If ord$(a)=m$, ord$(b)=n$ then does there exist  $c$ such that ord $(c)=lcm(m,n)$? 
Possible Duplicate:
Order of elements in abelian groups 


Let $G$ be an abelian group and suppose that $G$ has elements of orders $m$ and $n$, respectively. Prove that $G$ has an element whose order is the least common multiple of $m$ and $n$.

I've attempted this problem for quite some time, but didn't seem to get anywhere.
First, let $a$ and $b$ be the elements whose orders are $m$ and $n$, respectively. I guessed that we can find the element of order $lcm(m,n)$ explicitly, instead of simply proving its existence. Furthermore, I also guessed that the element can be expressed in the form $a^kb^l$, because the statement must also hold when $G$ is generated by $a$ and $b$.
Then I let $k$ be the smallest positive integer such that $a^k$ is a power of $b$, say $a^k=b^l$. Then I proved that $l$ is also the smallest positive integer such that $b^l$ is a power of $a$, and that $ml=nk$. I'm not sure whether it's correct though.
Then I tried to find the order of ab. I can prove that the order is divisible by $\frac{lcm(m,n)}{\gcd(m,n)}$, but I can't prove whether it is equal to $lcm(m,n)$. Apparently, taking any $a^ib^j$ won't be any better. And now, I'm at wits end.
Please tell me whether I'm on the right path. If not, please give me some adequate hints so I can work on it.
 A: First we consider the case where $(m,n)=1$.
Because $(m,n)=1$ the least common multiple of $m$ and $n$ is $mn$. 
Consider the element $ab$.
Because $G$ is Abelian, $$(ab)^{mn} = \overbrace{ab\ ab\ ab \ldots ab}^{mn} = \overbrace{aaa\ldots a}^{mn} \overbrace{bbb\ldots b}^{mn} = a^{mn}b^{mn} = ee =e$$
so the order of $ab$ is at most $mn$.
Say the order of $ab$ is $k$, then we just showed that $k \leq mn$.
We see that $$e= (ab)^k = (ab)^{km} = a^{km}b^{km} = eb^{km} = b^{km} $$
so $n$ divides $km$. But since $(m,n)=1$ it follows that $n$ divides $k$.
Similarly  $$e=(ab)^k = (ab)^{kn} = a^{kn}b^{kn} = a^{kn}e = a^{kn} $$ so $m$ divides $kn$, and so $m$ divides $k$.
Thus $mn$ divides $k$.
And since $k \leq mn$ it follows that $k = mn$, and so $ab$ has order $mn$, the lowest common multiple of $m$ and $n$.
We need to take from this the fact that the order of a product of elements that have relatively prime orders is the product of the orders of those elements.
Now we consider the case where $m$ and $n$ are not relatively prime.
Call $L$ the least common multiple of $m$ and $n$.
We write $L=p^{r_1}_1\cdots p^{r_s}_s$ for distinct primes $p_i$ and strictly positive powers $r_i$.
If we could find an element of $G$ with order $p^{r_i}_i$ for every $i$, then by the first half of this proof, the product of these elements would have order $L$ because prime powers are all relatively prime to prime powers of different primes.
Let $i$ with $1 \leq i \leq s$ be given.
We note that $p^{r_i}_i$ divides either $m$ or $n$.
Thus $a^{m/p^{r_i}_i}$ or $b^{n/p^{r_i}_i}$ (whichever one divides evenly) has order $p^{r_i}_i$.
Therefore we have shown that an element of order $L$, the least common multiple of $m$ and $n$, is in $G$.
The statement is not necessarily true if $G$ is not Abelian.
Consider $G=S_3$.
This group contains a cycle of order 3, and a transposition (of order 2), but no element of order 6.
