Let $G$ be a group of order $30$ and $A, B$ be two normal subgroups of $G$ of order $2$ and $5$ respectively. Show that $G/AB$ contains $3$ elements. The actual question is

Let $G$ be a group of order $30$ and $A, B$ be two normal subgroups of $G$ of order $2$ and $5$ respectively. Show that $G/AB$ contains $3$ elements.

In my approach:
Firstly I know that $AB$ is a normal subgroup in $G$.
And, since order of $B$ is $5$, a prime that implies it is a cyclic subgroup of $G$.
And, since order of $A$ is $2$, it contains the identity element and the elements of $G$ which are self inverse.
And I also know,
$$ G/AB = \{ gAB : g \in G\} $$
Now from this much information how can I conclude that $G/AB$ has only 3 elements? I know One element must be the identity. I am lost when trying to find the other 2 elements.
It would be great to know a hint for the proof!

Another approach:
Alternatively, you want to show that $|AB|=10$, then it will follow that $|G/AB|=|G|/|AB|=3$.
You may use the product formula $$|AB|=\frac{|A||B|}{|A\cap B|}.$$
Try to use Lagrange's Theorem and properties of $\gcd$ to show that $|A\cap B|=1$.
