The alternating group on five elements is simple.

Proof

Recall that $A_{5}$ has order $60$ , so Lagrange's theorem states that any subgroup of $A_{5}$ has order dividing $60$ .

Suppose $H$ is a normal subgroup of $A_{5}$ , which is not the trivial subgroup ${e}$ . If $H$ has order divisible by $3$ , then by Cauchy's theorem there is a $3$ -cycle in $H$ (because the $3$ -cycles are the only elements with order $3$ in $A_{5}$ ). Because $H$ is a union of conjugacy classes, and because the $3$ -cycles form a conjugacy class in $A_{n}$ for $n > 4$ , $H$ would therefore contain every $3$ -cycle; but then it would be the entire alternating group.

If instead $H$ has order divisible by $2$ , then there is a double transposition such as $(12) (34)$ in $H$ , since these are the only elements of order $2$ in $A_{5}$ . But then $H$ contains the entire conjugacy class so it contains every double transposition; in particular, it contains $(12) (34)$ and $(15) (34)$ , so it contains $(15) (34) (12) (34) = (125)$ . Hence as before $H$ contains every $3$ -cycle so is the entire alternating group.

So $H$ must have order exactly $5$ , by Lagrange's theorem; so it contains an element of order $5$ since prime order groups are cyclic.

The only such elements of $A_{n}$ are $5$ -cycles; but the conjugacy class of a $5$ -cycle is of size $12$ , which is too big to fit in $H$ which has size $5$ .