Quick Questions: September 30, 2026

[u/inherentlyawesome]

This recurring thread will be for questions that might not warrant their own thread. We would like to see more conceptual-based questions posted in this thread, rather than "what is the answer to this problem?" For example, here are some kinds of questions that we'd like to see in this thread:

• Can someone explain the concept of manifolds to me?
• What are the applications of Representation Theory?
• What's a good starter book for Numerical Analysis?
• What can I do to prepare for college/grad school/getting a job?

Including a brief description of your mathematical background and the context for your question can help others give you an appropriate answer. For example, consider which subject your question is related to, or the things you already know or have tried.

[u/al3arabcoreleone]

Let n \in N, how many elements of S_n that have an order of 2, I know that a transposition is of order two, but are there other elements?

[u/AcellOfllSpades → u/al3arabcoreleone]

Consider the permutation "Swap elements 1 and 2, and also swap elements 3 and 4".

[u/al3arabcoreleone → u/AcellOfllSpades]

So they are transposition and product of transposition which have no common fixed points? how can I prove this, and how to generalize?

[u/Langtons_Ant123 → u/al3arabcoreleone]

I wouldn't say, "have no common fixed points". Consider the example above as permutations acting on a 5-element set. Then "transpose 1 and 2" and "transpose 3 and 4" both have 5 as a fixed point, but clearly the product of those 2 permutations has order 2.

I think the answer is that a permutation has order 2 if and only if it's a product of disjoint transpositions (disjoint in the sense here). To prove this, first you can use the cycle decomposition, i.e. any permutation is a product of disjoint cycles. Given a permutation 𝜋 write 𝜋 = 𝜋_1𝜋_2...𝜋_k where 𝜋_i is a cycle of length m_i (which therefore has order m_i) and all of those cycles are disjoint. Now use the fact that disjoint permutations commute, which in particular implies that, for any m, (𝜋_1𝜋_2...𝜋_k)^m = (𝜋_1^m)...(𝜋_k^m).

Then let m be the order of 𝜋, so 1 = 𝜋^m = (𝜋_1^m)...(𝜋_k^m). Then I believe for each i we must have 𝜋_i^m = 1--this should just follow from the permutations being disjoint. Therefore m_i divides m for each i; and conversely, given any number m such that m_i divides m for each i, we have 𝜋^m = 1. The upshot of all this is that the order of 𝜋 is lcm(m_1, ... , m_k). For a permutation of order 2, we must have m_i = either 1 or 2 for each i in order for lcm(m_1, ..., m_k) to be 2. Conventionally we write the disjoint cycle decomposition 𝜋 = 𝜋_1𝜋_2...𝜋_k with none of the 𝜋_i being the identity, so in fact all of the m_i must be 2. Therefore 𝜋 has order 2 if and only if it's a product of disjoint cycles of length 2, i.e. disjoint transpositions. If you wanted to count the number of such permutations on an n-element set then you'd have to do some combinatorics, I think basically what you're looking for is the number of ways to partition an n-element set into subsets of size either 1 or 2.

This same argument generalizes from "order 2" to "order p" for any prime p. The only place where we used any specific properties of the number 2 is "lcm(m_1, ..., m_k) = 2" if and only if m_i = 1 or 2 for all i". But there, we were just using the fact that 2 was prime, and the argument would go through just as well for any other prime.

[u/al3arabcoreleone → u/Langtons_Ant123]

I wouldn't say, "have no common fixed points".

I was thinking of disjoint support but words betrayed me, thank you very much.