Can you find these sets of permutations?

Question

We have $n\geq2$ a natural number and $S_{n}$ the set of permutations with length n. Find the sets: $A=\left\{\sigma\in S_{n}\mid \exists\ r\geq1 \ and \ \sigma_{1},\sigma_{2},...,\sigma_{r}\in \ S_{n}\ such \ that \ \sigma=\sigma_{1}^2\sigma_{2}^2...\sigma_{r}^2\right\}$ $B=\left\{\sigma\in S_{n}\mid \exists\ r\geq1 \ and \ \sigma_{1},\sigma_{2},...,\sigma_{r}\in \ S_{n}\ such \ that \ \sigma=\sigma_{1}^3\sigma_{2}^3...\sigma_{r}^3\right\}$

Answer 1

Recall that any $\sigma$ can be written as the product of transpositions. Note also that for any transposition $\tau$, $\tau=\tau^3$. So if we take $\sigma$, write it as the product $\sigma=\sigma_1\ldots\ldots\sigma_r$ of transpositions $\sigma_1, \ldots, \sigma_r$, it's also true that $$\sigma=\sigma_1^3\ldots \sigma_r^3.$$ Therefore $B=S_n$.

Recall that $A_n\subset S_n$ is the alternating group, which is the set of all $\sigma\in S_n$ which can be written as the product of an even number of transpositions. If $$\sigma=\sigma_1^2\ldots \sigma_r^2,$$ then if we write $\sigma_i=\tau_{i,1}\ldots \tau_{i,r_i}$ for some transpositions $\tau_{i,1},\ldots, \tau_{i,r_i}$, then $$\sigma_i^2=\tau_{i,1}\ldots \tau_{i,r_i}\tau_{i,1}\ldots \tau_{i,r_i},$$ which is the product of $2r_i$ transpositions. Therefore $\sigma$ is the sum of $2\sum_{i=1}^r r_i$ transpositions, and $\sigma\in A_n$. Therefore $A\subset A_n$.

I claim that $A=A_n$. First consider two transpositions $(ab),(cd)$. If $\{a,b\}=\{c,d\}$, then $(ab)(cd)=\iota=\iota^2$, where $\iota$ is the identity permutation. If $|\{a,b\}\cap \{c,d\}|=1$, then without loss of generality we can assume $b=c$. Then $(ab)(cd)=(ab)(bd)=(bda)=(bad)^2$. Last, if $\{a,b\}\cap \{c,d\}=\varnothing$, then $(ab)(cd)=(acbd)^2$. So any product of two transpositions can be written as $\mu^2$ for some $\mu$. Therefore for $\sigma\in A$, we can write $$\sigma= \tau_1\ldots \tau_{2r}.$$ Note the $2r$ subscript, since we have an even number of transpositions. Then for some $\mu_1, \ldots, \mu_r$, we have $\tau_1\tau_2=\mu_1^2$, $\tau_3\tau_4=\mu_2^2$, etc. In general, $\tau_{2k-1}\tau_{2k}=\mu_k^2$. Then $$\sigma=(\tau_1\tau_2)(\tau_3\tau_4)\ldots (\tau_{2r-1}\tau_{2r})=\mu_1^2\ldots \mu_r^2.$$