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The problem is this: prove that the two permutations, $x=(1\ 2\ 5)(6\ 4)$ and $y=(2\ 3\ 4)(1\ 5) $ are conjugates in the group $S_6$.

I was trying to find a element $z$ such that $zxz^{-1} =y$, but am unsure how to start. How are you supposed to find $z$? Is there some kind of formula or systematic method?

A followup question was to find how many $ z$ exist such that $zxz^{-1} =y$.

Shaun
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kelvin
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  • In general in a group, if there is such a $z$, then the number of such $z$ is equal to the size of the centraliser of $x$ - ie the stabiliser of $x$ in the conjugation action of $G$ on itself. You can use the orbit-stabiliser theorem to find this number in terms of the size of the orbit of $x$ - in this case this is the number of permutations of cycle type $(\bullet \bullet \bullet)(\bullet \bullet{})$, which can be found by a simple combinatorial argument. There is also a direct formula for the size of the centraliser. – Izaak van Dongen Jul 06 '22 at 12:39

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Hint: For all $\sigma,\tau\in S_n$ for any $n$, with

$$\tau=(t_{i_1}t_{i_1+1}\cdots t_{i_1+k_1})\cdots (t_{i_j}t_{i_j+1}\cdots t_{i_j+k_j})$$

we have

$$\sigma\tau\sigma^{-1}=(\sigma(t_{i_1})\sigma(t_{i_1+1})\cdots\sigma (t_{i_1+k_1}))\cdots (\sigma(t_{i_j})\sigma(t_{i_j+1})\cdots\sigma( t_{i_j+k_j})),$$

where $\sigma(x)$ is $\sigma$ evaluate at $x$.

For example,

$$\begin{align} ((123)(45))\circ (\color{red}{12})\circ((123)(45))^{-1}&=(((123)(45))(\color{red}1)((123)(45))(\color{red}2))\\ &=(23), \end{align}$$

which checks out, since

$$\begin{align} 1&\xrightarrow{((123)(45))^{-1}}3\xrightarrow{(12)}3\xrightarrow{(123)(45)}1\\ 2&\xrightarrow{((123)(45))^{-1}}1\xrightarrow{(12)}2\xrightarrow{(123)(45)}3\\ 3&\xrightarrow{((123)(45))^{-1}}2\xrightarrow{(12)}1\xrightarrow{(123)(45)}2\\ 4&\xrightarrow{((123)(45))^{-1}}5\xrightarrow{(12)}5\xrightarrow{(123)(45)}4\\ 5&\xrightarrow{((123)(45))^{-1}}4\xrightarrow{(12)}4\xrightarrow{(123)(45)}5 \end{align}$$

Shaun
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    Wow thanks, it really does work! I am assuming because we have a 3 cycle and a 2 cycle, there are 6 values that z can take? Also, is there a name for this theorem? I am interested in the proof. – kelvin Jul 05 '22 at 13:41
  • You're welcome, @kelvin; I have included an example. – Shaun Jul 05 '22 at 13:43
  • It is a standard theorem for permutations, @kelvin. I don't think I has a name. If you search for "conjugation of permutations", though, you're likely to find a proof. – Shaun Jul 05 '22 at 13:46
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    This is interesting stuff, thanks once again @Shaun – kelvin Jul 05 '22 at 13:50
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    Because of this, two permutations are conjugate if and only if they have the same cycle structure. – suckling pig Jul 05 '22 at 15:00