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I'm trying to solve this question in Hungerford's algebra book:

and the solution I'm trying to give has a lot of calculations I'm a little confused, is there an easy way to solve it? I mean a more direct solution?

Thanks in advance

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2 Answers 2

up vote 5 down vote accepted

To write $\tau\sigma\tau^{-1}$ as a cycle, or a product of cycles, you just have to work out what each element is mapped to. So, start with $\tau(i_1)$ and apply $\tau\sigma\tau^{-1}$: that means apply $\tau^{-1}$ then $\sigma$ then $\tau$, and we get $$\tau(i_1)\mapsto i_1\mapsto \sigma(i_1)=i_2\mapsto \tau(i_2)\ .$$ And similarly for the rest.

If $j$ is any element other than $\tau(i_1),\ldots,\tau(i_r)$ then $\tau^{-1}(j)$ is fixed by $\sigma$, and so applying $\tau\sigma\tau^{-1}$ gives $$j\mapsto \tau^{-1}(j)\mapsto \tau^{-1}(j)\mapsto \tau(\tau^{-1}(j))=j\ .$$

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Great solution!!! simple and elegant, thanks a lot! –  user42912 Apr 22 '14 at 2:30

I think this is straightforward by direct computation:

Case (i): if $i\not\in \{\tau(i_1),\dots,\tau(i_r)\}$, then $\tau^{-1}(i)\not\in \{i_1,\dots,i_r\}$, whence $\sigma\tau^{-1}(i)=\tau^{-1}(i)$ by definition of $\sigma$. Therefore, $\tau\sigma\tau^{-1}(i)=i$.

Case (ii): if $i=\tau(i_j)$ for some $1\leq j\leq r$, then $\tau^{-1}(i)=i_j$, whence $\sigma\tau^{-1}(i)=i_{j+1}$ by the definition of $\sigma$ (if $j=r$, then replace $j+1$ by $1$ in this equation). Therefore, $\tau\sigma\tau^{-1}(i)=\tau(i_{j+1})$.

Is this the solution you had in mind which you felt had "lots of calculations"? I think it it is a perfectly reasonable direct proof if you write it out cleanly.

Exercise 1: The cycle structure of a permutation $\sigma\in S_n$ is defined as follows: if $\sigma = \sigma_1\cdots\sigma_i$ is the disjoint cycle product decomposition such that the lengths of the $\sigma_i$'s are monotonically increasing in $i$, then the cycle structure of $\sigma$ is the $i$-tuple $(l(\sigma_i))_{i}$ (where $l(\sigma_i)$ denotes the length of the cycle $\sigma_i$).

Prove that two permutations in $\sigma_n$ are conjugate if and only if they have the same cycle structure.

Exercise 2: Determine the number of $n$-cycles in $S_n$. (Hint: use the orbit-stabilizer formula #(centraliser of $\sigma$) x #(conjugacy class of $\sigma$) = $n!$ for any $\sigma\in S_n$.)

Hope this helps!

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thank you it helps a lot indeed, just one remark I think there is a typo in third line: I think $\tau\sigma\tau^{-1}(i)=i$ –  user42912 Apr 22 '14 at 2:38
Hi @user42912, you're right! Sorry about that. Is it fixed now? –  Amitesh Datta Apr 22 '14 at 2:47
yes, thanks again! –  user42912 Apr 22 '14 at 2:49
Great! No problem - I am very happy to have helped. –  Amitesh Datta Apr 22 '14 at 2:50

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