Q: Explain why there are $18$ elements of $S_6$ which commute with $\sigma = (1\ 2\ 3) (4\ 5\ 6)$ and why the centralizer $C (\sigma)$ is not abelian.
I got the first part fine, I think. I basically reasoned that elements that commute are the same elements that are in the centralizer. In that case, the question is how many elements are in the conjugacy class. $(1\ 2\ 3)$, $(2\ 3\ 1)$, $(3\ 1\ 2)$ all represent the same cycle, as do $(4\ 5\ 6)$, $(5\ 6\ 4)$, and $(6\ 4\ 5)$. So the number of elements with the $(1\ 2\ 3)/(2\ 3\ 1)/(3\ 1\ 2)$ first and the $(4\ 5\ 6)/(5\ 6\ 4)/(6\ 4\ 5)$ second is $3\times3 = 9$. Multiply this by $2$ to get the options where the order is reversed, and you get $18$.
For the second part, I know I could just pick two elements of the centralizer and explicitly show it is not abelian, although I feel like that is not what the question-setter is looking for. It seems I'm forgetting some fact about abelian subgroups or the symmetric group. Can anyone point out what it is I'm missing?
I think you can easily solve part two by making just a little bit explicit your solution of part one.
You have seen that in the centralizer of $\sigma$ there are elements (which actually generate it) $$ \sigma_{1} = ( 1 2 3 ), \sigma_{2} = (4 5 6), \tau = (1 4) (2 5) (3 6), $$ where $\tau$ conjugates $\sigma_{1}$ to $\sigma_{2}$, and thus centralizes their product $\sigma$.
But then you have shown that $\sigma_{1}$ and $\tau$ do not commute.