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I'm trying to understand where I'm wrong in my solution. I would like to find all $\tau \in S_9$ so $$\tau(1,2)(3,4)\tau^{-1}=(5,6)(1,3)$$

Meaning - $(\tau(1),\tau(2))(\tau(3),\tau(4))=(5,6)(1,3)$. Lets separated it:

  • If $(\tau(1),\tau(2))=(5,6)$ and also $(\tau(3),\tau(4))=(1,3)$:

    For $(\tau(1),\tau(2))=(5,6)$ we get two options:

    • if $\tau(1)=5,\tau(2)=6$ then $(1,5)(2,6)$.
    • if $\tau(1)=6,\tau(2)=5$ then $(1,6)(2,5)$.

    For $(\tau(3),\tau(4))=(1,3)$ we get two options:

    • if $\tau(3)=1,\tau(4)=3$ then $(1,4,3)$.
    • if $\tau(3)=3,\tau(4)=1$ then $(1,4)$.

    So we get:

$$ (1,5)(2,6)(1,4,3)=(5,1,4,3)(2,6)\\ (1,5)(2,6)(1,4)=(4,1,5)(2,6)\\ (1,6)(2,5)(1,4,3)=(6,1,4,3)(2,5)\\ (1,6)(2,5)(1,4)=(4,1,6)(2,5)$$

  • If $(\tau(1),\tau(2))=(1,3)$ and also $(\tau(3),\tau(4))=(5,6)$:

    For $(\tau(1),\tau(2))=(1,3)$ we get two options:

    • if $\tau(1)=1,\tau(2)=3$ then $(2,3)$.
    • if $\tau(1)=3,\tau(2)=1$ then $(2,1,3)$.

    For $(\tau(3),\tau(4))=(5,6)$ we get two options:

    • if $\tau(3)=5,\tau(4)=6$ then $(3,5)(4,6)$.
    • if $\tau(3)=6,\tau(4)=5=1$ then $(3,6)(4,5)$.

    So we get:

$$(2,3)(3,5)(4,6)=(2,3,5)(4,6)\\ (2,3)(3,6)(4,5)=(2,3,6)(4,5)\\ (2,1,3)(3,5)(4,6)=(2,1,3,5)(4,6)\\ (2,1,3)(3,6)(4,5)=(2,1,3,6)(4,5)$$

But the final solutions are different. For example, for $\tau(1)=5,\tau(2)=6,\tau(3)=1,\tau(4)=3$ they got $\tau=(1,5,4,3)(2,6)$ or $\tau=(1,5,2,6,4,3)$. Why is that?

I also tried to use a similar thread (link).

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  • $\begingroup$ I think your mistake is that you only look at $\;1\to 5\;$ , for example...but then what $\;5\;$ does?! Look at the answer below. $\endgroup$ – DonAntonio Jun 8 at 10:15
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For the case $(\tau(3),\tau(4)) = (1,3)$, the result should depend on the choice of $(\tau(1),\tau(2)) = (5,6)$ in your previous analysis, and depend on other elements in $\{1,\dots,9\}$. You cannot directly conclude "if $τ(3)=1,τ(4)=3$ then $(1,4,3)$".

You should discuss this way:

We have four cases:

  1. $\tau(1) = 5,\tau(3) = 1$. Then $\tau(2) = 6$, $\tau(4) = 3$ and hence $2\mapsto 6,4\mapsto 3\mapsto 1\mapsto 5$. Write \begin{equation*} \tau = \begin{pmatrix} 1&2&3&4&5&6&7&8&9\\ 5&6&1&3&a&b&c&d&e \end{pmatrix}, \end{equation*} where $a,b,c,d,e\in\{2,4,7,8,9\}$ are distinct. Any choice of $a,b,c,d,e$ satisfies your condition.

  2. $\tau(1) = 6,\tau(3) = 1$.

  3. $\tau(1) = 5,\tau(3) = 3$.

  4. $\tau(1) = 6,\tau(3) = 3$.

Try to do it yourself for cases 2,3,4, and for $(\tau(1),\tau(2)) = (1,3)$.

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You want $\;\tau\;$ such that

$$\tau:1\to 5,\,\tau:2\to 6,\,\tau:3\to1,\,\tau:4\to 3$$

Thus take for example $\;\tau=\;(1543)(26)$, and then

$$\tau(12)(34)\tau^{-1}=(1543)(26)(12)(34)(1345)(26)=(13)(56)$$

This solution isn't unique (can you see why?)

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    $\begingroup$ You are not answering the OP's question. He wants to find ALL solutions. $\endgroup$ – GreginGre Jun 8 at 10:35
  • $\begingroup$ @GreginGre It is, all permutations that obey the 4 conditions work. $\endgroup$ – Henno Brandsma Jun 8 at 10:36
  • $\begingroup$ But how did you understand that $\tau=(1,5,4,3)(2,6)$. It is different from the solution I got, which is $\tau=(5,1,4,3)(2,6)$. The question is why it's different. I just took $(1,5)(2,6)$ and $(1,4,3)$ combined them and got the answer $(5,1,4,3)(2,6)$ and not yours. $\endgroup$ – abuka123 Jun 8 at 10:38
  • $\begingroup$ $(5 1 4 3)(2 6)$ is not a solution, as $3$ is not mapped to $1$. $\endgroup$ – Henno Brandsma Jun 8 at 10:42
  • $\begingroup$ @GreginGre Where did you deduce that from? He wanted to know what's wrong with his work, I already explained in the comments below his question and added the answer for him to understand better. $\endgroup$ – DonAntonio Jun 8 at 12:12
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So you want $\tau(1 2)(3 4)\tau^{-1}=(5 6)(1 3)$ to hold, or multiplying both sides on the right by $\tau$:

$$\tau(1 2)(3 4) = (5 6)(1 3)\tau\tag{1}$$

What can $\tau(1)$ be? it's the image of $2$ under the LHS of $(1)$ (as $2 \rightarrow 1$ first and then $\tau$ gets applied), so also the image of $2$ of the RHS, so the image of $\tau(2)$ under $(5 6)(1 3)$. If $\tau(2) \notin \{5,6,1,3\}$ we would get $\tau(2)$ on the RHS, which cannot be as $\tau(1) \neq \tau(2)$ for a permutation. So $\tau(2) \in \{5,6,1,3\}$. A similar reasoning gives us that $\tau(1) \in \{5,6,1,3\}$ as well, and reasoning on:

And if $\tau(2)=5$ on the RHS we get $2 \to \tau(2)=5 \to 6$ and on the LHS we get $2 \to 1 \to \tau(1)$ so $\tau(1)=6$.

Similarly,

  • $\tau(2)=6 \implies \tau(1)=5$.
  • $\tau(2)=1 \implies \tau(1)=3$.
  • $\tau(2)=3 \implies \tau(1)=1$.

This gives us some first constraints on permutations $\tau$ that satisfy equation $(1)$.

Also, $\tau(3)$ is the LHS applied to $4$, so if $\tau(4) \notin \{5,6,1,3\}$ we get $\tau(3)=\tau(4)$ again etc. So also $\tau(4), \tau(3) \in \{5,6,1,3\}$. Now try to reason what the relations here are..

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