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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?

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All the permutations...in 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
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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

1 Answer 1

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, $\left|C(\omega)\right| = \left|S_{12}\right|/\left|O\right| = 2\cdot 3\cdot 7$.

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