Does every permutation, which is not an $n$-cycle, commute with some transposition?

Question

This question addresses how many permutations $σ∈S_n$ commute with a given transposition $(i \space j)$. What about the other way around, namely: How many transpositions $(i \space j)$ commute with a given permutation $σ∈S_n$? I know that if $\sigma$ in as $n$-cycle, then this number is zero (no one transpostition commutes with an $n$-cycle). I'm not particularly interested in the exact number, but rather

if it is nonzero for every permutation which is not an $n$-cycle.

Answer 1

$(123)(456)$ does not commute with any transpositions in $S_6$. It suffices to notice that:

$$ (123)(456)(12)=(13)(456)\\ (12)(123)(456)=(23)(456) $$

$$ (123)(456)(14)=(156423)\\ (14)(123)(456)=(123456) $$

and any transposition in $S_6$ will either have two elements in common with one of the two 3-cycles or one element in common with each, and so will behave analogously to one of the two above.

In general, a transposition $(ij)$ will commute with a permutation $\sigma$ if and only if one of the following two things is true:

• $i$ and $j$ are fixed points of $\sigma$, or
• the transposition $(ij)$ appears in the cycle structure of $\sigma$.

Answer 2

With $\theta \in S_n$ then $(i\space j)$ commutes with $\theta$ if and only if $\theta'= (i\space j)\theta (i\space j)=\theta$. We can get $\theta'$ "sweeping" $i$ and $j$ from $\theta$.

For example consider $\theta=\left(\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \\ \end{array}\right)$

Then "sweeping" 1 and 2 we get $\theta'=(1\space 2)\theta(1\space 2)= \left(\begin{array}{cccc} 2 & 1 & 3 & 4 \\ 3 & 4 & 1 & 2 \\ \end{array}\right)$

So $\theta \in S_n$ commutes with $(i\space j)$ if and only if $\theta(i)=i$ and $\theta(j)=j$