Show the commutator subgroup of $S_{n}$ is $A_{n}$ for $n \geq 5$

Show that for $n \geq 5$, the commutator subgroup of $S_{n}$ is $A_{n}$ for $n \geq 5$.

I'm not sure on how to handle this problem. I know that the commutator subgroup $G'$ will consist of products of commutators $[\sigma_{i}, \tau_{i}]$ for permutations $\sigma_{i}, \tau_{i} \in S_{n}$. However, I don't know where to start. What would be the best way to approach this?

• $A_n$ is generated by elements of the type $(abc)$ (i.e. 3-cycles)

• Each $3$-cycle is a commutator: $(abc)=(ab)(ac)(ab)(ac)$.

• $[S_n:A_n]=2$.

Can you complete the proof now?

Let us first prove the following lemma:

Lemma:
Let $n>2$. Then $A_{n}$ is generated by 3-cycles.

Proof:
The product $(1\,2)(2\, 3)$ is equal to the 3-cycle $(1\, 2\, 3)$. The product of two disjoint 2-cycles $(i\, j)$ and $(k\, l)$ is equal to $(i\, j)(j\, k)(j\, k)(k\, l)$ and is hence a product of two 3-cycles. Since any element of $A_{n}$ is a product of an even number of transpositions, it is therefore a product of 3-cycles.
$\square$

We are now ready to prove the theorem:
Proof:
Since $S_{n}/A_{n}$ is commutative, the commutator subgroup $S_{n}^{′}$ is contained in $A_{n}$. Conversely, we have

$\qquad \qquad \qquad \qquad \qquad \qquad (1\, 2)(1\, 3)(1\, 2)^{−1} (1\, 3)^{−1} = (1\, 2\, 3)$

showing that every 3-cycle is in $S_{n}^{′}$ . By the previous lemma, $A_{n}$ is generated by 3-cycles, so $S_{n}^{′} = A_{n}$ as required.
$\square$

Extra:
It is also useful to consider the identity

$\quad \qquad \qquad \qquad \qquad \qquad(1\,2\,3)(3\,4\,5)(1\,2\,3)^{−1}(3\,4\,5)^{−1} = (1\,4\,3)$

which shows that for $n ≥ 5$, every 3-cycle is a commutator of $A_{n}$, which implies that the commutator subgroup of $A_{n}$ is $A_{n}$ itself.

The fact that $A_n$ is simple for $n\geq 5$ can be used to answer this question in quite a different way from the other answers. Let $N=[S_n, S_n]$ denote the commutator subgroup. Then:

1. (General fact about groups.) Note that $G/H$ is abelian if and only if $[G, G]\leq H$.

2. $N\leq A_n$ as $S_n/A_n$ is abelian. Note that as $N$ is normal in $S_n$ it is also normal in $A_n$.

3. $N$ is non-trivial as $S_n$ is non-abelian.

4. Therefore, $N$ is a non-trivial, normal subgroup of $A_n$. As $A_n$ is simple we have that $N=A_n$, as required.