Let $S_{10}$ denote the permutation of $10$ symbols.Then the number of elements of $S_{10}$ that commutute to $(1 3 5 7 9 )$

Question

Let $S_{10}$ denote the permutation of $10$ symbols.Then the number of elements of $S_{10}$ that commutute to $(1 3 5 7 9 )$ is

a)$5!$

b)$5\cdot 5!$

c)$5! \cdot 5!$

d)$\frac{10!}{5!}$

Answer 1

HINT: $\tau(13579)\tau^{-1} = \left(\tau(1),\tau(3),\tau(5),\tau(7),\tau(9)\right)$. Now when is $(\tau(1),\tau(3),\tau(5),\tau(7),\tau(9)) = (13579)$?

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A more advanced method is to use the fact that $|a^{G}| = [G:C_g(a)]$, i.e. the number of conjugates of $a$ in a group $G$ is equal to the index of the centralizer of $a$ in $G$. So you're interested in $C_{S_{10}}(13579)$, which is equal to:

$$C_{S_{10}}(13579) = \frac{|S_{10}|}{|(13579)^{S_{10}}|}$$

Now using the fact all conjugates in a symmetric group have the same cycle we should be able to find $|(13579)^{S_{10}}|$, which is equal to $\frac{10!}{5\cdot 5!}$ and compute the wanted result.

Answer 2

As a simple approach, any permutation $\sigma$ which commutes with $\rho=(13579)$ can operate on the unaffected elements $\{2468A\}$ in any way ($5!$ options). It can certainly also include any power of $\rho$, ($5$ options, including $\rho^5=1$). Permutations which cross between the affected and unaffected elements of $\rho$ or disorder the cycle will not commute. This implies that the total available commuting permutations is just $5\cdot 5!$