# Permutation

• Sep 12th 2009, 06:46 PM
jackie
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 $\beta=(1)$
• Sep 12th 2009, 07:49 PM
ThePerfectHacker
Quote:

Originally Posted by jackie
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 $i$.
Quote:

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 $\alpha=\beta=\text{id}$.
• Sep 12th 2009, 10:39 PM
jackie
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, $sgn({\alpha}^{-1})=sgn(\alpha)^{-1}=sgn(\alpha)$
• Sep 13th 2009, 07:57 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 $sgn(\alpha)=sgn({\alpha}^{-1})$ where $\alpha$ is a permutation.
We have $sgn(\alpha {\alpha}^{-1})=sgn(1)=1$, so $sgn(\alpha)sgn({\alpha}^{-1})=1$. Hence, $sgn({\alpha}^{-1})=sgn(\alpha)^{-1}=sgn(\alpha)$

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