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

mdp
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1 Answers1

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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 ${\rm lcm} (|H|,|K|)$.

Boris Novikov
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    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
  • @Martin Brandenburg: Yes, you are right. – Boris Novikov Aug 12 '13 at 12:37
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    Or perhaps use my comment on the question. – Gerry Myerson Aug 12 '13 at 12:49
  • @Gerry How exactly would you use the structure theorem? – Leo Aug 12 '13 at 12:59
  • @Leo, too long for a comment. Post it as a question, and someone will show you how to do it. – Gerry Myerson Aug 12 '13 at 13:02
  • @Gerry Will do. – Leo Aug 12 '13 at 13:04
  • @Gerry: Your proof only works when $G$ is finite. Note that Boris' proof also works when $G$ is infinite. Actually Boris' proof reduces to the finite case by considerung $HK$. – Martin Brandenburg Aug 12 '13 at 13:19
  • @Leo: It suffices to consider finite abelian $p$-groups. These are isomorphic to $\mathbb{Z}/p^{k_1} \oplus \dotsc \oplus \mathbb{Z}/p^{k_s}$ for some natural numbers $k_1 \leq \dotsc \leq k_s$. Every divisor of the order $p^{k_1+\dotsc+k_s}$ has the form $p^{k'_1+\dotsc+k'_s}$ for some $k'_i \leq k_s$. Now it is easy to write down a subgroup of this order. – Martin Brandenburg Aug 12 '13 at 13:25
  • Since $\langle H,K \rangle$ is finite, and we can assume that $G = \langle H,K \rangle$, there is no need to assume that $G$ is finite. (In other words, we don't need to assume it because we can assume it!) – Derek Holt Aug 12 '13 at 17:28