Prove that if $|a|=m $ and $|b|=n$ with gcd(m, n)=1 then $\langle a \rangle \cap \langle b \rangle = \{e\}$

Question

$G$ is a group and $a, b, \in G$.

So summarizing the question, if the order of $a$ and $b$ are relatively prime, then the cyclic group generated by $a$ and $b$ will only have the identity element in common.

I'm not sure how to proceed. What should I start with?

Answer 1

Hint: $<a>\cap <b>$ is subgroup of both group and order of a subgroup dividis order of a group.

Answer 2

There exists $u,v\in Z$ such that $1=um+vn$ (By Bezout Theorem).

Let $x\in <a>\cap <b>$, then $x=a^k$ , and $x=b^l$ for some $k,l\in \Bbb N$.

$x=x^{1}=x^{mu+nv}=x^{mu}x^{nv}=(a^k)^{mu}(b^l)^{nv}=a^{muk}b^{nvl}=(a^m)^{uk}(b^n)^{vl}=e$.