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 $.