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## Homework Statement

Let ##G## be a finite group such that ##|G| = nm##.

Suppose ##x\in G## has order ##n## and let ##\sigma_x\in S_G## be the permutation such that ##\sigma_x(g)=xg## for every ##g \in G##. Prove that ##\sigma_x## is a product of ##m## disjoint n-cycles.

## Homework Equations

## The Attempt at a Solution

Here is my attempt. Note that the cycle decomposition will include every element of ##G##, as if it did not, we would have that for some ##g \in G##, ##xg = g \implies x=1##, which clearly is a contradiction.

Now, suppose that ##a \in G##. Then the cycle that contains ##a## must be ##(1,a,xa,x^2a,\dots, x^{n-1}a)##. Now, consider ##b\in G## s.t. ##b\not = a##. Then either ##b\in \langle x \rangle a## or ##b \not \in \langle x \rangle a##. Suppose the latter, since the former would just result in the same cycle. Then the unique cycle that contains ##b## must be ##(1,b,xb,x^2b,\dots, x^{n-1}b)##. The cycle with ##b## and the cycle with ##a## are disjoint, since if they weren't ##x^ia = x^jb## such that ##i,j\in [0,n)##, and so ##b = x^{j-i}a \implies b\in \langle x \rangle a##, which is a contradiction.

So, every cycle in the cycle decomposition must have n elements and must be disjoint with all other cycles. We also noted that ##\sigma_x## must map all elements, so there must be ##m## n-cycles for this to be the case, since ##|G| = mn## QED

Is this proof correct? I feel like in parts I am hand-wavy, but I don't see how else to argue it.

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