The alternating group \(A_5\) 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 \(5\).