Centralizer of an element in $S_{n}$

Question

I am trying to prove that the order of $C_{S_{n}}((12)(34))$ in $S_{n}$, $n\geq 4$ is $8(n-4)!$ and also find the elements of $C_{S_{n}}((12)(34))$. I tried to calculate the order by using the orbit-stabiliser lemma, so if $g=(12)(34)$ then $|C_{S_{n}}(g)|=\dfrac{n!}{|orb(g)|} $. Then since the cycle type of $(12)(23)$ is $[1^{n-4} \hspace{2mm} 2^{2}\hspace{2mm} 3^{0}\hspace{2mm} \dots \hspace{2mm}n^{0}]$ we have that all conjugate elements of $g$ are are of this form: $(a b)(c d)$. Now we have $n$ options for $a$, $n-1$ for $b$, $n-2$ for $c$ and $n-3$ for $d$. This should give us $|C_{S_{n}}(g)|=(n-4)!$. So where this $\frac{1}{8}$ comes from? Is this has something to do with the fact that $(12)(34)$ are disjoint circles and therefore they commute? Also how is this procedure going to give me the elements of $C_{S_{n}}(g)$? Should I try something else? We know that $$C_{S_{n}}((12)(34))=\{\sigma \in S_{n} : \sigma(12)(34)\sigma^{-1}=(12)(34)\}=\{\sigma \in S_{n} : (\sigma(1)\sigma(2))(\sigma(3)\sigma(4))=(12)(34)\}$$ but I'm not sure how to procede from here. Thank you in advance.

Answer 1

Let $\sigma=(12)(34)$; if $\tau$ commutes with $\sigma$, then $\tau$ must respect the partition of $\{1,2,\ldots,n\}$ into $\{1,2,3,4\}$ and $\{5,\ldots,n\}$. For otherwise, some element of $\{1,2,3,4\}$ is mapped to some $k\geq 5$, and then $\tau\sigma\tau^{-1}$ cannot equal $\sigma$ (as it will equal $(\tau(1),\tau(2))(\tau(3),\tau(4))$.

So in fact, the centralizer must be a subgroup of $S_4\times S_{n-4}$; the $S_{n-4}$ component can be anything, for it will certainly commute with $\sigma$; this is the source of the $(n-4)!$ in the order of the centralizer.

For the $S_4$ component, we are reduced to figuring out the order of the centralizer of $(12)(34)$ in $S_4$. The elements of the $4$-Klein group inside $A_4$ centralize $\sigma$: $e$, $(12)(34)$, $(13)(24)$, and $(14)(23)$. Call this subgroup $V$. So do $(12)$ and $(3,4)$, though you can realize this as the result of taking $V$ and $(1,2)$ in an internal semidirect product.

Is there any other element of $S_4$ that centralizes $(12)(34)$? Say $\tau$ does; then $(\tau(1),\tau(2))(\tau(3),\tau(4)) = (12)(34)$. Either $(\tau(1),\tau(2)) = (1,2)$, in which case $(\tau(3),\tau(4))=(3,4)$; or else $(\tau(1),\tau(2))=(3,4)$, in which case $(\tau(3),\tau(4))=(1,2)$.

In the former case, we can have any of: $e$, $(12)$, $(34)$, and $(12)(34)$; in the latter case we can have any of $(13)(24)$, $(14)(23)$; or $(1324)$, and $(1423)$. It is now not difficult to see that these elements, the only ones that can centralize $\sigma$, form a subgroup of order $8$ of $S_4$. This gives you the factor of $8$ in the computation.

Answer 2

For the first part, there are $8$ ways to write the same product of disjoint transpositions. For instance, $(12)(34),(21)(34),(12)(43),(21)(43),(34)(12),(34)(21),(43)(12),(43)(21)$ are all the same.

That's there are $2$ ways to write each transposition, and $2$ orders to put them in, to give $2^3=8$.

For part two, by the equation you wrote out, the only $\sigma$'s that work are those that take $\{1,2\}\to \{1,2\}$ and $\{3,4\}\to\{3,4\}$ or ones that map $\{3,4\}\to\{1,2\}$ and $\{1,2\}\to\{3,4\}$. That's $8$. Then as $n$ gets larger we can multiply cycles, or products of cycles, involving numbers from $5$ through $n$.