$N$ and $M$ are normal subgroups, whose sizes are coprime. Show $|MN|=|M||N|$

Question

Let $N,M$ be normal subgroups of $G$. Let $MN=\{mn \ | \ m\in M,n\in N\}$. Suppose $|N|,|M|$ are coprime integers. Show that $|MN|=|M||N|$.

I have shown that $MN$ is also a normal subgroup of $G$.

I know that $|N|,|M|$ and $|MN|$ divide $G$.

I observe that $N,M\subset NM$ so $|N|,|M|$ divide $|MN|$. Which leads to $$|MN|=k_1|N|=k_2|M| \ \text{for some } k_i\in \mathbb{Z}$$ so $|MN|=k|N||M|$ where $k\in \mathbb{N}$. I can see this intuitively but I am not entirely sure how to justify this.

Lastly I think $|MN|\leq|N||M|$ since $NM$ would be largest if every combination of $nm$ produces a distinct element. Which should combined with the above should produce the result.

Am I correct, how can I justify the last $2$ points more formally? Is there anything "obvious" or elegant that I have missed?

Answer 1

For the first one, it follows that $k_1 = k|M|$ as the cardinalities are coprime and this tells you that $|MN| = k_1|N| = k|M||N|.$

For the second, by definition $|MN| = |\{mn: m\in M, n\in N\}|\leq |M||N|.$

Answer 2

Let $|M|=m$ and $|N|=n$. We know that $$|MN| = \frac{|M| \cdot |N|}{|M \cap N|}$$ But if $\gcd(m,n) = 1$ then $|M \cap N| = 1 $ because $|M \cap N| \mid |M| = m$ and $|M \cap N| \mid |N| = n$ by Lagrange. So $$|MN| = |M| \cdot |N| = mn$$