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How can I go about showing that $f=(6\;9)(1\;3\;4)(2\;5\;7\;8)$ and $g=(1\;7)(2\;3\;5)(4\;9\;6\;8)$ are conjugate in $S_9$ (the set of permutations on 9 symbols)? I need to do this without using the Cauchy Theorem.

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Permutations are conjugate in the full symmetric group if and only if they have the same cycle structure. This is because conjugation corresponds to renumbering; that is, you get the cycle structure of $ghg^{-1}$ by permuting the numbers in the cycle structure of $h$ according to $g$. Since the full symmetric group always contains the required permutation $h$, any two permutations with the same cycle structure are conjugate in it.

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Could you go in a little more detail? If f and g are conjugate, wouldn't that give me the form (h^-1)fh=g? Also if we permute a cycle, would we have the same cycle structure as before? –  Mike Mar 1 '12 at 21:47
    
So one possibility for an element $h$ that conjugates $f$ to $g$ is the permutation that maps $6 \to 1$, $9 \to 7$, $1 \to 2$, $3 \to 3$, $4 \to 5$, $2 \to 4$, $5 \to 9$, $7 \to 6$, $8 \to 8$. (Or that might be $h^{-1}$, depending on your notation.) Geddit? Of course $h$ is not unique. –  Derek Holt Mar 1 '12 at 22:12
    
@Mike: Sorry, I didn't use the same letters as in your question. Yes, it would give you that form. I didn't speak of permuting a cycle but of permuting the numbers in the cycle structure. If you permute the numbers in $(13)(24)$ by exchanging $1$ and $2$, you still have the same cycle structure, just with other numbers: $(23)(14)$. Thus $(12)[(13)(24)](12)^{-1}=(23)(14)$. I'd recommend to check that explicitly to get a feel for how it works. –  joriki Mar 1 '12 at 22:12
    
@Mike: To explain Derek's comment about $h$ not being unique: You can write down the cycle structure of a permutation in different ways: You can choose any order of the cycles with the same length, and you can start each cycle with an arbitrary one of its elements. For a given way of writing the cycle structure of $f$, you get a different way of conjugating $f$ to obtain $g$ for each way of writing the cycle structure of $g$. –  joriki Mar 1 '12 at 22:20
    
Thanks guys! Thats very helpful. Derek, how did you find an h that works? –  Mike Mar 1 '12 at 22:31
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