Permutation group S8

Question

In symmetric group S8, there are two permutations namely a=(147)(258) and b=(14)(2578). We are to find the no of such permutations say 's' such that as=sb. I am trying to use the tool of orders of permutations a and b which I got 3 and 4 respectively. But have no idea how to proceed further to solve this problem.

Answer 1

If such a $s$ exist then $a$ and $b$ would be conjugates ($sas^{-1}=b$). But conjugates have the same cycle structure, which is not the case here. So no $s$ exists.

You can also argue based on orders. Conjugates have same order. But $a$ has order $3$ and $b$ has order $4$.

hint: let $|a|=t$ and $|b|=k$. Then \begin{align*} sas^{-1} &= b\\ (sas^{-1})^t &= b^t\\ sa^ts^{-1} &= b^t\\ e&=b^t. \end{align*} Thus $k$ divides $t$. Now show the converse to show conjugates have same order.

Answer 2

Another way is to use a parity argument. We see that $a$ is even and $b$ is odd. So for any $s$, $as$ and $sb$ will also have different parities. This means that $as$ cannot equal $sb$ for any $s$.