Help with groups

Question

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

Answer 1

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 $x^{m} = e = x^{n}$.

Answer 2

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

Answer 3

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 $|K|=o(b)$.

Answer 4

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?