Given a cycle $c \in S_n $ with $ ord(c) = s $ and $ s = kt $, prove that $c^k$ is a product of $k$ cycles of length $t$.

Question

I came across this question in a recent exam.

Given that $ ord(c) = s $, we assume that $c^s = c^{kt} = (id) \implies (c^{k})^t = (id)$.

That means that $c^k$ is a cycle of order $t$.

Can you please help me on the next step?

Answer 1

Power of a cycle may not be a cycle

Your sentence That means that $c^k$ is a cycle of order $t$. is not true.

Take $S_4$ as an example. $c = (1 \ 2 \ 3 \ 4)$ is a cycle of order 4. However, $c^2 = (1 \ 3) (2 \ 4)$ is not a cycle. It is a product of two cycles each one of order 2.

It can be proven that for a cycle $c$ of order $k$, $c^i$ is a cycle if and only if $i, k$ are coprime integers.

Coming back to your case

Using what I mentioned above, if $ord(c) = s$ and $s=kt$, $c^k$ is not a cycle if $1 < k <s$.

If $c = (a_0 \ \dots \ a_k \ a_{k+1} \ \dots \ a_{2k} \ \dots \ a_{(t-1)k} \ \dots \ a_{tk-1})$ is a cycle or order $s=kt$, then you'll be able to prove that $c^k$ is the product of the cycles

$$c= (a_0 \ a_k \ \dots \ a_{(t-1)k})(a_1 \ a_{k+1} \ \dots \ a_{(t-1)k+1}) \dots (a_{k-1} \ a_{2k-1} \ \dots \ a_{tk-1})$$

each one being of order $t$.