# Odd and Even Permutations

• Oct 15th 2011, 04:54 PM
Bernhard
Odd and Even Permutations
In Gallian Ch 5 Permutation Groups, Ex 12 is as follows:

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If a permutation $\alpha$ is even prove that ${\alpha}^{-1}$ is even.

Further, if a permutation $\alpha$ is odd prove that ${\alpha}^{-1}$ is odd.

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Can anyone help with this problem. Be grateful for some help

Peter
• Oct 15th 2011, 05:00 PM
Bernhard
Re: Odd and Even Permutations
Oh! Sitting here giving this more thought!

Is it something to do with the fact that the inverse would be obtained by reversing the symbols in each transposition and so the inverse would contain the same numberof transpositions???

Is it that simple?

Peter
• Oct 15th 2011, 05:01 PM
Deveno
Re: Odd and Even Permutations
what definition are you using for "even" and "odd"?

(hint: if $\alpha = \tau_1\tau_2\dots\tau_k$, where every $\tau_j$ is a transposition,

what is $\tau_k\dots\tau_2\tau_1$ ? use the fact that transpositions are of order 2).
• Oct 15th 2011, 05:22 PM
Bernhard
Re: Odd and Even Permutations
Thanks for the help!

I was taking a transposition to be a 2-cycle and was taking an even permutation as one that can be expressed as an even number of transpositions or 2-cycles.

Regarding your hint: not sure - if the 2-cycles were pairwise disjoint then reversing the order of them would make no difference?? Is that correct??

However, what you seem to be hinting is that the reverse order is the inverse - and in that case it would have the same number of 2-cycles. Must work on showing that reversing the order of 2-cycles gives the inverse.

Peter
• Oct 15th 2011, 05:54 PM
Deveno
Re: Odd and Even Permutations
no...what i have shown you IS the proof that reversing the order of the transpositions we write α as, is the inverse.

why? because $\tau_j\tau_j = 1$ for any transpositon $\tau_j$, so:

$(\alpha)(\tau_k\dots\tau_2\tau_1) = (\tau_1\tau_2\dots\tau_k)(\tau_k\dots\tau_2\tau_1)$

$= (\tau_1\tau_2\dots\tau_{k-1})(\tau_k\tau_k)(\tau_{k-1}\dots\tau_2\tau_1)$

$= (\tau_1\tau_2\dots\tau_{k-1})(\tau_{k-1}\dots\tau_2\tau_1)$

and....continuing....

$=(\tau_1\tau_2)(\tau_2\tau_1) = \tau_1(\tau_2\tau_2)\tau_1 = \tau_1\tau_1 = 1$

so, since $(\alpha)(\tau_k\dots\tau_2\tau_1) = 1$

$\alpha^{-1} = \tau_k\dots\tau_2\tau_1$

the transpositions don't need to be disjoint, that was never assumed. (in fact, in S3, you can't write a 3-cycle in terms of disjoint tranpositions).