Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm asked in this exercise to find all permutations that commute with $\omega$=(1 9 7 10 12 2 5)(4 11)(3 6 8) in $S_{12}$.

What I have so far: We could write $x$(1 9 7 10 12 2 5)(3 6 8)=(1 9 7 10 12 2 5)(3 6 8)$x$, which means (1 9 7 10 12 2 5)(4 11)(3 6 8)=$x^{-1}$(1 9 7 10 12 2 5)(4 11)(3 6 8)$x$=(1x 9x 7x 10x 12x 2x 5x)(4x 11x)(3x 6x 8x). This means $x$ commutes with $\omega$ iff $x$ 'transfers' one of the distinct cycles that construct $\omega$ unto itself.

There are $7\cdot 2\cdot 3$ ways to 'present' $\omega$ disregarding the order of multiplication of the cycle (should I disregard it?), each of which creates a distinct commuting permutation if constructed by the algorithm: 1x$\rightarrow$(1 9 7 10 12 2 5), 9x$\rightarrow$(1 9 7 10 12 2 5), et cetera. So all in all we end up with $7\cdot 2\cdot 3$ permutations.

I guess what I should ask is: (a) does this sound correct? (b) since the question wants us to find all permutations and not count them, perhaps there is a more general 'form' for the commuting permutations?

share|cite|improve this question
All the what group $\,S_n\,$?? – DonAntonio Nov 27 '12 at 22:32
$S_{12}$ - sorry! – ro44 Nov 27 '12 at 22:46
It does sound correct. And to find all permutations, I'd say you take each cycle $(a_1,...,a_k)$, map it to $(1,...,k)$ with permutation from $S_{12}$ we'll call $f$. Then you take a permutation $g$ of $S_k$ and extend it to $S_{12}$ by defining at identity on the remaining numbers and map it back top $(a_1,...,a_k)$. And your permutations are $f^{-1}\circ g\circ f$. Then you just compose one of those permutation per cycle and you get the general form. – xavierm02 Nov 27 '12 at 22:56
And I therefore think that the number of permutations is the product of the sizes of $S_7$, $S_2$ and $S_3$ which isn't $7*2*3$ from what I recall. – xavierm02 Nov 27 '12 at 23:00
The size of $S_n$ would be $n!$, by my count. But why would every permutation in say, $S_7$ be eligible? – ro44 Nov 27 '12 at 23:10

Your reasoning is fine, in my opinion. $\sigma\in S_{12}$ commutes with $\omega$ if and only if $\sigma\omega\sigma^{-1} = \omega$. Since there are $7\cdot 2\cdot 3$ ways to fix a representation of $\omega$ in the form $$ \sigma\omega\sigma^{-1} = (\sigma(1)\ \sigma(9)\ \sigma(7)\ \sigma(10)\ \sigma(12)\ \sigma(2)\ \sigma(5))\ (\sigma(4)\ \sigma(11))\ (\sigma(3) \ \sigma(6)\ \sigma(8)), $$ and each such representation uniquely determines $\sigma$, there are exactly $7\cdot 2\cdot 3$ elements $\sigma\in S_{12}$ which commute with $\omega$.

Here is an alternative solution using the theory of group actions to determine the number of elements of $S_{12}$ which commute with $\omega$:

The question asks for the centralizer $C(\omega)$ of $\omega$ in $S_{12}$. Now look at the group operation of $G$ on itself by conjugation. Then $C(\omega)$ is the stabilizer of $\omega$. The orbit $O$ of $\omega$ is the conjugacy class of $\omega$ in $S_{12}$, which is the set of all elements of the same cycle type. The standard counting method for permutations yields $$\left|O\right| = \frac{12!}{2\cdot 3\cdot 7}.$$

Now by the orbit-stabilizer-theorem, $\left|C(\omega)\right| = \left|S_{12}\right|/\left|O\right| = 2\cdot 3\cdot 7$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.