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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 $.

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

up vote 0 down vote accepted

Your proof is correct, but a bit repetitive. I'll throw in my version:

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 $p$.

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