you are correct I cant see any problem in your solution
prove the following property of cyclic
groups: Let G = <x> be a finite cyclic group.
Prove that <x^m> intersection <x^n> = <x ^ LCM(m,n)>
difficulty is that I have to account for situations where for example
When |G| = 12, we have for example x^44 is in <x^20> intersection <x^14>= <x^140>
attempt:
Prove that <x^m> intersection <x^n> = <x^ (LCM,m,n)>
let b be be in <x^m> intersection <x^n>
then b = x^mk=x^nt for some k, t in Z
then the exponent of b is a multiple of both m and n thus it's a multiple of Lcm(m,n), therefore b is in <x^LCM(m,n)
now let b be in <x^LCM(m,n)>
then b = x^LCM(m,n) *k for some k in Z
Now LCM(m,n) = mt=nr
then b = x^mtk=x^nrk thus b =x^m(tk)=x^n(rk) for some t,r in Z
exponent of b is a multiple of both m & n so b is in the intersection of <x^m> and <x^n>
I was told this is a quite difficult question
and this seems too easy , which makes me think I'm missing something..
thanks, Adam