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We investigate on an arbitrary $a_i$ : $c(a_1 \ a_2 \dots \ a_k)c^{-1}(a_i)$.

First step, $c(a_i)=a_k$. Second step, $(a_1 \ a_2 \dots \ a_k)(a_k)=a_{k+1}$, Third step, $c^{−1}(a_{k+1})=? $.

Any answer that I read in MSE was not helpful to understand. In the final step, they all imply $c^{−1}(a_{k+1})=c(a_i)$, but why $c^{−1}(a_{k+1})=a_k=(a_1 \ a_2 \dots \ a_k)^{−1}(a_{k+1})$? which may imply $c=(a_1 \ a_2 \dots \ a_k)$ !

I would appreciate any simple clear detailed explanation.

Frubiclé
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    The question is not really easy to understand either. Are you working in cyclic groups? Why you have $a_1,\ldots,a_k$ on your left but only $a_1,\ldots a_3$ on your right? Since all cyclic groups are abelian, should not this exactly equal to $a_1,\ldots,a_k$? In the right side it seems there are at least three $c's$. Are you thinking that $c^k=1$? – Darío G Mar 16 '16 at 13:52
  • @user27454 - thanks. I edited –  Mar 16 '16 at 13:53
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    I am afraid that you only have answered one of my questions. Are we working in cyclic groups? If so, note that cyclic groups are abelian and I am confident to tell you that the left hand side is simply $a_1\cdots a_k$ while the right hand side is $c^k\cdot a_1\cdots a_k$. Thus, the two expressions will be different unless $c^k=e$ where $e$ is the identity of the group. – Darío G Mar 16 '16 at 13:59
  • @user27454 - No, permutation group and $c$ is a $l$-cycle. $c=(a_1 \ a_2 \dots \ a_l)$ and not necessary $l=k$ –  Mar 16 '16 at 14:01
  • @user27454 - For example, in $S_p$, $\sigma (ab) \sigma^{-1} = (\sigma (a) \ \ \sigma (b))$ in which $\sigma = (12 \ldots p)$. –  Mar 16 '16 at 14:07
  • and what are $a_1,\ldots,a_k$? Elements of the set on which the permutation is defined? – Darío G Mar 16 '16 at 14:08
  • @user27454 - Yes. –  Mar 16 '16 at 14:08
  • Let us continue this discussion in chat. – Darío G Mar 16 '16 at 14:15
  • Read this

    http://math.stackexchange.com/questions/48134/why-are-two-permutations-conjugate-iff-they-have-the-same-cycle-structure

     – Frubiclé Mar 16 '16 at 14:31
  • @Frubiclé - Thanks, I read it. It doesn't answer my question. As you can see in comments there is my question by "Student" at Jun 22 '14 at 13:30. I couldn't find where it is asked. –  Mar 16 '16 at 14:39
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    @Liebe I just tried to do a quick proof and i realise that your $c$ and $c^{-1}$ are the wrong way round - just editing now – Frubiclé Mar 16 '16 at 14:45

1 Answers1

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Your question is about permutations and how conjugating a $k$-cycle by a permutation preserves the cycle type.

Claim: Let $c\in S_n$ and let $(a_1 \; a_2 \; \ldots \; a_k)$ be a $k$-cycle in $S_n$. Then $$ c (a_1 \; a_2 \; \ldots \; a_k) c^{-1} = (c(a_1) \; c(a_2) \; \ldots \; c(a_k))$$

Proof: Let's just consider how the left and right hand sides act on $c(a_1)$.

Then right hand side sends $c(a_1)$ to $c(a_2)$.

$c^{-1}$ sends $c(a_1)$ to $a_1$, which the $k$-cycle sends to $a_2$, which $c$ then sends to $c(a_2)$. Thus the left hand side as the composition of those three operations sends $c(a_1)$ to $c(a_2)$.

We chose $c(a_1)$ without loss of generality so we obtain the same agreement if we pick any $c(a_i)$. If a number in the set $\{ 1,\dots , n \}$ is not in the $k$-cycle, so it is not one of the $a_i$, then it is clearly fixed by both sides. So both sides are the same.

Frubiclé
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    Wow, You made my day! It's more than a year that I am deferring this problem as neither I can't prove it nor find a proof. Thank you very much. I will gift a bonus also (after it opens in 2 days). Sorry I don't have more than 50 though. Thank you. :):):) –  Mar 16 '16 at 15:08
  • No worries @Liebe :) – Frubiclé Mar 16 '16 at 15:56