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Are there multiple ways to translate a permutation $\alpha$ to a permutation $\beta$?

I came across the question below and found multiple solutions for $\gamma$. My approach to finding solutions is not very rigorous, but I did check my answers and they seem to be correct.

Is my work correct, and how would I go about finding all possible $\gamma$?

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

Find $\gamma$ such that $\alpha = \gamma^{-1}\beta\gamma$.

$$ \alpha = \gamma^{-1}\beta\gamma \iff \gamma\alpha\gamma^{-1} = \beta \iff \gamma \text{ translates } \alpha \text{ to } \beta $$ \begin{align*} \alpha &= (1)(2\ 7\ 3\ 8)(4\ 6)(5)(9) \\ \beta &= (1)(2)(3\ 5\ 6\ 7)(4)(8\ 9) \\ \gamma &= (1)(2\ 6\ 9\ 4\ 8\ 5)(3)(7) \end{align*} $$ \text{This result is obtained by noting the similarities between the cycles mean it's likely that } \\ \gamma \text{ translates } 1 \to 1, \ \ 3 \to 3, \ \ 7 \to 7 \ \implies \ 2 \to 6, \ \ 8 \to 5 \\\ \\ \text{After that, there are multiple ways of translating }(4\ 6) \to (8\ 9) \text{ and } (5),\ (9)\ \to \ (8),\ (4),\ e.g.: \\\ \\ (2\ 6\ 9\ 4\ 8\ 5), \ \ (2\ 6\ 8\ 5\ 4\ 9), \ \ (2\ 6\ 8\ 5)(4\ 9) $$

Zaz
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  • The term for this operation is "conjugate." "Translate" should be reserved for left or right multiplication, if used at all in this context. – Qiaochu Yuan Sep 16 '17 at 16:47
  • As a partial answer, notice that if $\gamma_1$ conjugates $\alpha$ to $\beta$, then $\gamma_2$ also conjugates $\alpha$ to $\beta$ if and only if $\gamma_1\gamma_2^{-1}$ commutes with $\beta$. Thus an element conjugating $\alpha$ to $\beta$, if it exists, is determined up to multiplication by an element commuting with $\beta$, so the number of such choices is the number of elements commuting with $\beta$. – Joey Zou Sep 16 '17 at 17:08

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Yes, if two permutations are conjugate they are conjugate in multiple ways. The cycle structure is preserved by conjugation but beyond that there are two kinds of freedom in the choice of the conjugating permutation: first, the cycles may be "spun" arbitrarily. And equal cycles may be permuted. In your case $\alpha$ and $\beta$ has cycle counts $c_1=3,$ $ c_2=1,$ $ c_4=1,$ (and the rest of the $c_i=0$ so you have $2\times 4$ ways to spin the 2 and 4 cycles and $3!=6$ ways to permute the 1 cycles, making $48=\prod_i (c_i!)i^{c_i}$ different conjugating maps.

kimchi lover
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