Alternating group

Let be the Permutation group. We define the alternating group as:

i.e. the group of even permutations.

Proposition

Every Conjugacy Class in is either the same as in or it splits into two. Moreover, splits if and only if the Cycle Type of has distinct odd numbers.

Proof

Let with Cycle Type . Note that:

Also and . So either the Centralizer splits into or the conjugacy class splits into 2. Centralizer splits iff there is some odd . Now if any are even then there is an odd cycle (note odd cycles have even length) in the unique factorisation of . This cycle has to be in . Furthermore, if there are two disjoint cycles of equal odd lengths, say then is odd and . Hence so it splits. Finally, if the cycle type consists of different odd numbers, let . If then for all , hence . Hence there are choices for so is odd and cannot split, so splits. Alternatively, is always a product of even cycles (odd length cycles) hence .

Lemma

Every is generated by 3-cycles.

Proof

Every is a product of an even number of transpositions. Now note (for ): Hence done.

Lemma

If then all 3-cycles are conjugate to each other (in )

Proof

Note that they are conjugate in (same cycle-type), so for for some . If is odd, take , hence done.

Theorem

is Simple for .

Proof

Let . If contains a 3-cycle, then it contains all 3-cycles so it has to be by previous lemmata. Otherwise find any Write it as a product of disjoint cycles:

Case 1

Long cycle where and fixes . Compute:

hence .

Case 2

Double 3-cycle Suppose where fixes . Compute:

so by Alternating group > Case 1. we are done

Case 3

If has at most one 3-cycle, no bigger cycles, and is not a 3-cycle itself it has to have at least 2 transpositions (otherwise its not even). WLOG . Consider first:

Now consider

so we are done by Alternating group > Case 1. (Note, we used again).