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- Thread starter Mr Davis 97
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- #2

fresh_42

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Let's say we have the following order: ##n = \sigma^r (n)= (c_1c_2 \dots c_k)^r(n)= (c_1^r(c_2^r( \dots (c_k^r)(n)\ldots )##. Now assume ##n## is one of the numbers, which are transformed by ##c_k##. What can you say about ##c_k^r(n)## and why?

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Well either ##c_k^r = 1## or ##c_k^r = (c_1^rc_2^r \dots c_{k-1}^r)^{-1} \ne 1##, but the latter is not possible because the cycles, their powers, and their inverses are all disjoint?Let's say we have the following order: ##n = \sigma^r (n)= (c_1c_2 \dots c_k)^r(n)= (c_1^r(c_2^r( \dots (c_k^r)(n)\ldots )##. Now assume ##n## is one of the numbers, which are transformed by ##c_k##. What can you say about ##c_k^r(n)## and why?

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fresh_42

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Well... If ##C_k^r = (c_1^rc_2^r \dots c_{k-1}^r)## and ##c_k^r## are disjoint, then if ##n## is transformed by ##c_k^r## then ##C_k^r## can't possibly transform ##c_k^r(n)## back to ##n## (since we know ##C^r_k c^r_k = 1##, ##C^r_k## would have to). So ##c_k^r = 1##.

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fresh_42

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You're thinking far too complicated. Say we have two disjoint permutations ##\sigma , \tau##. Now assume that ##\tau(n) = m##. So we have ##\sigma (\tau (n)) = \sigma (m)##. Do we already know the result of ##(\sigma \tau)(n)\,?##Well... If ##C_k^r = (c_1^rc_2^r \dots c_{k-1}^r)## and ##c_k^r## are disjoint, then if ##n## is transformed by ##c_k^r## then ##C_k^r## can't possibly transform ##c_k^r(n)## back to ##n## (since we know ##C^r_k c^r_k = 1##, ##C^r_k## would have to). So ##c_k^r = 1##.

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