Question

Abstract algebra proof of the order of cycles. Details are in attached document. Abstract algebra proof of the order of cycles. Details are in attached document. @Mathematics

Answer 1

\[Let \beta = (a _{1}, a _{2},...a _{n}) \] be a cyclic group. Prove that the order of \[\beta\] is n. Let \[\beta \] and \[\alpha\] be disjoint cycles of lengths n and m respectively. Prove that the order of beta times alpha is lcm(n,m)

Answer 2

not much for the first one. clearly
\[\beta^n=e\] and there is no smaller order since if
\[\beta^m\] sends
\[a_i\] to
\[a_{i+m}\] or more precisely
\[i+m(\text { mod }n)\] as for the second one, put
\[l=lcm(m,n)\] then since
\[m|l, n|l\] you have
\[(\alpha \beta)^l=\alpha^l\beta^l=e\] and now put order so order of
\[(\alpha \beta)\] divides l

now let the order of
\[(\alpha\beta)= j\] then
\[(\alpha \beta)^j=e\] so
\[\alpha ^j\beta^j=e\] and since
\[\alpha, \beta\] are disjoint, also
\[\alpha^j, \beta^j\] are disjoint, and since
\[(\alpha^j)=\beta^{-j}\]
this means
\[\alpha^j=\beta^{-j}=e\]
therefore
\[m|j, m|j\] and so
\[j=lcm(n,m)\]