Symmetric Group and its centralisers https://www.dropbox.com/s/te4uxryk82h9b23/Screenshot%202015-04-16%2000.38.42.png?dl=0
Let $S_7$ be the symmetric group of degree 7 (the group of permutations of the set {${1, 2,...,7}$}. Let $\sigma\in S_7$ be the following following permutation:
$\sigma = (123)(4567)$
Explain why the centraliser of $\sigma$ in $S_7$ consists precisely of the powers of $\sigma$; that is, $C(\sigma)$=$<\sigma>$.
State clearly any results you use.

I used the relation that $|G| = |cl(\sigma)||C(\sigma)|$
Also, I used the fact that $|G|=7!$
$cl(\sigma) = 2!\times\frac{7!}{4!\times3!}\times3! =12 $ (number of cycles with same cycle structure as $\sigma$
So $|C(\sigma)|=\frac{7!}{2!\times \frac{7!}{4!\times 3!}\times3!}=12 $
And I calculated $\sigma^i, \forall i\in \{1, ...,12 \}$
To realise the order of $\sigma$, $|\sigma|=12$
And because $C(\sigma)$ is a subgroup it is closed under composition so all powers of $\sigma$ must be in that group - there are twelve - so $|C(g)|\ge12$. But by the relation between $cl(\sigma)$ and $C(g)$ above, $C(g)=12$. So S_7 consists precisely of the powers of $\sigma$ and only these powers.

Is there not a quicker way?
Thanks
 A: This Exercise from Lang's Algebra should be of help.
1.36  Let $\sigma=[12\ldots n]$ in $S_n$.  Show that the conjugacy class of $\sigma$ has $(n-1)!$ elements.  Show that the centralizer of $\sigma$ is the cyclic group generated by $\sigma$.
Since, $\gamma\sigma\gamma^{-1}=[\gamma(1)\gamma(2)\ldots\gamma(n)]$, the conjugacy class of $\sigma$ consists of all $n$-cycles in $S_n$.  
Let $I_n=\{1,2,\ldots,n\}$, and $I_n^k=I_n\setminus\{k\}$ for $1\le k\le n$.
$\underline{\text{Claim: There are }(n-1)!\text{ distinct }n\text{-cycles in } S_n}$.
This is trivial for $n=1$.  Suppose, the claim is true for $n\ge1$.  Let $\gamma$ be an $n+1$-cycle in $S_{n+1}$. Let $\gamma(n+1)=k$ for some $1\le k\le n$.  Now, $\gamma|I_n$ is an $n$-cycle acting on $I_{n+1}^k$. So for each $k\in I_n$ there are $n!$, $n+1$-cycles.
So, the orbit of $\sigma$ under the action of conjugation has size $(n-1)!$. Since the isotropy subgroup of $\sigma$ is $Z(\sigma)$, we get
$$|S_n|/|Z(\sigma)|=(n-1)!$$
which implies
$$|Z(\sigma)|=n.$$
But clearly, $\langle\sigma\rangle\subseteq Z(\sigma)$, and $|\langle\sigma\rangle|=n$, so $Z(\sigma)=\langle\sigma\rangle$.
