# 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 $\operatorname{lcm}(n,m)$?

• Counterexample to the title: Consider $G=\mathbb{Z}_{10}$, $H=\{0\}$, $K=\{0,5\}$ – Amr Aug 12 '13 at 12:08
• @Amr What do you mean? $|H|=1$, so $K$ is a subgroup of order $\operatorname{lcm}(|H|,|K|)$. – user714630 Aug 12 '13 at 12:10
• @Amr I don't see the problem - $K$ is a subgroup of order $2=\operatorname{lcm}(1,2)$. – mdp Aug 12 '13 at 12:10
• @Matt Pressland I am sorry. I thought the title said "...., then $G$ has order $lcm(m,n)$" – Amr Aug 12 '13 at 12:11
• Oops, I left a comment/flag saying this question is a duplicate of something it is not a duplicate of. – user714630 Aug 12 '13 at 12:24

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 ${\rm lcm} (|H|,|K|)$.
• Easier: $H$ is a subgroup of $HK$, hence $|H|$ divides $|HK|$, and likewise $|K|$ divides it, hence also $\mathrm{lcm}(|H|,|K|)$ divides it. – Martin Brandenburg Aug 12 '13 at 12:35