If primes $p_1 \neq p_2$ divide $n$, then $\exists$ $x, y \in G \setminus \{e\}$ such that $\langle{x}\rangle \cap \langle{y}\rangle = \{e\}$

Question

Q: If $G$ is a group with order $n$, and $p_1$ and $p_2$ are distinct primes that divide $n$, show that there exists $x, y \in G \setminus \{e\}$ such that $\langle{x}\rangle \cap \langle{y}\rangle = \{e\}$, where $e$ is the identity element.

I know that since $p_1$ and $p_2$ both divide $n$, i.e. $p_1 \mid n$ and $p_2 \mid n$, and if for some $x, y \in G \setminus \{e\}$ is such that $x = k_1^{n/p_1}$ and $y = k_2^{n/p_2}$ where $x$ would then have an order of $p_1$ and $y$ have order $p_2$ and as the two primes are distinct, then $\langle{x}\rangle \cap \langle{y}\rangle = \{e\}$.

I am not 100% sure if the method I am using to prove this statement is correct. Suggestions would be great.

Answer 1

That is correct. Let $H = \langle x \rangle \cap \langle y \rangle,$ then $H$ is a subgroup of both $\langle x \rangle$ and $\langle y \rangle.$ Thus, $|H|$ divides both $p$ and $q$ by Lagrange's Theorem, implying $|H| = 1.$

Answer 2

It's simpler than that. Since $p_1 \mid n, \exists x \in G ~(x^{p_1}=e)$. Similarly, since $p_2 \mid n, \exists y \in G ~(y^{p_2}=e)$.

Note that $\vert \langle x \rangle \vert = p_1, \vert \langle y \rangle \vert = p_2$, so $\vert \langle x \rangle \cap \langle y \rangle \vert$ must divide both $p_1$ and $p_2$. Since $p_1$ and $p_2$ are distinct primes, that can only mean $\vert \langle x \rangle \cap \langle y \rangle \vert = 1$, so that $\langle x \rangle \cap \langle y \rangle = \{ e \}$.