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 $H_0$ and $H_1$ be subgroups of size $m,n$ and let $H_2=H_0\cup H_1=\{h_1,h_2\dots h_s\}$ and suppose the elements have orders $o_1,o_2\dots o_s$ respectively .Then the set $H_3=\{h_1^{e_1}h_2^{e_2}\dots h_2^{e_s}| e_1,e_2\dots e_s\in \mathbb Z\}$ is a subgroup, because $G$ is abelian. It is also finite as it has at most $o_1o_2\dots o_s$ elements.
By Lagrange's theorem $m$ and $n$ divide $|H_3|$, so $lcm(m,n)$ divides $|H_3|$, so by the following lemma $H$ has a subgroup of order $lcm(m,n)$ (Notice $H_3$ is clearly abelian also).
Lemma: Let $G$ be a finite abelian group and $d$ a divisor of $|G|$, then $G$ has a subgroup of order $d$ (in other words, every abelian group is a converse lagrange group).
Proof:
The proof is by strong induction over $n$
If $|G|=1$ it is trivial. The inductive case is as follows:
If $d=1$ it is trivial, otherwise take $p$ a prime that divides $d$. By cauchy's theorem there is a subgroup $N$ of order $p$, because $G$ is abelian $N$ is normal, so we can consider the canonical projection $\varphi:G\rightarrow \frac{G}{N}$.
Now notice that $d/p$ divides the order of $\frac{G}{N}$, so by the induction hypothesis there is a subgroup $K\leq \frac{G}{N}$ of order $d/p$.
Consider the subgroup $H=\varphi^{-1}(K)$ what is it's order ? Consider the restriction of $\varphi$ to $H$, the kernel is $N$ and the image is $K$, we conclude $\frac{H}{N}\cong K$, so $H$ has order $p\frac{d}{p}=d$.
We have found a subgroup of order $d$.
