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..
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 ${\rm lcm}(r,s)$.