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Thread: Permutations

  1. #1
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    Permutations

    Hey, i'm having a hard time seeing how to do this problem. Any hints?

    $\displaystyle \alpha \in S_{n}$ a cycle $\displaystyle (a_{1},a_{2},...,a_{k}) (a_{1},a_{2},...,a_{k} \:distinct \:elements \:from \:the \:set \:\{1,2,..n\})$. Let $\displaystyle \beta \in S_{n} $ a permutation. Show that

    $\displaystyle \beta\alpha \beta ^{-1} = (\beta (a_{1}),\beta (a_{2}),...,\beta (a_{k}))$.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Permutations

    Quote Originally Posted by gordo151091 View Post
    Hey, i'm having a hard time seeing how to do this problem. Any hints?

    $\displaystyle \alpha \in S_{n}$ a cycle $\displaystyle (a_{1},a_{2},...,a_{k}) (a_{1},a_{2},...,a_{k} \:distinct \:elements \:from \:the \:set \:\{1,2,..n\})$. Let $\displaystyle \beta \in S_{n} $ a permutation. Show that

    $\displaystyle \beta\alpha \beta ^{-1} = (\beta (a_{1}),\beta (a_{2}),...,\beta (a_{k}))$.
    Hi gordo151091!

    Can you apply $\displaystyle \beta\alpha \beta^{-1}$ to $\displaystyle \beta (a_{1})$?
    What do you get?
    Thanks from gordo151091
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  3. #3
    Junior Member Nehushtan's Avatar
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    Re: Permutations

    Consider the effect of $\displaystyle \beta\alpha\beta^{-1}$ on $\displaystyle \beta(a_i)$ for $\displaystyle i=1,\ldots,k-1$. We have

    $\displaystyle \beta\alpha\beta^{-1}(\beta(a_i))=\beta\alpha(a_i)=\beta(a_{i+1})$

    So $\displaystyle \beta\alpha\beta^{-1}$ maps $\displaystyle \beta(a_i)$ to $\displaystyle \beta(a_{i+1})$ for $\displaystyle i=1,\ldots,k-1$, and it maps $\displaystyle \beta(a_k)$ to $\displaystyle \beta(a_1)$ since $\displaystyle \alpha$ maps $\displaystyle a_k$ to $\displaystyle a_1$.

    Now suppose $\displaystyle b\in\{1,\ldots,n\}$ and $\displaystyle b\notin\{\beta(a_1),\ldots,\beta(a_k)\}$. Then $\displaystyle b=\beta(a)$ for some $\displaystyle a\notin\{a_1,\ldots,a_k\}$. Thus $\displaystyle \alpha$ fixes $\displaystyle a$ and

    $\displaystyle \beta\alpha\beta^{-1}(b)=\beta\alpha\beta^{-1}(\beta(a))=\beta\alpha(a)=\beta(a)=b$

    so $\displaystyle \beta\alpha\beta^{-1}$ fixes $\displaystyle b$.

    Hence $\displaystyle \beta\alpha\beta^{-1}=(\beta(a_1)\,\beta(a_2)\,\cdots\,\beta(a_k))$.
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  4. #4
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    Re: Permutations

    Hi! Thanks to you two for your help
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