In an abelian Group in which there are 2 subgroups of order m and n there is a subgroup of order mcm(m,n), how to prove? This is an Herstein's exercise


  If an abelian group has two subgroups of order m and n, prove that it
  also has a subgroup of order lcm(m,n).


I've solved the exercise right before this, which is an analogue statement for elements


  If an abelian group has two elements of order m and n, prove that it
  also has a element of order lcm(m,n).


I would like some hints (just give me half an hour) before a complete answer.

I'm trying to demonstrate that, given $H&lt;G,K&lt;G; m=\#H, n=\#K$ then $\#HK=r=$ lcm$(m,n)$

I know (/ I can prove) that $$\#HK={{\#H\#K}\over{\#(H\cap K)}}$$; so I would need to show, if my intuition is right this time, that $\#(H\cap K)=$ gcd$(m,n)$.

@Simon

Yes, I know that $\#H\cap K|\#H $ and $\#H\cap K|\#K $, but I don't know how to prove maximality. It sort of involves a converse of lagrange theorem, maybe. But Herstein implicitly wrote it was not needed.



@Qil

Ok, I believe I've proved what it's called Abelian groups fundamental theorem. I'll just sketch it here, so you can tell we if I'am using the concepts correctly.

Step 1 - Let $G$ be a group, and $H$ a normal subgroup. Then we can define an homomorphism to $G/H$ and consequently the isomorphism $$G\rightarrow H \times G/H.$$
Step 2 - Now, let G be an abelian group, all subgroups are normal. Let us consider $$\alpha_1=\max_G \{k:o(g)=k\} $$$$g_1\in G : o(g_1)=\alpha_1$$We know that $\alpha_1|n$ for Lagrange; we can now write, using step 1, the following isomorphism $$G\rightarrow \langle g_1\rangle \times G/\langle g_1\rangle$$
Step 3 - We repeat step two, substituting $G$ with $G/\langle g_1\rangle$ and find a non-increasing sequence $\alpha_1,\cdots,\alpha_k$ such that $n=\alpha_1\cdots\alpha_k$; each $\alpha_i$ corresponding to a cyclic group of order $\alpha_i.$

Step 4 - We know that every cyclic group of order $\beta$ has a subgroup of order $\beta_k$ iff $\beta_k | \beta$. We've obtained a decomposition of $G$ into cyclic groups of prime order.

Merging those bricks together leads us to any subgroup of order $d$ when $d|n$.

But would Herstein want us to prove this for that exercise?
If $G$ is an abelian group of order $d$, then for any positive divisor $k$ of $d$, there is a subgroup of $G$ of order $k$. 
Apply (1) to $HK$: let $e=\# (H\cap K)$. Then $\# (HK)=mn/e$. As $e \mid \gcd(m,n)$ and $mn=\gcd(m,n)\mathrm{lcm}(m,n)$, we see that $\mathrm{lcm}(m,n) \mid \# (HK)$. So (1) implies that $HK$ has a subgroup of order $\mathrm{lcm}(m,n)$.
If $G$ is an abelian group of order $d$, then for any positive divisor $k$ of $d$, there is a subgroup of $G$ of order $k$. 
Apply (1) to $HK$: let $e=\# (H\cap K)$. Then $\# (HK)=mn/e$. As $e \mid \gcd(m,n)$ and $mn=\gcd(m,n)\mathrm{lcm}(m,n)$, we see that $\mathrm{lcm}(m,n) \mid \# (HK)$. So (1) implies that $HK$ has a subgroup of order $\mathrm{lcm}(m,n)$.