# Order of a permutation is the lcm of the orders of cycles

Let ##\sigma \in S_n##, and ##|\sigma| = r##. Then, we'll assume that ##\sigma## can be decomposed into a product of disjoint cycles, which commute with each other. So ##\sigma = c_1c_2 \dots c_k##. Then ##\sigma^r = (c_1c_2 \dots c_k)^r## = 1. Since the cycles commute, we have ##c_1^rc_2^r \dots c_k^r = 1##. Now, I am a little stumped. How can I conclude that ##c_1^r = c_2^r = \cdots = c_k^r = 1##?

Related Linear and Abstract Algebra News on Phys.org
fresh_42
Mentor
Let ##\sigma \in S_n##, and ##|\sigma| = r##. Then, we'll assume that ##\sigma## can be decomposed into a product of disjoint cycles, which commute with each other. So ##\sigma = c_1c_2 \dots c_k##. Then ##\sigma^r = (c_1c_2 \dots c_k)^r## = 1. Since the cycles commute, we have ##c_1^rc_2^r \dots c_k^r = 1##. Now, I am a little stumped. How can I conclude that ##c_1^r = c_2^r = \cdots = c_k^r = 1##?
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?

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?
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?

fresh_42
Mentor
Concentrate on ##n##, a number which is transformed by ##c_k##. What do the consecutive ##c_i\, , \,i<k ## to ##c_k(n)## resp. ##c_k^r(n)\,?## Can they reverse the transformation done by ##c_k\,?##

Concentrate on ##n##, a number which is transformed by ##c_k##. What do the consecutive ##c_i\, , \,i<k ## to ##c_k(n)## resp. ##c_k^r(n)\,?## Can they reverse the transformation done by ##c_k\,?##
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##.

fresh_42
Mentor
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##.
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)\,?##