111100/Simplicity of the Alternating Group

The following is the proof for the simplicity of . It is available individually in pdf format, can be found in the course notes, or on another user page.
Theorem: The alternating group is simple for .
Note that for this is trivial. For we have that which is an abelian group of primeorder, and hence simple.
We know that for is not simple. Indeed, we have seen that there is a nontrivial homomorphism By restricting to we still have a nontrivial homomorphism whose kernel is nontrivial. This kernel is a normal subgroup.
We will need the following Lemmas for our proof.
Lemma: Every element of is a product of 3cycles.
Proof
 Every is a product of an even number of 2cycles. Without loss of generality, we will demonstrate this on the following "cycles. Indeed,
Lemma: If contains a 3cycle, then .
Proof
 Without loss of generality, we can consider We want to show that for all we must have that . If then this is clear since is normal in; otherwise, take with . Since we have that So contains all three cycles.
Proof [Proof of Theorem]
 Let . By the previous two lemmas, it is sufficient to show that contains a three cycle.
 "Case 1:" contains an element with cycle length at least 4.
 Let . Now is normal in so . Similarly, multiplying by will keep the element in , so and contains a three cycle.
 Case 2: contains an element with two cycles of length 3.
 Let . Then by the same reasoning as before and can be computed as . We can now use Case 1 to deduce that has a three cycle.
 Case 3:
 contains an element that is a threecycle and a product of disjoint transpositions. Write and note that
 Then but the elements of are disjoint with , and we conclude that and commute; that is, . Thus and contains a three cycle.
 Case 4: Finally, consider the case when the element of is a product of disjoint 2cycles.
 Write . By previous rationale, we know that and can be computed as We can view this as a sort of purification, in that we have simplified a product of disjoint 2cycles of arbitrary length into the case of two disjoint 2cycles. Applying this procedure again, we get and we again refer to Case 1 to conclude contains a three cycle.
Note that this last case is the only case in which we did not assume "5" was part of the hypothesis, but needed to use it. Hence we need for this case to hold.