Existence of an Element with a particular order I've solved the following problem in what seems to me to be a very inelegant way:

Let $G$ be an abelian group containing elements $a$ and $b$ of orders $m$ and $n$, respectively, where $m$ and $n$ are not necessarily relatively prime.  Show that $G$ must contain an element $c$ whose order $\mathcal{O}(c)$ equals the least common multiple of $m$ and $n$.

My solution is the following:

Let $d=\gcd(m,n)$ and note that $m=dk$, $n=dl$ where $k$ and $l$ are relatively prime. Now, define
      $$
      r=\prod_{p\mid d, p\mid k}p^i,
    $$
      where $p^i\mid d$ but $p^{i+1}\nmid d$. Now, by construction, since $k$ and $l$ are relatively prime, $r$ and $l$ are relatively prime. Likewise, since $r$ takes all of the primes in $d$ that are in common with $k$, we know $r$ and $\frac{d}{r}$ are relatively prime. Now we have
      \begin{align}
        &m=dk=r\left(\frac{d}{r}\right)k\\
        &n=dl=r\left(\frac{d}{r}\right)l
    \end{align}
      Clearly the element $a^{d/r}$ has order $rk$ and the element $b^{r}$ has order $\frac{dl}{r}$. Now the orders are relatively prime, so from a previous problem (where we showed that elements with rel. prime orders has the property that their product has the order of the product of their individual orders), we know the order of $a^{d/r}b^{r}$ is $dkl$, which by definition is the least common multiple of $m$ and $n$.

This doesn't strike me as being an elegant proof, so I'm wondering if there are any other ways to prove the statement?
 A: An alternative way is to do the prime fiddling inductively as follows.
LEMMA $\ $ A finite abelian group $\rm\:G\:$ has an  lcm-closed  order set, i.e. 
$$\rm X,Y \in G\ \ \Rightarrow\ \ \exists\ Z \in G:\  o(Z) = lcm(o(X),o(Y)),\,\ \ o(X) =\: order\ of\ X$$
Proof $\ \ $  By induction on  $\rm\, o(X)\ o(Y).\ $ If it is $\,1\,$ then $\rm\:Z = 1\, $ works. $\ $ Otherwise
write  $\rm\ o(X)\ =\ AP,\: \ \ o(Y) = BP',\ \ \ P'\mid P = p^m > 1,\ \ $  prime $\rm\: p\:$ coprime to $\rm\: A,B.$
Then  $\rm\: o(X^P) = A,\ \ o(Y^{P'}) = B.\ $  By induction there is a $\rm\: Z\:$ with $\rm \: o(Z) = lcm(A,B)$
so  $\rm\ \ o(X^A\: Z)\: =\: P\ lcm(A,B)\: =\: lcm(AP,BP')\: =\: lcm(o(X),o(Y)).\ \ $ QED
A: Let $m=dm'$ and $n=dn'$ with $gcd(m',n')=1$.
Define 
$$f: Z \times Z \to G \,;\, f(k,l)=a^{kd}b^{ld} \,.$$
We claim $\ker(f)=m'Z \times n'Z$. Since $\frac{Z \times Z}{m'Z \times n'Z} \sim Z_{m' n'}$ pick a generator $a^{id}b^{jd}$ for $Im(f)$ and then look to $a^ib^j$.
We prove now the claim
$$f(k,l)=0 \Rightarrow a^{kd}b^{ld}=e \,.$$
Thus, since $m',n'$ are relatively prime
$$b^{n'dl}=e \Rightarrow dm'|dln' \Rightarrow m'|ln' \Rightarrow m'|l \,,$$
$$a^{m'dk}=e \Rightarrow dn'|dkm' \Rightarrow n'|km' \Rightarrow n'|k \,,$$
A: Order of $ab$ need not be equal to the lcm of $|a|$ and $|b|$ if the gcd is not relatively prime. For example consider $G=\langle x\rangle$ with $a=x$ and $b=x^3$. Here $|a|=4, |b|=4$ but $|ab|=1$.
Unless $\gcd(|a|, |b|)=1$ is given we can't say that order of $ab$ is same as lcm$(|a|, |b|)$
