Showing that an element has order 2 if product of 2-cycles

[Mr Davis 97]

Homework Statement

Show that an element has order 2 in $S_n$ if and only if its cycle decomposition is a product of commuting 2-cycles.

Homework Equations

The Attempt at a Solution

Suppose that $\sigma \in S_n$ and $| \sigma | =2$. Let the cycle decomposition of $\sigma$ be the following: $\sigma = c_1 c_2 \dots c_m$. Then $| \sigma | = lcm (|c_1|, |c_2|, \dots , |c_m|) = 2$. This is the case only if the $|c_i| = 1$ or $|c_i| = 2$ with at least one such that $|c_i| = 2$. Hence $\sigma$ is a product of disjoint 2-cycles.
\Now, suppose that the cycle decomposition of $\sigma$ is a product of commuting 2-cycles: $\sigma = c_1 c_2 \dots c_m$, where $|c_1| = \cdots = |c_m| = 2$. Then $| \sigma | = lcm (|c_1|, |c_2|, \dots, |c_m|) = lcm (2,2, \dots, 2) = 2$.

Is this at all a correct proof? Is there a way to do this without assuming a permutation can be decomposed uniquely into disjoint cycles, or that the order of a permutation is the least common multiple of the orders of the cycles in is decomposition?

 

[Orodruin]

This works if you have previously shown that $|\sigma| = \operatorname{lcm}(|c_i|)$. The cleaner way is to just use that the $c_i$ have order 2 (which is trivial to show) and that, since they commute, $\sigma^2 = c_1^2 c_2^2 \ldots c_m^2 = e^m = e$.

 

[fresh_42]

Homework Statement

Show that an element has order 2 in $S_n$ if and only if its cycle decomposition is a product of commuting 2-cycles.

Homework Equations

The Attempt at a Solution

Suppose that $\sigma \in S_n$ and $| \sigma | =2$. Let the cycle decomposition of $\sigma$ be the following: $\sigma = c_1 c_2 \dots c_m$. Then $| \sigma | = lcm (|c_1|, |c_2|, \dots , |c_m|) = 2$. This is the case only if the $|c_i| = 1$ or $|c_i| = 2$ with at least one such that $|c_i| = 2$. Hence $\sigma$ is a product of disjoint 2-cycles.

Why do they have to be disjoint, resp. why is $(12)(23)$ no cycle decomposition? And you can eliminate the identities by simply demanding $c_i \neq 1$.

\Now, suppose that the cycle decomposition of $\sigma$ is a product of commuting 2-cycles: $\sigma = c_1 c_2 \dots c_m$, where $|c_1| = \cdots = |c_m| = 2$. Then $| \sigma | = lcm (|c_1|, |c_2|, \dots, |c_m|) = lcm (2,2, \dots, 2) = 2$.

As @Orodruin has said, a simple calculation of $\sigma^2$ is easier. It also has the big advantage, that you actually see, why the commutation part is essential!

 

[Mr Davis 97]

Why do they have to be disjoint, resp. why is $(12)(23)$ no cycle decomposition? And you can eliminate the identities by simply demanding $c_i \neq 1$.

As @Orodruin has said, a simple calculation of $\sigma^2$ is easier. It also has the big advantage, that you actually see, why the commutation part is essential!

Don't they have to be disjoint 2-cycles since I demanded to being with that the cycle decomposition of $\sigma$ is a product of disjoint cycles?
So I see how the backward direction is easier by just assuming they commute and raising the product to a power of 2. But is the forward direction okay? How could I show the forward direction without the use of the result $|\sigma| = \operatorname{lcm}(|c_i|)$?

 

[fresh_42]

Don't they have to be disjoint 2-cycles since I demanded to being with that the cycle decomposition of $\sigma$ is a product of disjoint cycles?

Yes, but only implicit in the word "cycle decomposition". Since you haven't filled out part two of the template which I personally consider even more important than part three, although our rules lay emphasis on own effort, you have to rely on the fact that everybody knows, that $(12)(23)$ isn't a decomposition whereas $(123)$ is. I think you should have mentioned this somewhere because it is crucial for the proof.

So I see how the backward direction is easier by just assuming they commute and raising the product to a power of 2. But is the forward direction okay? How could I show the forward direction without the use of the result $|\sigma| = \operatorname{lcm}(|c_i|)$?

Both directions are correct, since you already have proven $|\sigma \tau|=\operatorname{lcm}\{\,|\sigma | , |\tau | \,\}$ in case $[\sigma,\tau]=1$ and disjoint permutations commute. However, both facts are necessary for your way to prove it - plus a proper definition of a cycle decomposition. You see, there are many hidden statements, which you all assumed to be known.