If there is a non-trivial normal subgroup of the alternating group $A_{5}$ on five elements, then it is a union of conjugacy classes. Additionally, by Lagrange's theorem, the order of a subgroup must divide the order of the group, so the total size of $H$ must divide $60$ .

We can list the conjugacy classes of $A_{5}$ ; they are of size $1, 20, 15, 12, 12$ respectively.

By a brute-force check, no sum of these containing $1$ can possibly divide $60$ (which is the size of $A_{5}$ ) unless it is $1$ or $60$ .

The brute-force check

We first list the divisors of $60$ : they are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60$

Since the subgroup $H$ must contain the identity, it must contain the conjugacy class of size $1$ . If it contains any other conjugacy class (which, as it is a non-trivial subgroup by assumption, it must), then the total size must be at least $13$ (since the smallest other class is of size $12$ ); so it is allowed to be one of $15$ , $20$ , $30$ , or $60$ . Since $H$ is also assumed to be a proper subgroup, it cannot be $A_{5}$ itself, so in fact $60$ is also banned.

The class of size $20$

If $H$ then contains the conjugacy class of size $20$ , then $H$ can only be of size $30$ because we have already included $21$ elements. But there is no way to add just $9$ elements using conjugacy classes of size bigger than or equal to $12$ .

So $H$ cannot contain the class of size $20$ .

The class of size $15$

In this case, $H$ is allowed to be of size $20$ or $30$ , and we have already found $16$ elements of it. So there are either $4$ or $14$ elements left to find; but we are only allowed to add classes of size exactly $12$ , so this can't be done either.

The classes of size $12$

What remains is two classes of size $12$ , from which we can make $1 + 12 = 13$ or $1 + 12 + 12 = 25$ . Neither of these divides $60$ , so these are not legal options either.

This exhausts the search, and completes the proof.