# Help needed in proving a theorem on a permutation that is the product of dis. cycles of prime length.

So we were given the following proof to do:

Let $p$ and $q$ be distinct primes. Suppose $\alpha$ is a permutation of $S_n$ and suppose $\alpha = \gamma_1 \gamma_2$ where $\gamma_1$ and $\gamma_2$ are disjoint cycles of length $p$ and $q$, respectively. Prove that if $\alpha^m$ is a cycle of length $t$ and gcd$(m, p) = 1$, then $p$ divides $t$.

I think I have come up with one that works, but I don't use a part of the hypothesis so I'm concerned I'm missing something.

If anyone could help me correct my proof it would be much appreciated:

$i.$ As $\gamma_1$ is a cycle of length $p$, it is likewise a cycle of order $p$. Therefore, because the gcd$(m,p)=1$, the order and length of $\gamma_1^m$ must also be $p$.

$ii.$ $\alpha^m$ may be written as a product of the disjoint cycles $\gamma_1^m$ and $\gamma_2^m$ as $\alpha^m = (\gamma_1 \gamma_2)^m = \gamma_1^m \gamma_2^m$. So the order of $\alpha^m$ is the least common multiple of the lengths of $\gamma_1^m$ and $\gamma_2^m$. From $i.$ we know that the length of $\gamma_1^m$ is $p$, and so the least common multiple of $\gamma_1^m$ and $\gamma_2^m$ must be divisible by $p$.

If $\alpha^m$ is a cycle of length $t$, then it is a cycle of order $t$, and so $t =$ lcm($\gamma_1^m$, $\gamma_2^m$).

Therefore $t$ is divisible by $p$.

-
Thanks for letting me know. I didn't know we were supposed to. –  mkeachie Oct 12 '12 at 22:25

We are given that the $p$- and $q$-cycles are disjoint in $\alpha$, so $\alpha^m=(\alpha_p\alpha_q)^m=\alpha_p^m\alpha_q^m$. Since $p$ is prime, $\alpha_p^m=1$ if and only if $p$ divides $m$. In particular, $\alpha_p^m$ is a $p$-cycle whenever $p$ does not divide $m$, whence $\alpha^m=\alpha_p^m\alpha_q^m$ has order divisible by $p$.