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Can you help me finding an odd permutation which commutates with $(1,2,3)(4,5,6)$ in $S_6$ ?

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I've changed the title of your question because the word "commutator" means something different (namely, an element of the form $aba^{-1}b^{-1}$. –  Noah Snyder Oct 7 '12 at 18:30
    
yes thats better. thanks –  André Oct 7 '12 at 18:32
    
I think the right word is "commutes." –  Rasmus Oct 7 '12 at 18:36

4 Answers 4

up vote 3 down vote accepted

$(14)(25)(36)$ commutes with $(123)(456)$ and is odd.

If $\sigma$ is a permutation of $\{1,2,4,5,6\}$, $\sigma \circ (123) \circ \sigma^{-1}=(\sigma(1) \sigma(2) \sigma(3))$.

Similarly, $\sigma \circ (123)(456) \circ \sigma^{-1}=(\sigma(1) \sigma(2) \sigma(3))(\sigma(4) \sigma(5) \sigma(6))$.

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+1 ;) ${}{}{}{}$ –  Rasmus Oct 7 '12 at 18:34
    
It appears from the phrasing of this question that Andre is looking for some understanding as to where he should look for such an element, not just the answer. –  Noah Snyder Oct 7 '12 at 18:35
1  
I think the right word is "commutes." –  Rasmus Oct 7 '12 at 18:35
    
@Rasmus: Thanks. –  francis-jamet Oct 7 '12 at 18:36

This seems to be a good candidate: $(14)(25)(36)$

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indeed. how did you figure it out ? just by trying ? –  André Oct 7 '12 at 18:39
    
To be honest: yes, just by trying. I might try to post a more enlightening answer later if no one else does. –  Rasmus Oct 7 '12 at 18:45

You have two valid answers already, but I wil try to explain how they could be derived. In my answer, I will let permutations act from the right. Notice that $\sigma^{-1} (123)(456)\sigma = ( 1\sigma 2\sigma 3\sigma) (4 \sigma 5\sigma 6\sigma)$. Hence if $\sigma$ is chosen to interchange $1$ and $4$, and $2$ and $5$ and $3$ and $6,$ we will still have $\sigma^{-1} (123)(456)\sigma = (456)(123) = (123)(456).$ Hence the permutation $\sigma = (14)(25(36)$ commutes with $(123)(456),$ and is an odd permutation, as it is a product of an odd number of transpositions.

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Saying that x and y commute is the same as saying that $yxy^{-1} = x$, that is that conjugation by y preserves x. Conjugation by a permutation is just renaming the elements. E.g. conjugating by $(1 2)$ just means change all 1's to 2's and vice-versa. So you want to find a way of renaming the elements between 1 and 6 which doesn't change $(1 2 3)(4 5 6)$. One way to do that would be to rename the elements within a give triple, e.g. conjugate by $(1 2 3)$. But that only gives you even permutations. The other way to do it is to interchange the roles of the two triples. This gives you the example in the other answers.

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