Show that A(4) is not simple

I.e., show that A₄ has a normal subgroup.

I did it by brutal force and would like to know if there is a more elegant way.

My solution:

A₄ consists of 4!/2=12 permutations:

e = identity,

8 permutations of the kind (1 2 3)=(1 2)(2 3), (1 3 2)=(1 3)(3 2), etc., and

3 permutations with separated cycles:

a = (1 2)(3 4)

b = (1 3)(2 4)

c = (1 4)(2 3)

Because of the separation, the cycles in a, b, and c commute, and thus a² = b² = c² = e.

I've checked manually that ab = ba = c, ac = ca = b, and bc = cb = a.

So, {e, a, b, c} is an abelian group. An abelian subgroup is normal. Thus, A₄ is not simple.

Let me know if any of the above need elaboration.