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\begin{document}% \tiny \scriptsize \footnotesize \small \normalsize \large \Large \LARGE \huge \Huge 
\begin{center}{\Large Simplicity of $A_n$}\bigskip\\ Pierre-Yves Gaillard\footnote{ORCID https://orcid.org/0000-0002-7960-1698} \end{center} 

\noi In \emph{Simplicity of $A_n$}, \href{https://kconrad.math.uconn.edu/blurbs/grouptheory/Ansimple.pdf}{https://kconrad.math.uconn.edu/blurbs/grouptheory/Ansimple.pdf}, Keith Conrad gives five proofs of the simplicity of $A_n$ for $n\ge5$. We use below some of the ideas contained in Conrad's text. (A short proof of the fact that the sign of a permutation is well defined is given at the end of the text \emph{The dimension of a finitely generated vector space} \href{https://zenodo.org/records/13740597}{https://zenodo.org/records/13740597}.)

Let $n$ be an integer $\ge5$. 

\begin{thm}\label{T1}
The alternating group $A_n$ is simple. 
\end{thm} 

Write $e$ for the identity permutation of $S_n$, and call the product of two disjoint transpositions a \emph{bi-transposition}. 

\begin{lem}\label{L2}
The 3-cycles generate $A_n$. The bi-transpositions also generate $A_n$. The 3-cycles are conjugate in $S_n$. The bi-transpositions also are conjugate in $S_n$. 
\end{lem} 

\begin{lem}\label{L3} 
The only normal subgroup $G$ of $S_n$ contained in $A_n$ such that $\{e\}\ne G$ is $A_n$. 
\end{lem} 

\begin{proof}[Proof of Lemma \ref{L2}] 
The last two statements follow from the fact that permutations with the same cycle structure are conjugate in $S_n$. Clearly the 3-cycles and the bi-transpositions, taken together, generate $A_n$. Hence it suffices to show that a 3-cycles is a product of bi-transpositions, and that a bi-transposition is a product of 3-cycles. But we have 
$$
(1\ 2\ 3)=(1\ 2)(4\ 5)\cdot(4\ 5)(2\ 3)
$$ 
(here we use the assumption $n\ge5$) and 
$$
(1\ 2)(3\ 4)=(1\ 2)(2\ 3)(2\ 3)(3\ 4)=(1\ 2\ 3)(2\ 3\ 4).
$$ 
\end{proof} 

\begin{proof}[Proof of Lemma \ref{L3}] 
For $s$ in $S_n$ we define $f(s)$ as the number of those $i$ in $\{1,\ldots,n\}$ which satisfy $s(i)\ne i$. Let $g_0\in G$ be such that $f(g_0)$ is minimum among the $f(g)$ with $g\in G\setminus\{e\}$. Claim: $f(g_0)\in\{3,4\}$. Cleary $f(g_0)\ge3$. To show $f(g_0)\le4$, note that, if $g$ is in $G\setminus\{e\}$, if $t$ is a transposition in $S_n$, and if we set $c:=tgtg^{-1}$, then we get $c=(tgt)g^{-1}\in G$, and $f(c)\le4$ because $c=t(gtg^{-1})$ is a product of two transpositions. This proves the claim. Now, if $f(g_0)=3$, then $g_0$ is a 3-cycles, and if $f(g_0)=4$, then $g_0$ is a bi-transposition. In both cases the result follows from Lemma~\ref{L2}. 
\end{proof} 

\begin{proof}[Proof of Theorem \ref{T1}] 
Assume by contradiction that there is a normal subgroup $G$ of $A_n$ such that $\{e\}\ne G\ne A_n$. Let $t\in S_n$ be a transposition, and set $G':=tGt$. It is easily seen that $G'$ does not depend on $t$ and is again a normal subgroup $G$ of $A_n$ such that $\{e\}\ne G'\ne A_n$. Then Lemma~\ref{L3} implies $G'\ne G$. The sets $G\cap G'$ and $GG'=G'G$ are normal subgroups of $A_n$. We claim that they are normal even in $S_n$. Indeed, if $t\in S_n$ is a transposition, we have 
$$
t(G\cap G')t=tGt\cap tG't=G'\cap G=G\cap G',
$$ 
$$
t(GG')t=(tGt)(tG't)=G'G=GG'.
$$ 
This proves the claim, and, in view of Lemma~\ref{L3}, this implies $G\cap G'=\{e\}$ and $GG'=A_n$. Moreover the order $|G|$ of $G$ satisfies $|G|^2=n!/2$, so $|G|$ is even, and $G$ has an order two element $g$. We can assume 
$$
g=(1\ 2)(3\ 4)\cdots(2k\!\!-\!\!1\ 2k)
$$ 
with $2\le2k\le n$. Setting $t:=(1\ 2)$ we get $tgt=g$, that is $g\in G\cap G'$, contradiction. 
\end{proof} 

%\noi[1] Keith Conrad, \emph{Simplicity of $A_n$},\\ \href{https://kconrad.math.uconn.edu/blurbs/grouptheory/Ansimple.pdf}{https://kconrad.math.uconn.edu/blurbs/grouptheory/Ansimple.pdf} 

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