# Difficulty with a lemma needed to prove $A_n$ is a simple group for $n>4$

The theorem is:

For $n \geq 5$, every normal subgroup $N$ of $A_n$ contains a $3$- cycle.

The proof starts like this:

Let $\sigma$ be an arbitrary element in a normal subgroup $N$. There are several possible cycle structures for $\sigma$:

• $\sigma$ is a $3$-cycle.
• $\sigma$ is the product of disjoint cycles, $\sigma = \tau(a_1 a_2 \cdots a_r) \in N$, where $r > 3$.
• $\sigma$ is the product of disjoint cycles, $\sigma = \tau(a_1 a_2 a_3)(a_4 a_5 a_6)$.
• $\sigma = \tau(a_1a_2a_3)$, where $\tau$ is the product of disjoint $2$-cycles.
• $\sigma = \tau(a_1a_2)(a_3a_4)$, where $\tau$ is the product of an even number of disjoint $2$-cycles.

I don't understand why $\sigma$ must be one of the forms. Is is possible they are just saying that $A_n$ must contain at least one of these and not that every element is one of these? If anyone could explain what I am missing that would be great.

## 1 Answer

It's worded ... not optimally, I agree. But, they are saying all even permutations fall into one of those 5 categories:

1. $\sigma$ is a $3$-cycle.
2. $\sigma$ is the product of disjoint cycles, $\sigma = \tau(a_1 a_2 \cdots a_r) \in N$, where $r > 3$.
3. $\sigma$ is the product of disjoint cycles, $\sigma = \tau(a_1 a_2 a_3)(a_4 a_5 a_6)$.
4. $\sigma = \tau(a_1a_2a_3)$, where $\tau$ is the product of disjoint $2$-cycles.
5. $\sigma = \tau(a_1a_2)(a_3a_4)$, where $\tau$ is the product of an even number of disjoint $2$-cycles.

It's really a classification based on the number of $2$- and $3$-cycles the permutation has.

Does the permutation have only $2$- or $3$-cycles, and no longer cycles?

• If yes:

• A single $3$-cycle, and nothing else? Case 1.
• At least two $3$-cycles? Case 3.
• A single $3$-cycle, and the rest $2$-cycles? Case 4.
• No $3$-cycles, and only $2$-cycles?
• If no, the permutation has at least one cycle of length $\ell \geq 4$: Case 2.

Based on that consideration, I think a more revealing description for each case (using the original numbering) might be:

1. A single $3$-cycle
2. Contains a "long" cycle, of length at least $4$
3. At least two $3$-cycles, potentially with $2$-cycles
4. Has a single $3$-cycle, and the rest $2$-cycles, with no larger cycles
5. An even number of $2$-cycles, and nothing else

I've stared at this far too long and changed my mind several times, so I'll leave it up to you to check that these really partition the possibilities with respect to $2$- and $3$-cycles. But that should, at least, give you the idea of the classification that's being used.

• I see. It is important here that cycles can't be broken down into disjoint smaller cycles. The cycles would have to have elements in common. In particular you can't break a 3-cycle down into disjoint 2-cycles. – Maxwell May 9 '15 at 5:58
• Indeed, I'm assuming we were working with the cycle structure of $\sigma$, that it's been broken down into disjoint cycles for us. – pjs36 May 9 '15 at 11:39