11-1100/Simplicity of the Alternating Group
Theorem: The alternating group is simple for .
Note that for this is trivial. For we have that which is an abelian group of prime-order, and hence simple.
We know that for is not simple. Indeed, we have seen that there is a non-trivial homomorphism By restricting to we still have a non-trivial homomorphism whose kernel is non-trivial. This kernel is a normal subgroup.
We will need the following Lemmas for our proof.
Lemma: Every element of is a product of 3-cycles.
- Every is a product of an even number of 2-cycles. Without loss of generality, we will demonstrate this on the following "cycles. Indeed,
Lemma: If contains a 3-cycle, then .
- 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 three-cycle 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 2-cycles.
- 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 2-cycles of arbitrary length into the case of two disjoint 2-cycles. 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.