Group theory, conjugation of permutations

Question

I have a past exam question that says...

Decompose the following permutations into a product of disjoint cycles. Are the two permutations conjugate?

$$\alpha= \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\ 7 & 4 & 5 & 3 & 8 & 6 & 9 & 1 & 2\\ \end{bmatrix}$$

and

$$\beta= \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\ 6 & 7 & 8 & 9 & 1 & 2 & 3 & 4 & 5\\ \end{bmatrix}$$

I think that you have to find $x^{-1}\alpha x=\beta$, however I do not know how to find this $x$. Any help with this would be much appreciated. Thank you!

Answer 1

Easily we have

$$\alpha=(17924358)$$ and $$\beta=(162738495)$$

Now assume that we can find $x$ such that $$x^{-1}\alpha x=\beta$$ and since $o(\beta)=9 $ and $o(\alpha)=8$ then $$\beta^9={\rm{id}}=x^{-1}\alpha x\Rightarrow \alpha=\rm{id}$$ which is a contradiction.

Answer 2

Write down specifically the cycle decomposition of each permutation:

$$\begin{cases}\alpha=(17924358)\\{}\\\beta=(162738495)\end{cases}$$

Thus we have an $\;8-$ cycle and a $\;9-$ cycle...so are they conjugate?

Answer 3

Decomposition in cycle is fairly easy, and it yields

$$\alpha = (1\ 7\ 9\ 2\ 4\ 3\ 5\ 8)$$

$$\beta = (1\ 6\ 2\ 7\ 3\ 8\ 4\ 9\ 5)$$

Notice $6$ does not appear in the cycle decomposition of $\alpha$ (it would be a cycle of length 1, usually they are not written).

Two permutations of $S_n$ are conjugate iff they have the same cycle structure (see here): basically you only rename elements. Hence $\alpha$ and $\beta$ cannot be conjugate.

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Where $\alpha$ and $\beta$ of same "cycle type", it would be easy to find a permutation to transform one into the other.

Say

$$\alpha=(1\ 7\ 4)(3\ 2\ 6\ 5)(8\ 9)$$ $$\beta=(3\ 4\ 7\ 6)(5\ 2\ 9)(1\ 8)$$

Then send each cycle to a cycle of same length: $1 \rightarrow 5$, $7 \rightarrow 2$...

So this one would do (it's not unique, just try to rotate any cycle) $$x= \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9\\ 5 & 4 & 3 & 9 & 6 & 7 & 2 & 1 & 8\\ \end{bmatrix}$$

Or $x=(1\ 5\ 6\ 7\ 2\ 4\ 9\ 8)$

Then $x\alpha x^{-1}=\beta$ if you use the convention $(xy)(i) = x(y(i))$, or $x^{-1}\alpha x=\beta$ with the other convention $(xy)(i)=y(x(i))$. In France it's usually the former, and in english-speaking countries, the latter.

For example, $x^{-1}({\color{red}{5}}) = 1, \alpha(1)=7, x(7)={\color{red}{2}}$, and $\beta({\color{red}{5}})={\color{red}{2}}$.

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To answer mayi's comment below, let's try with $\alpha=(1\ 3)(4\ 7\ 6)$ and $\beta=(1\ 5)(2\ 6\ 4)$.

They have the same cycle structure, hence they are conjugate. The trick is you have to work in $S_7$, the group of permutations of $\{1,2,3,4,5,6,7\}$, so that $\alpha$ and $\beta$ share the same range. You may then use the permutation

$$\pi= \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & - & 5 & 2 & - & 4 & 6\\ \end{bmatrix}$$

The images of $2$ and $5$ can be chosen freely, since they do not appear in $\alpha$, and the only remaining elements are $3$ and $7$. So, either $2\to3$ and $5\to7$, either $2\to7$ and $3\to5$. For instance:

$$\pi= \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 1 & 3 & 5 & 2 & 7 & 4 & 6\\ \end{bmatrix}$$

Or in cycle notation, $\pi=(2\ 3\ 5\ 7\ 6\ 4)$. Then, with the "french" convention of permutation composition, $\beta=\pi\alpha\pi^{-1}$, and with the "english" convention, $\beta=\pi^{-1}\alpha\pi$.

But $\pi$ is of course not the only possible permutation that transforms $\alpha$ into $\beta$. If you write $\beta=(1\ 5)(4\ 2\ 6)$, it's easy to see that $\pi=(2\ 7)(3\ 5)$ is also valid. To find your $x=(1\ 5\ 3)(2\ 7)$, write instead $\beta=(5\ 1)(4\ 2\ 6)$ (still with $\alpha=(1\ 3)(4\ 7\ 6)$), and

$$\pi= \begin{bmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 5 & - & 1 & 4 & - & 6 & 2\\ \end{bmatrix}$$

Again, the images of $2$ and $5$ can be chosen freely. With $2\to7$ and $5\to3$, you get $\pi=(1\ 5\ 3)(2\ 7)$, your $x$. With the other choice, you would get $\pi=(1\ 5\ 7\ 2\ 3)$.