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\parbox[b]{2in}{\tiny
  \href{http://www.math.toronto.edu/~drorbn}{Dror Bar-Natan}: \href{http://www.math.toronto.edu/~drorbn/classes}{Classes}:
  \newline \href{http://www.math.toronto.edu/~drorbn/classes/\#2526}{2025-26}: \href{http://drorbn.net/25-347}{MAT 347 Groups, Rings, Fields}:
}~{\LARGE\bf The Simplicity of $A_n$}
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\begin{multicols}{2}

{\olive This handout is to be read twice: First read {\red red} only, to ascertain that everything in {\red red} is easy and boring. Then read {\black black} and {\red red}, to actually understand the proof.}

\noindent{\bf Theorem.} The alternating group $A_n$ is simple for $n\neq 4$.

\noindent{\bf Remark.} Easy for $n\leq 3$ and false for $n=4$ as there is a $\phi\colon A_4\to A_3$ (see below). So we assume that $n\geq 5$.

\noindent{\bf Reminder} (from HW3). Two permutations in $S_n$ are conjugate iff the sequences of lengths of cycles in their cycle decompositions are
the same (up to a permutation of these lengths).

\noindent{\bf Lemma 1.} Every element of $A_n$ is a product of 3-cycles.

\noindent{\em Proof.} Every element of $A_n$ is a product of an even number of 2-cycles, and {\red $(12)(23)=(123)$} and {\red $(12)(34)=(123)(234)$}. \qed

\noindent{\bf Lemma 2.} If $N\lhd A_n$ contains a 3-cycle, then $N=A_n$.

\noindent{\em Proof.} WLOG, $(123)\in N$. Then for all $\sigma\in S_n$, $(123)^\sigma\in A_n$. Indeed, if $\sigma\in A_n$, this is clear. Otherwise $\sigma=(12)\sigma'$ with $\sigma'\in A_n$, and then as {\red $(123)^{(12)} = (123)^2$}, we have that $(123)^\sigma = (123)^{(12)\sigma'} = ((123)^2)^{\sigma'} \in N$. And so $N$ contains all the 3-cycles, and so by Lemma~1, $N=A_n$. \qed

\noindent{\em Proof of the Theorem.} We now assume that $N\lhd A_n$ is not trivial, and check a few cases. In each case we find that $N=A_n$:

\noindent{\bf Case 1.} $N$ contains an element whose cycle decomposition has a cycle of length $\geq 4$.

\noindent{\em Resolution.} {\red $\sigma = (123456)\sigma'$} (with $\sigma'$ fixing 1,2,3,4,5,6) implies {\red $\sigma^{-1}\sigma^{(123)} = (236)$}$\in N$. \qed

\noindent{\bf Case 2.} $N$ contains an element with two cycles of length 3.

\noindent{\em Resolution.} {\red $\sigma = (123)(456)\sigma'$} implies {\red $\sigma^{-1}\sigma^{(124)} = (12436)$}$\in N$. \qed

\noindent{\bf Case 3.} $N$ contains {\red $\sigma = (123)\cdot(\text{disjoint 2-cycles})$}.

\noindent{\em Resolution.} {\red $\sigma^2 = (132)$}$\in N$. \qed

\noindent{\bf Case 4.} $N$ contains a disjoint product of 2-cycles.

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\noindent{\em Resolution.} {\red $\sigma = (12)(34)\sigma'$}$\in N$ implies {\red $\sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau$}$\in N$ implies {\red $\tau^{-1}(125)\tau(125)^{-1} = (13452)$}$\in N$. \linebreak[1]\null\hfill$\Box\ \Box$

\[ \includegraphics[width=0.6\linewidth]{ColouredTetrahedron.png} \]

\end{multicols}

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