11-1100/Simplicity of the Alternating Group

DBN: Classes: 2011-12: 11-1100/Navigation # Week of... Notes and Links Additions to the MAT 1100 web site no longer count towards good deed points 1 Sep 12 About This Class, Tuesday - Non Commutative Gaussian Elimination, Thursday - NCGE completed, the category of groups, images and kernels. 2 Sep 19 I'll be in Strasbourg, class was be taught by Paul Selick. See my summary. 3 Sep 26 Class Photo, HW1, HW1 Submissions, HW 1 Solutions 4 Oct 3 The Simplicity of the Alternating Groups 5 Oct 10 HW2, HW 2 Solutions 6 Oct 17 Groups of Order 60 and 84 7 Oct 24 Extra office hours: Monday 10:30-12:30 (Dror), 5PM-7PM @ Huron 1028 (Stephen). Term Test on Tuesday. Summary for the midterm exam. Term Test - Sample Solutions 8 Oct 31 HW3, HW 3 Solutions 9 Nov 7 Monday-Tuesday is November Break, One Theorem, Two Corollaries, Four Weeks 10 Nov 14 HW4, HW 4 Solutions 11 Nov 21 12 Nov 28 HW5 and last week's schedule 13 Dec 5 Tuesday - the Jordan form; UofT Fall Semester ends Wednesday; our Final Exam took place on Friday Register of Good Deeds Add your name / see who's in! See Non Commutative Gaussian Elimination

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.

Proof

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 $N=A_{n}$.

Proof

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.