gaillard\footnote{ORCID https://orcid.org/0000-0002-7960-1698} \end{center} %\centerline{Jordan-Hölder Theorem} 

\noi Let $G$ be a finite group, and $S_1,\ldots,S_q$ a sequence of simple finite groups such that any simple quotient of the form $H/N$, where $H$ is a subgroup of $G$, and $N$ is a normal subgroup of $H$, is isomorphic to $S_j$ for a unique $j\in Q:=\{1,\ldots,q\}$. For any $j\in Q$ and any increasing sequence $G_1\subset\cdots\subset G_k$ (with $k\ge2$) of subgroups of $G$, with $G_i$ normal in $G_{i+1}$ for $i=1,\ldots,k-1$, let $f_j(G_1,\ldots,G_k)$ be the number of quotients $G_{i+1}/G_i$ isomorphic to $S_j$: 
$$
f_j(G_1,\ldots,G_k)=|\{i\ |\ 1\le i\le k-1\text{ and }G_{i+1}/G_i\simeq S_j\}|.
$$ 
Recall that $(G_1,\ldots,G_k)$ as above is a \emph{composition series} for $G_k$ if $G_1=1$, and if all the quotients $G_{i+1}/G_i$ are simple. In this case we clearly have 

\begin{equation}\label{E1}
\sum_{j\in Q}f_j(G_1,\ldots,G_k)=k-1.
\end{equation} 

\begin{thm}[Jordan-Hölder Theorem]\label{JHT}
If $(G_1,\ldots,G_k)$ and $(H_1,\ldots,H_m)$, with $k,m\ge2$, are composition series for $G_k=G=H_m$, then we have  $f_j(G_1,\ldots, G_k)=f_j(H_1,\ldots,H_m)$ for all $j\in Q$. In particular \eqref{E1} implies $k=m$.  
\end{thm} 

It suffices to prove: 

\begin{lem}\label{L2}
If $G$ is a finite group admitting a composition series $(G_1,\ldots,G_k)$ with $k\ge2$, then Theorem~\ref{JHT} holds for $G$. 
\end{lem}  

\begin{proof}[Proof of Lemma~\ref{L2}]
We prove Lemma~\ref{L2} by induction on $k\ge2$. The case $k=2$ is obvious. Assume that $k\ge3$ and that the Lemma holds for $n$ with $2\le n\le k-1$. Let $(G_1,\ldots,G_k)$ and $(H_1,\ldots,H_m)$ be two composition series for $G$. If $G_{k-1}=H_{m-1}$ we can apply the induction hypothesis to this group, and the result follows. Hence we can assume $G_{k-1}\ne H_{m-1}$. Setting $K:=G_{k-1}\cap H_{m-1}$, and letting $(K_1,\ldots,K_p)$ be a composition series for $K=K_p$, we get 
$$ 
f_j(G_1,\ldots,G_k)=f_j(G_1,\ldots,G_{k-1})+f_j(G_{k-1},G_k)
$$ 
for all $j\in Q$. Applying the induction hypothesis to $G_{k-1}$ yields $p=k-2$ and 
\begin{equation*} 
\begin{split}
f_j(G_1,\ldots,G_k)
%
&=f_j(K_1,\ldots,K_{k-2},G_{k-1})+f_j(G_{k-1},G_k)\\ 
%
&=f_j(K_1,\ldots,K_{k-2})+f_j(K_{k-2},G_{k-1})+f_j(G_{k-1},G_k)\\ 
%
&=f_j(K_1,\ldots,K_{k-2})+f_j(K_{k-2},H_{m-1})+f_j(H_{m-1},H_m)\\ 
%
&=f_j(K_1,\ldots,K_{k-2},H_{m-1})+f_j(H_{m-1},H_m),
\end{split}
\end{equation*} 
the third equality following from the Diamond Isomorphism Theorem. Arguing as above, we conclude that $m=k$ and that 
\begin{equation*} 
\begin{split}
f_j(G_1,\ldots,G_k) 
%
&=f_j(H_1,\ldots,H_{k-1})+f_j(H_{k-1},H_k)\\ 
%
&=f_j(H_1,\ldots,H_k).
\end{split}
\end{equation*} 
\end{proof} 

\end{document}