Prove that if $A, B $ are two sub-groups of $G$, and $\gcd(|A|,|B|)=1$ then $A \cap B = ${${e_G}$}

Question

I need some perspective to know if my solution is correct.

I suppose there exist $a \neq e \in A \cap B$, then $a^{|A|}=e $ and $a^{|B|}=e $.

Considering that $|A| \neq 1$ and $|B| \neq 1$ (otherwise it's obvious), then $gcd(|A|,|B|)\neq 1$ $\longrightarrow$ contradiction.

What do you say?

Answer 1

From $|A|\neq1\wedge|B|\neq1$ it cannot be concluded that $\gcd(|A|,|B|)\neq1$. For instance take $|A|=2$ and $|B|=3$

Contradiction is not necessary.

If $a\in A\cap B$ then $a$ has an order that divides $|A|$ and $|B|$, hence divides $\gcd(|A|,|B|)=1$.

So $a$ must have order $1$ and consequently $a=e$.

Proved is now that $A\cap B=\{e\}$.

(not $A\cap B=e$ as you write).

Answer 2

As I mentioned in a comment, it's simpler to do this proof as a direct proof instead of a contradiction. Here's how it would go:

Suppose that $g\in A\cap B$. Then, $g\in A$ and $g\in B$. By Lagrange, $o(g)\mid|A|$ and $o(g)\mid|B|$. Therefore, $o(g)|\gcd(|A|,|B|)$. Since $\gcd(|A|,|B|)=1$, $o(g)\mid 1$. Therefore, $o(g)=1$ and so $g^1=e$. Therefore, $g=e$. Hence $A\cap B=\{e\}$.