Find the number of permutations in $S_6$ which commute with $\sigma = (1 \ 2 \ 3) (4 \ 5)$
This was my attempt:
A permutation $\tau \in S_6$ commutes with $\sigma \iff \sigma =\tau^{-1} \sigma \tau = (\tau(1) \ \tau(2) \ \tau(3)) ( \tau (4) \ \tau(5))$.
Using the theorm on the uniqueness of decomposition to cycles we want the following to hold: $$ (\tau(1) \ \tau(2) \ \tau(3)) = (1 \ 2 \ 3)$$ $$ ( \tau (4) \ \tau(5)) = (4 \ 5)$$
Now comes my confusion:
The way I see it the only valid values for $ (\tau(1) \ \tau(2) \ \tau(3))$ are $(1 \ 2 \ 3), (3 \ 1 \ 2), (2 \ 3 \ 1)$ since these are all the ways to write the same cycle. and similarly $(4 \ 5), (5 \ 4)$. so we should have $3*2=6$ commuting permutations in total.
However, when I check manually I find that $ (\tau(1) \ \tau(2) \ \tau(3)) = (1 \ 3 \ 2)$ also commutes:
$$ \sigma \tau = (1 \ 2 \ 3) (4 \ 5) (1 \ 3 \ 2) (4 \ 5) = (1 \ 2 \ 3) (1 \ 3 \ 2) = I = (1 \ 3 \ 2)(1 \ 2 \ 3) = (1 \ 3 \ 2) (4 \ 5) (1 \ 2 \ 3) (4 \ 5) = \tau \sigma$$
Can someone please explain my mistake?
$\tau=(1\ 3\ 2)$ does indeed commute with $\sigma$, and $(\tau(1)\ \tau(2)\ \tau(3))=(3\ 1\ 2)$ which is one of your valid values, so there is no problem. The mistake is that you've treated the statements "$(\tau(1)\ \tau(2)\ \tau(3))=(a\ b\ c)$" and "$\tau=(a\ b\ c)$" as the same, which they aren't.