Number of elements of $S_{10}$ commuting with element (1 3 5 7 9)

Question

Find Number of elements of $S_{10}$ commuting with element (1 3 5 7 9)

I think we need to find order of centralizer of given permutation but how to find it?

score 3 · Answer 1 · answered Jan 21 '15 at 16:54

3

I expect that any interaction with the odd numbers must involve a power of the given element, so there are 5 options. Then the even numbers have complete freedom to interact with each other, $S_5$, another 120 options, giving 600 in all.

answered Jan 21 '15 at 16:54

Empy2

52,372

score 1 · Answer 2 · answered Jan 21 '15 at 16:59

If $\sigma(1\,3\,5\,7\,9)=(1\,3\,5\,7\,9)\sigma$, then: $$\sigma(1\,3\,5\,7\,9)\sigma^{-1} = (1\,3\,5\,7\,9)$$ where the LHS is a conjugate of the initial $5$-cycle. So $\sigma$ is free to act as it likes on the set $\{2,4,6,8,10\}$, but once it sends $1$ into an element of $\{1,3,5,7,9\}$, the images of $3,5,7,9$ are fixed. So there are $5\cdot 5!$ permutations in $S_{10}$ that commutes with $(1\,3\,5\,7\,9)$.

Number of elements of $S_{10}$ commuting with element (1 3 5 7 9)

2 Answers2