Permutation

**jackie** · September 12th 2009, 05:46 PM

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 $\beta=(1)$

**ThePerfectHacker** · September 12th 2009, 06:49 PM

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 $i$ .

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 $\alpha=\beta=\text{id}$ .

**jackie** · September 12th 2009, 09:39 PM

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, $sgn({\alpha}^{-1})=sgn(\alpha)^{-1}=sgn(\alpha)$

**ThePerfectHacker** · September 13th 2009, 06:57 AM

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, $sgn({\alpha}^{-1})=sgn(\alpha)^{-1}=sgn(\alpha)$

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

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