11-1100/Simplicity of the Alternating Group

From Drorbn
Revision as of 17:14, 6 October 2011 by Tholden (Talk | contribs)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to: navigation, search

The following is the proof for the simplicity of  A_n . It is available individually in pdf format, can be found in the course notes, or on another user page.

Theorem: The alternating group A_n is simple for n\neq 4.

Note that for n=1,2 this is trivial. For n=3 we have that A_3 = \mathbb Z/3 which is an abelian group of prime-order, and hence simple.

We know that for n=4, A_n is not simple. Indeed, we have seen that there is a non-trivial homomorphism \phi:S_4 \to S_3. By restricting \phi to A_4 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 A_n is a product of 3-cycles.


Every \sigma \in A_n is a product of an even number of 2-cycles. Without loss of generality, we will demonstrate this on the following "cycles. Indeed,
(12)(23) = (123) \qquad (123)(234) = (12)(34)

Lemma: If N \triangleleft A_n contains a 3-cycle, then N=A_n.


Without loss of generality, we can consider (123) \in N. We want to show that for all \sigma \in S_n we must have that (123)^\sigma \in N. If \sigma \in A_n then this is clear since N is normal in A_n; otherwise, take \sigma = (12)\sigma' with \sigma' \in A_n. Since (123)^{(12)} = (123)^2 we have that (123)^\sigma = \left((123)^2 \right)^{\sigma'} \in N. So N contains all three cycles.

Proof [Proof of Theorem]

Let n \geq 5. By the previous two lemmas, it is sufficient to show that N contains a three cycle.
"Case 1:" N contains an element with cycle length at least 4.
Let \sigma = (123456)\sigma' \in N. Now N is normal in  A_n so (123) \sigma (123)^{-1} \in N. Similarly, multiplying by \sigma^{-1} will keep the element in N, so \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N and N contains a three cycle.
Case 2: N contains an element with two cycles of length 3.
Let \sigma = (123)(456) \sigma' \in N. Then by the same reasoning as before \sigma^{-1}(124)\sigma(124)^{-1} \in N and can be computed as \sigma^{-1}(124)\sigma(124)^{-1} = (14263). We can now use Case 1 to deduce that N has a three cycle.
Case 3:
N contains an element that is a three-cycle and a product of disjoint transpositions. Write \sigma = (123)\sigma' and note that {\sigma'}^2 = e.
Then \sigma^2 = (123)\sigma'(123)\sigma' but the elements of \sigma' are disjoint with (123), and we conclude that \sigma' and (123) commute; that is, (123)\sigma' = \sigma'(123). Thus \sigma^2 = (123)^2 \sigma'^2 = (132) \in N, and N contains a three cycle.
Case 4: Finally, consider the case when the element of N is a product of disjoint 2-cycles.
Write \sigma = (12)(34) \sigma'. By previous rationale, we know that \sigma^{-1} (123) \sigma (123)^{-1} \in N and can be computed as \sigma^{-1} (123) \sigma (123)^{-1} = (13)(24) = \tau \in N. 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 \tau^{-1}(125)\tau(125)^{-1} = (13452) \in N and we again refer to Case 1 to conclude N 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 n \geq 5 for this case to hold.