# show that $\operatorname{cent}_{S_n}(h) \neq \langle h\rangle$

After counting the number of conjugates of $h=(1 2 \ldots n-2)$ I get $n(n-1)(n-3)!$ which when plugged in to the orbit stabilizer theorem gives the centralizer to have order $n-2$ but this would show the opposite of what is asked?

Is it mine or the books mistake?

Never mind. Think I should have divided number of conjugates by two which would show what is required

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If all you want is to show that the centralizer of $h=(1\ 2\ \dots\ n-2)$ in $S_n$ is not the subgroup of $S_n$ generated by $h$, then you are working much too hard if you try to count conjugates and use theorems. All you need to do is find one element of $S_n$ which commutes with $h$ but is not a power of $h$, and that's easy! Think about an element of $S_n$ that moves the elements of $\{{1,2,\dots,n\}}$ that $h$ doesn't move.
Hint. Disjoint permutations commute. Can you find a nontrivial permutation in $S_n$ that is disjoint from $h$?