Let $G$ be abelian, $H$ and $K$ subgroups of orders $n$, $m$. Then G has subgroup of order $\operatorname{lcm}(n,m)$.

Let $G$ be abelian, $H$ and $K$ subgroups of orders $n$, $m$. Then G has subgroup of order $\operatorname{lcm}(n,m)$.

This is a statement that my lecturer mentioned in my (beginners') Abstract Algebra class. I'm not sure I understand why it's true.

What I have so far: Use the abelian group structure theorem on $\langle H, K\rangle$ (finite group generated by $H$ and $K$). Then $\langle H,K\rangle=C_{a_1}C_{a_2}\dotsm$ and $n, m |\langle H,K\rangle$. Which means it's also true that $\operatorname{lcm}(n,m)|\langle H,K\rangle$. Can I leverage this to say there's a subgroup of order $\operatorname{lcm}(n,m)$?

Solution 1:

Since $|H\cap K|$ divides $|H|$ and $|K|$, it divides ${\rm gcd} (|H|,|K|)$, so ${\rm gcd} (|H|,|K|)=a|H\cap K|$ for some integer $a$. Further, $$ |HK|=\frac{|H||K|}{|H\cap K|}=\frac{|H||K|a}{{\rm gcd} (|H|,|K|)}={\rm lcm} (|H|,|K|)a. $$ Now one must use the assertion: if $G$ is abelian and $n$ divides $|G|$ then $G$ has a subgroup of order $n$. Therefore $HK$ has a subgroup of order ${\rm lcm} (|H|,|K|)$.