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

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

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    $\begingroup$ Easier: $H$ is a subgroup of $HK$, hence $|H|$ divides $|HK|$, and likewise $|K|$ divides it, hence also $\mathrm{lcm}(|H|,|K|)$ divides it. $\endgroup$ – Martin Brandenburg Aug 12 '13 at 12:35
  • $\begingroup$ @Martin Brandenburg: Yes, you are right. $\endgroup$ – Boris Novikov Aug 12 '13 at 12:37
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    $\begingroup$ Or perhaps use my comment on the question. $\endgroup$ – Gerry Myerson Aug 12 '13 at 12:49
  • $\begingroup$ @Gerry How exactly would you use the structure theorem? $\endgroup$ – Leo Aug 12 '13 at 12:59
  • $\begingroup$ @Leo, too long for a comment. Post it as a question, and someone will show you how to do it. $\endgroup$ – Gerry Myerson Aug 12 '13 at 13:02

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