How many elements of $S_9$ commute with $(123)(4567)$? I know that I need to find a $\theta \in S_9$ such that $\theta \sigma \theta^{-1}=(\theta(1),\theta(2),\theta(3))\cdot (\theta(4),\theta(5),\theta(6),\theta(7))=(1,2,3)(4,5,6,7)$.
Kind of stuck here. How do I check which elements of $S_9$ will satisfy it? 
 A: Consider the action of $ S_9 $ on itself, given by conjugation. The stabilizer of an element $ x \in S_9 $ is its centralizer, and the orbit-stabilizer theorem gives
$$ 9! = |S_9| = |C(x)| \cdot |S| $$
where $ S $ denotes the set of conjugates of $ x $, and $ C(x) $ is the centralizer. Conjugate elements in $ S_n $ are precisely those which have the same cycle type, therefore, $ S $ is the set of all $ 3, 4 $ cycles when $ x = (123)(4567) $. By elementary combinatorics, there are exactly
$$ \binom{9}{3} \cdot 2! \cdot \binom{6}{4} \cdot 3! = 15120 $$
such cycles. Thus,
$$ C((123)(4567)) = \frac{9!}{15120} = 24 $$
elements of $ S_9 $ commute with $ (123)(4567) $.
A: Hint
$ O(Class(a))=\frac{O(S_n)}{O(C(a))}$
$C(a)$ is the centralizer of $a$
You need $C(a)$ as your answer. Can you proceed?
In general if you have to find out the $O(C(a))$ of any element $a$ then first express the element into a product of disjoint cycles .say this representation has cycles of length 
$n_1,n_2,......n_n$ with occurence $r_1,r_2,.....r_n $ times respectively. Then $C(a)$=$(r_1×(n_1)^{r_1}).(r_2×(n_2)^{r_1})......$
A: The centralizer (also called the normalizer by some) of $\sigma$ is the intersection of the centralizer of $(1,2,3)$ and the centralizer of $(4,5,6,7)$. The centralizer of a cycle in general is given by the conjugation by the permutations that relabels the cycle (e.g. for $(1,2,3)$ giving $(2,3,1)$ and $(3,1,2)$). For a given cycle these permutations form the group generated by the cycle itself and all permutations of the elements that are not moved by the cycle (for $(1,2,3)$ this is $S_{\{4,5,6,7,8,9\}}$). So this intersection consists of the group generated by $(1,2,3),(4,5,6,7)$ and $(8,9)$ giving the group $C_3 \times C_4 \times C_2$ of order 24.
A: Now the thing you have computed states that  the three length cycle is only possible to be $(1,2,3)$, the four length cycle is $(4,5,6,7)$ in a similar manner. The only possiblity is $(8,9)$ is possible.
So the permutations that shall commute are $\{(1,2,3)(4,5,6,7) , (1,2,3)(4,5,6,7)(8,9) \} $
