My Math Forum prove permutation inverse

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 November 25th, 2009, 06:12 AM #1 Guest   Joined: Posts: n/a Thanks: prove permutation inverse This should be rather straight forward, but I thought I would ask for some inout from those better an group theory than me. Let ${\sigma}=(a_{1} a_{2}....a_{k})\in S_{n}$. Prove that $(a_{1}a_{2}...a_{k})^{-1}=(a_{1}a_{k}a_{k-1}...a_{3}a_{2})$ What would be a good way to go about this?. I am just learning some group theory and thought this was a good problem involving cycles. I was thinking Let ${\sigma}={\sigma}_{1}{\sigma}_{2}...{\sigma}_{k}$, where ${\sigma}_{i}$ Then, ${\sigma}^{-1}={\sigma}_{k}^{-1}{\sigma}_{k-1}^{-1}....{\sigma}_{1}^{-1}$ because $({\sigma}_{1}{\sigma}_{2}...{\sigma}_{k})({\sigma} _{k}^{-1}{\sigma}_{k-1}^{-1}....{\sigma}_{1}^{-1})$ $=({\sigma}_{1}{\sigma}_{2}...{\sigma}_{k-1})({\sigma}_{k}{\sigma}_{k}^{-1})({\sigma}_{k-1}^{-1}....{\sigma}_{1}^{-1})$ and keep going until we get to e. I have more, but it is too much Latexing with all those sigmas
 November 25th, 2009, 08:26 AM #2 Newbie   Joined: Aug 2008 Posts: 10 Thanks: 0 Re: prove permutation inverse To start with $(\tau \rho)^{-1}= \rho^{-1} \tau^{-1}$, holds for all bijections (not just permutations -- on the other hand it holds for all groups, either way works). Splitting any cycle into a product of 2-cycles, and knowing that 2-cycles and their own inverse will be useful -- but it's important to know which 2-cycles a cycle is split into. You should be able to prove the theorem just using these three facts: If so, then it's worth going back to prove those lemmas.
 November 26th, 2009, 04:50 AM #3 Guest   Joined: Posts: n/a Thanks: Re: prove permutation inverse Thanks.

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