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Let $\sigma \in S_n$. Calculate the conjugate of $(ab)$ with $\sigma$.

Any ideas?

I understand we can represent $S_n$ as $(1 2)(2 3)\cdots(n-1 n )$, and so

$$\sigma (ab)\sigma^{-1} =(1 2)(2 3)\cdots(n-1 n ) (ab) (n-1 n)$$

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up vote 1 down vote accepted

For this particular problem, You are interested in $\sigma (a~b) \sigma^{-1}$ for $\sigma \in S_n$. (Naively, you assume $a,b \le n$.)

Any element in $S_n$ can be written as a product of disjoint cycles. So, let $$\sigma=(a_{11}a_{12} \cdots)(a_{21}a_{22}\cdots)\cdots(a_{r1}a_{r2}\cdots)$$

If $a,b \notin \{a_{11},a_{12},\cdots,a_{21},a_{22}, \cdots,a_{r1},a_{r2}, \cdots\}$, then the cycles $\sigma$ and $(a~b)$ are disjoint and hence they commute. Note that, in this case, $\sigma(a)=a; ~~\sigma(b)=b.$ So, we have that $$\sigma(a~b) \sigma^{-1}=(a ~b)=(\sigma(a)~\sigma(b))$$

Suppose this is not the case. Then, $$\sigma(a)=a_k\\\sigma(b)=b_l$$

Now, complete the argument as above by considering the effect of this conjugate on $\sigma(a)$ and on $\sigma(b)$ and you'll be through.

For a general discussion about conjugacy in $S_n$, you may find this answer of mine here useful.

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You can show more generally that for any $r$-cycle $(a_1,\ldots,a_r)$ and any $\sigma \in S_n$, $$\sigma\circ (a_1,\ldots,a_r)\circ \sigma^{-1}=(\sigma(a_1),\ldots,\sigma(a_r))$$ to do this, observe that if $i\notin \{ $$ \sigma(a_1),\ldots,\sigma(a_n) $$\}$, then $\sigma\circ (a_1,\ldots,a_r)\circ \sigma^{-1}(i)=i$.

Also observe that if $i=\sigma(a_j)$, $\sigma\circ (a_1,\ldots,a_r)\circ \sigma^{-1}(i)=\sigma(a_{j+1})$ (and $\sigma\circ (a_1,\ldots,a_r)\circ \sigma^{-1}(\sigma(a_r))=\sigma(a_{1})$, of course).

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