Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Calculus and Beyond Homework Help
Is m the smallest order for an m-cycle in the symmetric group?
Reply to thread
Message
[QUOTE="Mr Davis 97, post: 6047717, member: 515461"] [h2]Homework Statement [/h2] Let ##n## be a natural number and let ##\sigma## be an element of the symmetric group ##S_n##. Show that if ##\sigma## is an m-cycle ##(a_1a_2 \dots a_m)##, then ##|\sigma|=m## [h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2] First, we want to show that ##\sigma ^m = id##. To this end, we claim that for all ##i \in \mathbb{N}##, ##\sigma ^i (a_k) = a_{(k+i) \bmod m}##. We proceed by induction. The base case holds by the fact that ##\sigma## is an m-cycle. Next, suppose that for some ##j## we have ##\sigma ^j (a_k) = a_{(k+j) \bmod m}##. Then ##\sigma ^{j+1} (a_k) = \sigma (\sigma ^{j} (a_k)) = \sigma (a_{(k+j) \bmod m}) = a_{(k+(j+1)) \bmod m}##. So we have shown that for all ##i \in \mathbb{N}##, ##\sigma ^i (a_k) = a_{(k+i) \bmod m}##, and if we let ##i=m##, we see that ##\sigma ^m = id##. Second, we want to show that ##m## is the smallest positive integer such that ##\sigma ^m = id##. To the contrary, suppose that there is a ##p \in [1, m)## such that ##\sigma ^p = id##. Then ##\sigma ^p (a_p) = a_p## and also ##\sigma ^p (a_p)= a_{(p+p) \bmod m} = a_{(2p) \bmod m}##. So ##p \equiv 2p \bmod m \implies p \equiv 0 \bmod m##, which contradicts the assumption that ##p \in [1, m)##. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Is m the smallest order for an m-cycle in the symmetric group?
Back
Top