# How to solve equation $\tau(123)(45)(67)\tau^{-1}=(765)(43)(21)$ and alike?

Let $\sigma\in S_7$ be $(123)(45)(67)$. And Find $\tau\in S_7$ such that $\tau\sigma\tau^{-1}=(765)(43)(21)$. I understand that in symmetric group conjugate elements have the same cycle structure. Hence, $\tau$ should share the structure of $\sigma$. Then I guess I could trial and error to find the solution for $\tau$. However, it is not very nice to do so. I am wondering whether there is a good way to solve such equation in general case. Thank you!

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I am assuming that you multiply permutations from right to left like function composition.

Let $\sigma,\tau\in S_{n}$, and Let $\sigma(i)=j$. Then $\tau\sigma\tau^{-1}\tau(i)=\tau(j)$. Therefore $\tau$ acts on the "display" of $\sigma$. So and say $\sigma=(a_{1}\cdots a_{r})$ then $\tau\sigma\tau^{-1}=(\tau(a_{1})\cdots\tau(a_{r}))$

For example, if $\sigma=(1 \ 2)(3 \ 4)$ and $\tau=(2 \ 4)$ then $\tau \sigma \tau^{-1}= (\tau(1) \ \tau(2))(\tau(3) \ \tau(4))=(1 \ 4)(3 \ 2)$.

In your case $\tau(1) =7$. I will leave the rest to you.

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This is very clear. Thank you! –  20824 Nov 17 '13 at 23:18
You're welcome! I made a slight edit changing "left to right" to "right to left"-- such things are important to keep straight. –  Eric Nov 17 '13 at 23:27

There are more solutions than just $\tau=(17)(26)(35)$. Your task was only to find one such $\tau$, so that's fine, but it is not too hard to find them all. And in the process, see that the cycle structure of $\tau$ is unrelated to the cycle structure of your permutations.

The cycles in your given permutations are all disjoint, and that is important. $\tau$ must map the three elements of the triple $(123)$ to $7$, $6$, and $5$, but not necessarily in that order, since $(765)=(657)=(576)$. $\tau(1)$ could be $5$, $6$, or $7$. Based on the choice made, $\tau(2)$ and $\tau(3)$ are determined.

The other two cycles are both 2-cycles, and they are disjoint so they permute. So $\tau$ could be anything that takes $4$ and $5$ to either $4$ and $3$ or $2$ and $1$, and further there is not need to respect the internal order of these pairings ($(43)=(34)$).

Considering the three-cycle, we have one of the following: \begin{align} \tau:&1\mapsto7&\tau:&1\mapsto6&\tau:&1\mapsto5\\ \tau:&2\mapsto6&\tau:&2\mapsto5&\tau:&2\mapsto7\\ \tau:&3\mapsto5&\tau:&3\mapsto7&\tau:&3\mapsto6\\ \end{align}

Considering the two-cylces, we have one of the following: \begin{align} \tau:&4\mapsto4&\tau:&4\mapsto4&\tau:&4\mapsto3&\tau:&4\mapsto3\\ \tau:&5\mapsto3&\tau:&5\mapsto3&\tau:&5\mapsto4&\tau:&5\mapsto4\\ \tau:&6\mapsto2&\tau:&6\mapsto1&\tau:&6\mapsto2&\tau:&6\mapsto1\\ \tau:&7\mapsto1&\tau:&7\mapsto2&\tau:&7\mapsto1&\tau:&7\mapsto2\\ \end{align}

This gives $12$ possible maps for $\tau$. Calculating each one's cycle structure, the possibilities for $\tau$ are: $$(17)(26)(35), (1726)(35), (17)(26)(354), (1726)(354), (162537), (16)(2537), (1625437), (16)(25437), (153627), (1536)(27), (1543627), (15436)(27)$$

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Conjugation permutes elements "in place" in their cycle structure. Hence you want $\tau$ to map 1 to 7, 2 to 6, 3 to 5, 4 to 4, 5 to 3, 6 to 2, and 7 to 1. That is, $$\tau=(17)(26)(35)$$

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I am afraid that your answer is not correct. As I mentioned above, $\tau$ should have the same cycle structure at least. Right? –  20824 Nov 17 '13 at 23:07
It is $(123)(45)(67)$ and $(765)(43)(21)$ that are conjugates, and that have the same cycle structure. –  vadim123 Nov 17 '13 at 23:08
Oh, that is right. Thank you! –  20824 Nov 17 '13 at 23:17