Abelian group and their subgroups Is it true that If an abelian group has subgroups of order m and n respectively then it has a subgroup whose order is the least common multiple of m and n? If it is then can anyone explain it with a valid proof?
 A: Let $G$ be an abelian group with subgroups $H, K$ of orders $m, n$ respectively. Then $HK$ is a finite abelian group with $H$ and $K$ as subgroups, so its order is divisible by both $m$ and $n$, hence it is a mulitple of $(m, n)$.
The result will now follow from the following fact: If $G$ is a finite abelian group of order $k$ and $l | k$, then $G$ has a subgroup of order $l$.
Proof: We proceed by induction on $k$. If $k=1$ there is nothing to prove. Now suppose that the claim holds whenever $|G| < k$. If $l=1$, there is again nothing to prove (since the trivial group is a subgroup of every group), so suppose that $l >1$, and let $p$ be some prime dividing $l$. By Cauchy's theorem, there exists some $x \in G$ of order $p$. Since $G$ is abelian, $\langle x\rangle$ is a normal subgroup of $G$, and so we may consider the abelian group $G/\langle x \rangle$. This has strictly smaller order than $G$, so by the inductive hypothesis, it has a subgroup $\tilde{H}$ of order $l/p$.
Lastly, I will use without proof that there is a natural one-to-one correspondence $$\phi: \{K \leq G: x \in K\} \to \{\tilde{K} : \tilde{K} \leq G/\langle x\rangle\}$$ given by $\phi(K) = K/\langle x\rangle$ (it's very easy to check that $\phi$ is a bijection). Thus $\tilde{H}$ lifts (i.e. corresponds) to a subgroup $H = \phi^{-1}(\tilde{H})$ of $G$ containing $x$, and indeed, we have that $|H| = |\tilde{H}| \cdot |\langle x\rangle| = l$. This completes the proof.
