# 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 $H<G,K<G; m=\#H, n=\#K$ 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?

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Yes, I was using the italian acronyms. Sorry. –  Temitope.A Nov 2 '12 at 21:55
$H\cap K$ is a subgroup of both $H$ and $K$, don't know whether it helps, just a first idea. –  Simon Markett Nov 2 '12 at 21:57
Let $G=\mathbb Z/p\mathbb Z \times \mathbb Z/p\mathbb Z$, $H=\mathbb Z/p\mathbb Z \times \{ 0\}$ and $K= \{ 0\}\times \mathbb Z/p\mathbb Z$. What is the order of $H\cap K$ and what is $\gcd(\# H, \# K)$ ? –  user18119 Nov 2 '12 at 22:14
1,p. Ok, this is frustrating:D –  Temitope.A Nov 2 '12 at 22:25

1. 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 $k$.
2. 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 $\mathrm{lcm}(m,n)$.
Wait, isn't your example above a counterexample to using $HK$? –  Temitope.A Nov 3 '12 at 9:57
@Temitope.A: no, I don't think so. I meant apply (1) with $G=HK$. I don't say $HK$ has order lcm$(m,n)$. –  user18119 Nov 3 '12 at 10:45
@Temitope.A: your step 1 is incorrect. Look at the example $G=\mathbb Z/4\mathbb Z$ and $H=2\mathbb Z/4\mathbb Z$. The result you can use is any finite abelian group is a finite product of $\mathbb Z/p^r\mathbb Z$ for various prime numbers $p$ and various positive integers $r$. –  user18119 Nov 4 '12 at 17:11