The alternating group on five elements is simple: Simpler proof
If there is a non-trivial normal subgroup \(H\) 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 \(1, 20, 15, 12, 12\) respectively.; they are of size
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 \(60\): they are \($1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60\)$of
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.