Help with groups let $G$ be a finite  group with $e$ Identity element and let $a$ and $b$ belong to $g$
prove that if: $\gcd(o(a),o(b)) =1$
then $\langle a \rangle \cap \langle b \rangle = \{ e \}$.
if someone can give me any direction
or and idea or something it will help me a lot
thanks
 A: Hint 1

If $o(a) = m$ and $o(b) = n$, then there are $u, v$ such that $m u + n v = 1$.

Hint 2

$o(a) = \lvert \langle a \rangle \rvert$ and $o(b) = \lvert \langle b \rangle \rvert$.

Hint 3

If $x \in \langle a \rangle \cap \langle b \rangle$, then $x^{m} = e = x^{n}$.

A: Hint: Take $x \in \langle a \rangle \cap \langle b \rangle $. Consider $o(x)$ and use Lagrange's theorem twice.
Full solution:

 $o(x)$ divides both $o(a)$ and $o(b)$ and so must divide $\gcd(o(a),o(b)) =1$

Actually, you don't need Lagrange's theorem:

 You just need $x^k=e \implies o(x) \mid k$

A: If you have learned Lagrange's Theorem (the order $|H|$ of a subgroup $H$ must divide the order of the whole group $G$), then observe this: if $H$ and $K$ are subgroups of $G$, with $\text{gcd}(|H|,|K|)=1$, then $H \cap K=\{e\}$. Why? $H \cap K$ is a subgroup of both $H$ and $K$, so its order must divide the order of $H$ and that of $K$. In your special case take $H=\langle a\rangle$ and $K=\langle b\rangle$ and note that $|H|=o(a)$ and $|K|=o(b)$.
A: Use Lagrange's theorem to prove the following useful fact:

The order of $x^n$ divides the order of $x$.

Now imagine some element of $\langle a \rangle$ is equal to some element of $\langle b \rangle$. What happens?
