Lets say we have a finite group $G$ of order $m = m_{1} \cdot m_{2}$. Suppose we have elements $g_1, g_2 \in G$ such that $o(g_1) = m_1$ and $o(g_2) = m_2$. Let $N = \langle g_1 \rangle$ be a normal subgroup of $G$.
What is the condition on $m_1$ and $m_2$ for $o(g_2) = o(g_2 N)$? I basically want to preserve the order of $g_2$ in the quotient group $G/N$. I have some intuition as to why this will be the case for $\gcd(m_1, m_2) = 1$, but I can't arrive at a proof for the same.
Moreover, how can I extend this to prove that $\langle g_1 \rangle \cap \langle g_2\rangle = e$ i.e. they have trivial intersection?
It is clear from the definition that $$o(g_2) = o(g_2 N)\Longleftrightarrow \langle g_1 \rangle \cap \langle g_2\rangle = \{e\}$$
If you want a condition on the orders of $g_1$ and $g_2$, there is no necessary and sufficient one: indeed, in the products $\Bbb Z/n\times \Bbb Z/m$ you will find infinitely many situations where $\langle g_1 \rangle \cap \langle g_2\rangle = \{e\}$ with all possible couples of orders.
However, there is a sufficient condition:
$$\gcd(o(g_1),o(g_2))=1\Longrightarrow \langle g_1 \rangle \cap \langle g_2\rangle = \{e\}$$ This is because any element in the intersection has an order that divides both $o(g_1)$ and $o(g_2)$.