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Before I start, please note that this post is not duplicate

Let $G$ be an abelian group.

Let $H,K$ be finite cyclic subgroups of $G$ such that $|H|=r,|K|=s$.

Then, how do I prove that there exists a cyclic subgroup of $G$ of order $lcm(r,s)$, using only basic techniques?

I could prove this by applying the Fundamental theorem of finitely generated abelian group, but I got this question from a freshman and this exercise is at the chapter introducing finding orders of subgroups of a cyclic group.

Let $d=gcd(r,s)$ and set $r=da, s=db$ and $h,k$ be generators of $H,K$ respectively.

Then, the order of $h^s$ is $a$.

I chekced up a solution, and it was written there that the order of $h^s k$ is $lcm(r,s)$, but I don't get this and I think this is false.

Why is $(h^s k)^l\neq e$ when $l≦lcm(r,s)$?

I have tried several ways to prove this, but I am stuck at showing $|H\cap K|=gcd(r,s)$ in every way if not using Fundamental theorem of finitely gerated abelian geoups..

Rubertos
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  • Can this have something related to the fact that Converse of Lagrange's Theorem holds for Finite Abelian Groups? I mean; G has an element of order lcm(r,s) (since r,s both divides the order of G and lcm will also divide it) and so it has a cyclic subgroup of order lcm(r,s). – S.B. Apr 01 '15 at 07:38
  • @smyr I don't get you. The converse of Lagrange's theorem is that there exists a subgroup $H$ of $G$ such that $|H|=lcm(r,s)$, not that $H$ is cyclic. – Rubertos Apr 01 '15 at 07:43
  • I mean if there are elements g,h in the group, there is an element of order lcm(|g|,|h|), which generates a cyclic group of order lcm(|g|,|h|). – S.B. Apr 01 '15 at 07:50
  • @smyr Well, isn't that what I asked for? Anyway, I don't know a way to prove that without FTFGAG. If you have one, would you please write that as an answer? – Rubertos Apr 01 '15 at 07:53
  • You can certainly prove this without using FTFGAG. As a hint, first consider the case when $r$ and $s$ are coprime. In that case, it is true that ${\rm Ord}(rs) = {\rm Ord}(r){\rm Ord}(s)$. Can you prove that? When you have done that, you can do the general case by replacing $h$ by a suitable power with order coprime to $g$. – Derek Holt Apr 01 '15 at 08:01
  • @DerekHolt Yes, it can be easily proven if $gcd(r,s)=1$, but I don't know how to prove in general. Let $d=gcd(r,s)$ and $r=da, s=db$. Note that the order of $h^s$ is $a$. Since $abd=lcm(r,s)$, following your argument, the suitable power of $h$ is $s$. However, there is no reason that $gcd(a,s)=1$. I have tried to prove in your way, but as I said in my post, it seemed inevitable to show that $|H\cap K|=gcd(r,s)$ – Rubertos Apr 01 '15 at 08:38
  • @Derek Holt, if you take $r:=18$ and $s:=12$ then $d=6$, $a=3$ and $b=2$ but neither $a$ is coprime to $db$ nor $b$ coprime to $da$, so you cannot get the order directly from the coprime orders lemma... – Clément Guérin Apr 01 '15 at 09:59
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    Sorry, I oversimplified things! See my answer below. – Derek Holt Apr 01 '15 at 11:17

1 Answers1

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Let $p_1,\ldots,p_k$ be the distinct primes dividing $r$ or $s$, and let $r=p_1^{r_1}\cdots p_k^{r_k}$, $s = p_1^{s_1}\cdots p_k^{s_k}$ (where some of the $r_i$, $s_i$ may be $0$). So ${\rm lcm}(r,s) = p_1^{m_1}\cdots p_k^{m_k}$, where $m_i = \max(r_i,s_i)$.

Reorder the primes if necessary so that $r_i \ge s_i$ for $1 \le i \le j$ and $s_i > r_i$ for $i< j \le k$. Then $h^a$ has order $p_1^{r_1}\cdots p_j^{r_j}$ (where $a=p_{j+1}^{r_{j+1}}\cdots p_k^{r_k}$) and similarly some power $k^b$ of $k$ has order $p_{j+1}^{s_{j+1}}\cdots p_k^{s_k}$, so $h^a$ and $k^b$ have coprime orders, and their product $h^ak^b$ has the required order ${\rm lcm}(r,s)$.

Derek Holt
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  • Could you elaborate a little bit on why it follows that $h^ak^b$ has the desired order? At first I thought it's clear, but then I started wondering why some relation of sort $h^2=k^3$ can't break the result. – Wojowu Apr 06 '15 at 06:26