# Permutations

• Mar 9th 2013, 01:34 PM
gordo151091
Permutations
Hey, i'm having a hard time seeing how to do this problem. Any hints?

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

$\beta\alpha \beta ^{-1} = (\beta (a_{1}),\beta (a_{2}),...,\beta (a_{k}))$.
• Mar 9th 2013, 02:19 PM
ILikeSerena
Re: Permutations
Quote:

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

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

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

Hi gordo151091! :)

Can you apply $\beta\alpha \beta^{-1}$ to $\beta (a_{1})$?
What do you get?
• Mar 9th 2013, 02:30 PM
Nehushtan
Re: Permutations
Consider the effect of $\beta\alpha\beta^{-1}$ on $\beta(a_i)$ for $i=1,\ldots,k-1$. We have

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

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

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

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

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

Hence $\beta\alpha\beta^{-1}=(\beta(a_1)\,\beta(a_2)\,\cdots\,\beta(a_k))$.
• Mar 9th 2013, 02:48 PM
gordo151091
Re: Permutations
Hi! Thanks to you two for your help