let $S_{10}$ denote the group of permutations on ten symbols ${1,2,3,....,10}$.Then how do we calculate number of elements of $S_{10}$ commuting with the element $\sigma=(1\ 3\ 5\ 7\ 9)$?
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2Quick beginner guide for asking a well-received question + please avoid "no clue" questions. Anyway, this answered your question: Order of the centralizer of a permutation – Anne Bauval Jun 30 '23 at 12:50
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Given $\tau \in S_{10}$, the condition that $\tau$ should commute with $\sigma$ is equivalent to $\tau \sigma \tau^{-1} = \sigma$. But $\tau \sigma \tau^{-1} = (\tau(1) \ \tau(3) \ \tau(5) \ \tau(7) \ \tau(9))$. The condition can therefore be rewritten as $(\tau(1) \ \tau(3) \ \tau(5) \ \tau(7) \ \tau(9)) = (1 \ 3 \ 5 \ 7 \ 9)$.
So $\tau(2)$, $\tau(4)$, $\tau(6)$, $\tau(8)$, $\tau(10)$ can be freely chosen among $2$, $4$, $6$, $8$, $10$, yielding $5!$ possibilities.
On the other hand, once $\tau(1)$ has been chosen from among $1$, $3$, $5$, $7$, $9$, the remaining values of $\tau$ (i.e., $\tau(3)$, $\tau(5)$, $\tau(7)$, $\tau(9)$) are entirely determined.
Thus there are $5 \cdot 5! = 600$ possible permutations commuting with $\sigma$.
Dave
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The size of the conjugacy class of an element $g$ in a group $G$ is the index of the subgroup consisting of the elements commuting with $g$. In the symmetric group, it's easy to work out the size of the conjugacy class of $g$, since two elements are conjugate in the symmetric group if and only if they have the same cycle structure. So all you have to do for your problem is figure out how many 5-cycles there are in $S_{10}$, and then divide the order of $S_{10}$ by that number.
Gerry Myerson
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