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

• Mr Davis 97
This is of course legitimate in a professional conversation, but not in a homework. That's why I asked you for the definition of cycle decomposition.

## 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.

## 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?

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##.

Mr Davis 97 said:

## 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.

## 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!

fresh_42 said:
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|)##?

Mr Davis 97 said:
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.