# Permutation

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• Sep 12th 2009, 05:46 PM
jackie
Permutation
I'm stuck on this problem. I really have no clue. Would someone give me a hand?
1)Show $\displaystyle \forall i$ $\displaystyle \alpha \in S_n$ moves $\displaystyle i$ iff $\displaystyle {\alpha}^{-1}$ moves $\displaystyle i$
2) Show that if $\displaystyle \alpha \beta=(1)$ and $\displaystyle \alpha , \beta$ are disjoint, then $\displaystyle \alpha=(1)$ and $\displaystyle \beta=(1)$
• Sep 12th 2009, 06:49 PM
ThePerfectHacker
Quote:

Originally Posted by jackie
1)Show $\displaystyle \forall i$ $\displaystyle \alpha \in S_n$ moves $\displaystyle i$ iff $\displaystyle {\alpha}^{-1}$ moves $\displaystyle i$

Let $\displaystyle \alpha(i) = j$, note that $\displaystyle i\not = j$ since it is moved. If $\displaystyle \alpha^{-1} (i) = i$ then $\displaystyle i = \alpha(i) = j$ which is a contradiction. Thus, $\displaystyle \alpha^{-1}$ must move $\displaystyle i$.
Quote:

2) Show that if $\displaystyle \alpha \beta=(1)$ and $\displaystyle \alpha , \beta$ are disjoint, then $\displaystyle \alpha=(1)$ and $\displaystyle \beta=(1)$
Notice $\displaystyle \beta = \alpha^{-1}$, by above $\displaystyle \alpha,\beta$ must move the same elements. However, they are disjoint, therefore this forces $\displaystyle \alpha=\beta=\text{id}$.
• Sep 12th 2009, 09:39 PM
jackie
Thank you very much for your help TPH. Can you check if my reasoning is correct on this problem? Show that $\displaystyle sgn(\alpha)=sgn({\alpha}^{-1})$ where $\displaystyle \alpha$ is a permutation.
We have $\displaystyle sgn(\alpha {\alpha}^{-1})=sgn(1)=1$, so $\displaystyle sgn(\alpha)sgn({\alpha}^{-1})=1$. Hence, $\displaystyle sgn({\alpha}^{-1})=sgn(\alpha)^{-1}=sgn(\alpha)$
• Sep 13th 2009, 06:57 AM
ThePerfectHacker
Quote:

Originally Posted by jackie
Thank you very much for your help TPH. Can you check if my reasoning is correct on this problem? Show that $\displaystyle sgn(\alpha)=sgn({\alpha}^{-1})$ where $\displaystyle \alpha$ is a permutation.
We have $\displaystyle sgn(\alpha {\alpha}^{-1})=sgn(1)=1$, so $\displaystyle sgn(\alpha)sgn({\alpha}^{-1})=1$. Hence, $\displaystyle sgn({\alpha}^{-1})=sgn(\alpha)^{-1}=sgn(\alpha)$

$\displaystyle \text{sgn}(\alpha)\text{sgn}(\alpha^{-1}) = \text{sgn}(\alpha \alpha^{-1}) = \text{sgn}(\text{id}) = 1 \implies \text{sgn}(\alpha) = \frac{1}{\text{sgn}(\alpha^{-1})} = \text{sgn}(\alpha^{-1})$.