How to show the existence of a subgroup of an abelian group?? If an abelian group has subgroups of orders m and n, respectively, then show that it has a subgroup whose order is the least common multiple of m and n.
I tried solving this on assumption that if an abelian group has a subgroup of order, say, N then it has an element whose order is also N. This leads me to the point where I arrive at two different elements having orders m and n as in the question, whereby I obtain a third element whose order is the l.c.m of m and n. 
Now, was my assumption correct in the first place?? If it's wrong, please let me know and help me get this solution right.

Let $U$ and $V$ be the subgroups of order $m$ and $n$ respectively, where the prime decomposition of $m$ and $n$ is given by, 
$$
m = \prod_{i=1}^K p_i^{\alpha_i}\:\: 
\text{and}\:\: 
n = \prod_{j=1}^L p_j^{\beta_j}
$$ 
For any prime $p$  such that $p|m$ or $p|n$ or both, consider the maximum power of $p$ that divides either $m$ or $n$ or both,   let that be $\gamma(p)$. Let $S$ be set of all such primes. Then,
$$lcm(m,n) = \prod _{p \in S} p^{\gamma(p)}$$ 
For any  $ p \in S$, there will be a corrsponding  p-syllow  subgroup either in $U$ or $V$ or both, let that subgroup be $H_p$ and $o(H_p) = p^{\gamma(p)}$. Now the desired subgroup of order $lcm(m,n)$ is direct product of all such $H_p$,
$$H = \prod _{p\in S} H_p$$
