Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question

2 Answers 2

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.

share|improve this answer

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.