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I need help with d) here.

Let $2 \le r \le n$ be two natural numbers. Assume that $\rho \in S_n$ is a permutation of the set $I_n=\{1,2,...,n\}$. Let $x_i \in I_n$ for $1 \le i\le r$ be $r$ different numbers.

a) Show that $\rho(x_1,...,x_i,...,x_r)\rho^{-1}=(\rho(x_1),...,\rho(x_i),\rho(x_{i+1}),...,\rho(x_r))$. HINT: For $i < r$ feed the left side with $\rho(x_i)$ and check that you get $\rho(x_{i+1})$.

b) Show that: $(2, 4)(1 ,5)(1, 2 ,3 ,4, 5)(2, 4)(1, 5)=(5,4,3,2,1)$.

c) Let $\sigma$ be the $r$-cycle $(1,2,3,...,r)$ in $S_n$. Show that $\sigma$ is conjugate to its own inverse; that is, there is a permutation such that $\rho\sigma\rho^{-1}=\sigma^{-1}$.

d) Show that in c) one may take for $\rho$ a permutation that fixes any of the numbers that $\sigma$ moves. (This means: Pick one $1 \le i\le r$, then one may find a $\rho$ with $\rho(i)=i$).

If $r$ is odd I can just generalize b), but what if $r$ is even? I don't see how to solve the exercise then.

user119615
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  • Do you compose permutations left-to-right or right-to-left (note that both conventions are common)? – A.P. Mar 20 '15 at 11:37
  • @A.P. From left to right I think. That is $(1,2,3)(3,6)=(1,2,3,6)$, not $(1,2,6,3)$. – user119615 Mar 20 '15 at 11:42
  • Re: But what if r is even? $(14)(23)(1234)(23)(14)=(4321)$. It is good to learn what happens in general when you create a conjugate of cycle. See, for example: http://math.stackexchange.com/questions/48134/why-are-two-permutations-conjugate-iff-they-have-the-same-cycle-structure and https://unapologetic.wordpress.com/2010/09/10/conjugates/ – Martin Sleziak Mar 20 '15 at 13:05
  • @MartinSleziak In your case does $\rho$ satisfy that it leaves an element fixed? – user119615 Mar 20 '15 at 15:10
  • That's true. This was intended to help you with the answers of c). In any case, the answers posted so far and the links which you were given seems to be sufficient to help you with the additional condition, too. – Martin Sleziak Mar 20 '15 at 15:17
  • @MartinSleziak I understand, thanks! – user119615 Mar 20 '15 at 18:59

2 Answers2

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To answer the d) question. Fix $1\leq i\leq r$ such an integer then you can write :

$$\sigma:=(1,2,...,r)=(i,i+1,...,r,1,...,i-1)$$

This is just another way of writing your permutation.

Now you know that :

$$\sigma^{-1}=(1,r,r-1,...,2)=(i,i-1,...,1,r,...,i+1) $$

And you want $\rho$ such that :

$$\rho\sigma\rho^{-1}=\sigma^{-1} \text{ and } \rho(i)=i$$

From a) you get :

$$\rho\sigma\rho^{-1}=(\rho(i),\rho(i-1),...,\rho(1),\rho(r),...,\rho(i+1))=(i,\rho(i-1),...,\rho(1),\rho(r),...,\rho(i+1))$$

Finally you just want :

$$(i,\rho(i+1),...,\rho(r),\rho(1),...,\rho(i-1))=(i,i-1,...,1,r,...,i+1) $$

You can now see that such a $\rho$ always exists. You just want to send the $r-1$ tuple:

$$(i+1,...,r,1,...,i-1)\text{ on } (i-1,...,1,r,...,i+1)$$

with a permutation of the set $\{1,...,i-1,i+1,...,r\}$ and this can be done because the action of the group permutation is $r-1$ transitive.

Clément Guérin
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Hint: consider the permutation $(1,2,3,4)$. What can you say of the permutation $(1,4,3,2)$?

Suppose you want a $\rho$ that fixes $i$. Since you can always write $\sigma^{-1}$ with $i$ as the leading number, without loss of generality we may assume that $i = 1$. Now, by a) you know that $$ \rho \sigma \rho^{-1} = (\rho(1),\rho(2),\dotsc,\rho(r)) $$ so it is enough to choose a $\rho$ that sends the sequence $2,3,\dotsc,r$ into the sequence $r,r-1,\dotsc,2$ and fixes every other number in $\{1,\dotsc,n\}$.

A.P.
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