Abstract Algebra 1 about permutation I have a question from exam and I really don't know how to solve it, and tried a lot.
Say I have a permutation $a=(1 6)(2 5 7)(3 8 4 9)$ so that $a \in S_9$ and I want to find how many permutations $b\in S_9$ exist that commute with $a$ with respect to the composition operation: $a\star b=b \star a$
Thanks in adavance!
 A: Consider, just for testing, $b=(124)(35)(6789)$ and compute
\begin{align}
b\star a\star b^{-1}
&=
(124)(35)(6789)(16)(257)(3849)(142)(35)(6987)\\
&=(1659)(27)(384)\\
&=\bigl(b(1)\,b(6)\bigr)\,
  \bigl(b(2)\,b(5)\,b(7)\bigr)\,
  \bigl(b(3)\,b(8)\,b(4)\,b(9)\bigr)
\end{align}
This holds in general (and you should be able to prove it): if $b$ is any permutation in $S_9$, then $b\star a\star b^{-1}$ is the permutation obtained by applying $b$ to every element in the disjoint cycle decomposition of $a$.
Now, $a\star b=b\star a$ is the same as $b\star a\star b^{-1}=a$.
A: Note that instead of computing the size of the centralizer $C_{S_9}(a)$ of $a$ directly you can compute the size of the conjugacy class $C_a$ of $a$, because $|C_a| |C_{S_9}(a)| = |S_9| = 9!$. But the conjugacy class of $a$ just consists of all the permutations of the form $(s_1,s_2)(s_3,s_4,s_5)(s_6,s_7,s_8,s_9)$ with $\{s_1,\dots,s_9\} = \{1,\dots, 9\}$, so this reduces your problem to a question of combinatorics, namely counting the number of permutations of this form.
