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Math Help - Permutation

  1. #1
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    Permutation

    I'm stuck on this problem. I really have no clue. Would someone give me a hand?
    1)Show \forall i \alpha \in S_n moves i iff {\alpha}^{-1} moves i
    2) Show that if \alpha \beta=(1) and \alpha , \beta are disjoint, then \alpha=(1) and \beta=(1)
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  2. #2
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    Quote Originally Posted by jackie View Post
    1)Show \forall i \alpha \in S_n moves i iff {\alpha}^{-1} moves i
    Let \alpha(i) = j, note that i\not = j since it is moved. If \alpha^{-1} (i) = i then i = \alpha(i) = j which is a contradiction. Thus, \alpha^{-1} must move i.
    2) Show that if \alpha \beta=(1) and \alpha , \beta are disjoint, then \alpha=(1) and \beta=(1)
    Notice \beta = \alpha^{-1}, by above \alpha,\beta must move the same elements. However, they are disjoint, therefore this forces \alpha=\beta=\text{id}.
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  3. #3
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    Thank you very much for your help TPH. Can you check if my reasoning is correct on this problem? Show that sgn(\alpha)=sgn({\alpha}^{-1}) where \alpha is a permutation.
    We have sgn(\alpha {\alpha}^{-1})=sgn(1)=1, so sgn(\alpha)sgn({\alpha}^{-1})=1. Hence, sgn({\alpha}^{-1})=sgn(\alpha)^{-1}=sgn(\alpha)
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  4. #4
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    Quote Originally Posted by jackie View Post
    Thank you very much for your help TPH. Can you check if my reasoning is correct on this problem? Show that sgn(\alpha)=sgn({\alpha}^{-1}) where \alpha is a permutation.
    We have sgn(\alpha {\alpha}^{-1})=sgn(1)=1, so sgn(\alpha)sgn({\alpha}^{-1})=1. Hence, sgn({\alpha}^{-1})=sgn(\alpha)^{-1}=sgn(\alpha)
    \text{sgn}(\alpha)\text{sgn}(\alpha^{-1}) = \text{sgn}(\alpha \alpha^{-1}) = \text{sgn}(\text{id}) = 1 \implies \text{sgn}(\alpha) = \frac{1}{\text{sgn}(\alpha^{-1})} = \text{sgn}(\alpha^{-1}).
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