Why $c(a_1 \ a_2 \dots \ a_k)c^{-1}=(c(a_1) c(a_2)… c(a_k))$? [duplicate]

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.

• 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 – Liebe Mar 16 '16 at 13:53
• 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 $l=k$ – Liebe Mar 16 '16 at 14:01
• @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

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.

• 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. :):):) – Liebe Mar 16 '16 at 15:08
• No worries @Liebe :) – Frubiclé Mar 16 '16 at 15:56