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.

Question

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!

Answer 1

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 $|A\cap B|=1$.

Answer 2

Hint

It suffices to prove $|AB|=10$. But by the second isomorphism theorem (diamond theorem), $AB/B\cong A$.