TITLE: Conjugate permutations in $S_n$ and / or $A_n$
QUESTION [20 upvotes]: Given two permutations , I'm asked to answer is they are conjugate permutations .
The two permutations are : $ \alpha=(12)(345)(78)$, $\beta=(162)(35)(89)$.
Definition: Two permutations  $ \sigma,\sigma'\in S_n$ are conjugate if exists $\tau \in S_n $ such that:
$\sigma'=\tau\sigma\tau^{-1} = (\tau(a_0),\tau(a_1)\ldots \tau(a_k)) $, where $ \alpha=(a_0a_1\ldots a_k)$ . 
It took me a long time to find the correct $\tau$ that would compute the exact $\beta$ , which is: 
$$\tau=(13)(25)(46)(789)$$
So if we want to produce are $\beta$ we can do the following : 
$$(\tau(1),\tau(2))(\tau(3)\tau(4)\tau(5))(\tau(7)\tau(8))=(35)(162)(89)$$
and indeed , they ($α$ and $β$) are conjugate. 
My question, after this "long" post, is rather simple: is there a simple way to compute the $\tau$ ?
Regards

REPLY [13 votes]: Yes.
Note, however, that in general there are many $\tau$ that work. You found one; here's another:
$$\tau= (1,8,5,2,9,4,6,7,3).$$
Indeed,
$$\tau\alpha\tau^{-1} = (\tau(1)\tau(2))(\tau(7)\tau(8))(\tau(3)\tau(4)\tau(5)) = (8,9)(3,5)(1,6,2)=\beta.$$
Theorem. Two permutations $\alpha$ and $\beta$ are conjugate in $S_n$ if and only if they have the same cycle structure.
The proof provides an algorithm for finding a $\tau$ that works.
If $\alpha$ and $\beta$ are conjugate, then they have the same cycle structure. This, because the conjugate of an $n$-cycle is an $n$-cycle, and the conjugate of a product is a product of conjugates. 
Conversely, suppose that $\alpha$ and $\beta$ have the same cycle structure. Say 
$$\alpha = \sigma_1\cdots\sigma_m,\quad \beta=\sigma'_1\cdots \sigma'_m$$
where $\sigma_i$ and $\sigma'_i$ are $n_i$-cycles. Write $\alpha$ and $\beta$ one above the other, so that cycles of the same length coincide; interpret this as a $2$-line description of a permutation, completing it to an element of $\sigma_n$ any way you want.
For example, with the two permutation you have above, $\alpha=(1,2)(7,8)(3,4,5)$, $\beta=(3,5)(8,9)(1,6,2)$, we have:
$$\left(\begin{array}{ccccccccc}
1 & 2 & 7 & 8 & 3 & 4 & 5 & & &\\
3 & 5 & 8 & 9 & 1 & 6 & 2 & & &
\end{array}\right)$$
Note that the top line is missing $6$ and $9$, and the bottom line is missing $4$ and $7$; we can add them any way we want; for example,
$$\left(\begin{array}{cccccccc}
1 & 2 & 7 & 8 & 3 & 4 & 5 &6 & 9\\
3 & 5 & 8 & 9 & 1 & 6 & 2 & 7 & 4 
\end{array}\right)$$
Now viewing this as a 2-line decription of a permutation, we get
$$\tau = (1,3)(2,5)(7,8,9,4,6)$$
as one possibility. If you order the cycles differently (so long as they match), or reorder the terms within the cycle, you may get a different $\tau$. For instance, writing instead
$$\begin{align*}
\alpha &= (12)(78)(453)\\
\beta &= (89)(53)(162)
\end{align*}$$
we get
$$\left(\begin{array}{ccccccccc}
1 & 2 & 7 & 8 & 4 & 5 & 3 & &\\
8 & 9 & 5 & 3 & 1 & 6 & 2 & &
\end{array}\right)$$
with $6$ and $9$ in the top line going to $4$ and $7$ in some way; randomly choosing one,
$$\left(\begin{array}{ccccccccc}
1 & 2 & 7 & 8 & 4 & 5 & 3 & 6 & 9 \\
8 & 9 & 5 & 3 & 1 & 6 & 2 & 4 & 7
\end{array}\right)$$
we get $\tau = (1,8,3,2,9,7,5,6,4)$.
To find out if the permutations are conjugate in $A_n$, however, is a bit trickier: it is necessary that they have the same cycle structure, but in general it is not sufficient. For instance, in $S_3$ the permutations $(123)$ and $(132)$ are conjugate (for example, conjugate via $(23)$); but in $A_3$ they are not (since $A_3$ is abelian, different elements cannot be conjugate).

REPLY [12 votes]: You want to make $\alpha=(12)(345)(78)$ ‘look like’ $\beta=(162)(35)(89)$. First write them with matching cycle structures; there are several ways to do this, and it doesn’t matter which you choose, so I’ll rewrite $\beta$ as $(35)(162)(89)$. Now you want $\tau$ to ‘translate’ between the two permutations. The way I’ve lined them up, $$\begin{align*}&(12)(345)(78)\\&(35)(162)(89)\;,\end{align*}\tag{1}$$ you can see that $\tau$ needs to translate $1$ to $3$ and $2$ to $5$, so that $\alpha$’s $(12)$ turns into $\beta$’s $(35)$, and so on. As Jack Schmidt suggested, this amounts to writing $$\tau=\pmatrix{1&2&3&4&5&7&8\\3&5&1&6&2&8&9}\tag{2}$$ in two-line notation. However, this is clearly incomplete: it’s missing $6$ and $9$ on the top line and $4$ and $7$ on the bottom line. Thus, $(1)$ needs to be completed to $$\begin{align*}&(12)(345)(78)(6)(9)\\&(35)(162)(89)(4)(7)\;,\end{align*}$$ thereby allowing $(2)$ to be completed to
$$\tau=\pmatrix{1&2&3&4&5&7&8&6&9\\3&5&1&6&2&8&9&4&7}\;.$$
In cycle notation this is $\tau=(13)(25)(46)(789)$, the one that you actually found.
This works because $\tau^{-1}$ in effect just renames the integers $1$ through $9$, so that $3$ is temporarily called $1$, $5$ is temporarily called $2$, and so on. Thus, when you apply $\tau^{-1}$, you’re translating from $\beta$’s language to $\alpha$’s; then you perform $\alpha$ and apply $\tau$, thereby translating back from $\alpha$’s language to $\beta$’s. The net result is that you’ve performed $\beta$. Consider $6$, for instance: $\tau^{-1}$ translates it to $4$, which $\alpha$ then carries to $5$, which $\tau$ translates back to $2$, exactly where $\beta$ would have put it in the first place.