Lemma 9.3.1.
For \(n \geq 3\text{,}\) \(A_n\) is generated by \(3\)-cycles.
Section 9.3 Simplicity of \(A_n\)
We now show that \(A_n\) has no proper nontrivial normal subgroups when \(n \geq 5\text{.}\) The proof proceeds by showing that any nontrivial normal subgroup of \(A_n\) must contain a \(3\)-cycle, and that a normal subgroup containing a single \(3\)-cycle must already be all of \(A_n\text{.}\)
Proof.
Every element of \(A_n\) is a product of pairs of transpositions, and every such pair reduces to \(3\)-cycles:
\begin{align*}
(a,b)(a,b) & = \identity\\
(a,b)(c,d) & = (a,c,b)(a,c,d)\\
(a,b)(a,c) & = (a,c,b)\text{.}
\end{align*}
Lemma 9.3.2.
Let \(n \geq 3\) and let \(N \trianglelefteq A_n\text{.}\) If \(N\) contains a \(3\)-cycle, then \(N = A_n\text{.}\)
Proof.
Every \(3\)-cycle is a product of \(3\)-cycles of the special form \((i,j,k)\) with \(i, j\) fixed and \(k\) varying, since
\begin{align*}
(i, a, j) & = (i, j, a)^2\\
(i, a, b) & = (i, j, b) (i, j, a)^2\\
(j, a, b) & = (i, j, b)^2 (i, j, a)\\
(a, b, c) & = (i, j, a)^2 (i, j, c) (i, j, b)^2 (i, j, a)\text{.}
\end{align*}
Suppose \(N\) contains a \(3\)-cycle \((i, j, a)\text{.}\) By Theorem 7.4.1, \(N\) is closed under conjugation, so
\begin{equation*}
[(i, j)(a, k)](i, j, a)^2 [(i, j)(a, k)]^{-1} = (i, j, k)
\end{equation*}
is in \(N\) for every \(k\text{.}\) So \(N\) contains every \((i,j,k)\text{,}\) which by Lemma 9.3.1 generate \(A_n\text{;}\) hence \(N = A_n\text{.}\)
Lemma 9.3.3.
For \(n \geq 5\text{,}\) every nontrivial normal subgroup \(N\) of \(A_n\) contains a \(3\)-cycle.
Proof.
Let \(\sigma \neq \identity\) be in \(N\text{,}\) and write it as a product of disjoint cycles (Theorem 9.1.2). There are four cases, beyond \(\sigma\) already being a \(3\)-cycle.
A cycle of length \(\gt 3\): \(\sigma = \tau(a_1, a_2, \ldots, a_r)\text{,}\) \(r \gt 3\text{.}\) Since \(N\) is normal, \((a_1,a_2,a_3)\sigma(a_1,a_2,a_3)^{-1} \in N\text{,}\) hence so is
\begin{align*}
\sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1}
& = (a_1, a_r, \ldots, a_2 )(a_1, a_2, a_3) (a_1, a_2, \ldots, a_r)(a_1, a_3, a_2)\\
& = (a_1, a_3, a_r)\text{,}
\end{align*}
a \(3\)-cycle in \(N\text{.}\)
Two \(3\)-cycles: \(\sigma = \tau(a_1,a_2,a_3)(a_4,a_5,a_6)\text{.}\) As above, \(N\) contains
\begin{equation*}
\sigma^{-1}(a_1, a_2, a_4) \sigma(a_1, a_2, a_4)^{-1} = (a_1, a_4, a_2, a_6, a_3)\text{,}
\end{equation*}
a \(5\)-cycle, reducing to the previous case.
A \(3\)-cycle and transpositions: \(\sigma = \tau(a_1,a_2,a_3)\text{,}\) \(\tau\) a product of disjoint transpositions. Then \(\sigma^2 = \tau^2(a_1,a_2,a_3)^2 = (a_1,a_3,a_2) \in N\text{,}\) a \(3\)-cycle.
Only transpositions: \(\sigma = \tau(a_1,a_2)(a_3,a_4)\text{,}\) \(\tau\) a product of an even number of further disjoint transpositions. As before, \(N\) contains
\begin{equation*}
\sigma^{-1}(a_1, a_2, a_3)\sigma(a_1, a_2, a_3)^{-1} = (a_1, a_3)(a_2, a_4)\text{.}
\end{equation*}
Since \(n \geq 5\text{,}\) choose \(b \notin \{a_1,a_2,a_3,a_4\}\) and let \(\mu = (a_1,a_3,b)\text{.}\) Conjugating again, \(N\) contains
\begin{equation*}
\mu^{-1} (a_1, a_3)(a_2, a_4) \mu (a_1, a_3)(a_2, a_4) = (a_1, a_3, b)\text{,}
\end{equation*}
a \(3\)-cycle.
Every case produces a \(3\)-cycle in \(N\text{.}\)
Theorem 9.3.4. Simplicity of \(A_n\).
The alternating group \(A_n\) is simple for \(n \geq 5\text{.}\)
Proof.
Let \(N\) be a nontrivial normal subgroup of \(A_n\text{.}\) By Lemma 9.3.3, \(N\) contains a \(3\)-cycle. By Lemma 9.3.2, \(N = A_n\text{.}\) So \(A_n\) has no proper nontrivial normal subgroups.
