# Finding $\tau\in S_9$ for $\tau(1,2)(3,4)\tau^{-1}=(5,6)(1,3)$

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).

• 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. – DonAntonio Jun 8 at 10:15

## 3 Answers

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

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?)

• You are not answering the OP's question. He wants to find ALL solutions. – GreginGre Jun 8 at 10:35
• @GreginGre It is, all permutations that obey the 4 conditions work. – Henno Brandsma Jun 8 at 10:36
• 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. – abuka123 Jun 8 at 10:38
• $(5 1 4 3)(2 6)$ is not a solution, as $3$ is not mapped to $1$. – Henno Brandsma Jun 8 at 10:42
• @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. – DonAntonio Jun 8 at 12:12

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..