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I have a question regarding permutations.

If $\gamma = (123) (45) (6)$ and $\gamma ' = (1)(23) (456)$ in $S_{6}$ how do I then determine the numbers of permutations $\sigma$ in $S_{6}$ so $\gamma ' = \sigma \gamma \sigma ^{-1}$?

How do I do this in general? and when I have determine the number, how can I then find them? (I suppose bruteforce is a way).

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Is this a homework question? – Donkey_2009 Apr 2 '12 at 15:11
I'm trying to solve old exam problems (4 years old) in order to prepare for an exam. – bemyguest Apr 2 '12 at 17:15
up vote 1 down vote accepted

If you use the rule that $\sigma\gamma\sigma^{-1}=\left (\sigma(1)\;\sigma(2)\;\sigma(3)\right )\;\left (\sigma(4)\;\sigma(5)\right )\;\left (\sigma(6)\right )$ then the question becomes easy.

Of course, you'll have to prove this rule, but I'll leave that to you.

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To find $\sigma$: $\gamma = (123) (45) (6)$ and $\gamma ' = (456)(23) (1)$, so you can take $\sigma=(123456\to456231)$ (or its inverse, depending on the notation). Such a $\sigma$ exists iff $\gamma$ and $\gamma'$ have the same cycle structure (they do in this case). The number of $\sigma$'s is equal to the number of $\pi$'s s.t. $\pi\gamma\pi^{-1}=\gamma$, i.e. to the number of permutations preserving the cycle structure. In your case it is $3!\times 2!\times1!=12$. In general, if you have $k_i$ $i$-cycles in your permutation, the number would be $\prod_i k_i!(i!)^{k_i}$.

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How do you know that the numbers of $\sigma $ 's is equal to the number of $\pi$'s ? – bemyguest Apr 2 '12 at 17:05

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