Order of a Group Based on Their Exponents 
Let $G$ be a finite commutative group. The exponent of $G$ is the least common multiple of all orders of all the elements of $G$. Show that $G$ has an element whose order equals the exponent of $G$.

Here is the question that I have been stuck on for the past 2 days. I haven't been able to come up with anything useful that would help show this result is true. Even my two good math friends, through a collaborative effort, weren't able to figure this question out... What do the orders of $G$ have to do with the least common multiple of the exponent of $G$?
 A: Okay, I think I finally got it. There may be some things for you to fill in, depending on how much you know, but here's an outline.
Suppose $|G|=p_1^{n_1}\ldots p_k^{n_k}$, then there exists subgroups $P_1,\ldots,P_k$ each of order $p_i^{n_i}$ (Sylow subgroups), and furthermore $G=P_1\times\ldots\times P_k$. Now for each $i$ take an element $g_i\in P_i$ with the highest order $p_i^{m_i}$, then we claim that the element $g_0=g_1\ldots g_k$ has order $p_1^{m_1}\ldots p_k^{m_k}=\text{lcm}(|G|)$.
A: Here is an argument that reduces the problem to when the exponent is a prime power.
Let $e$ be the exponent of $G$. If $e=ab$ with $\gcd(a,b)=1$ and $a,b >1$, let $A=\{ g \in G : g^a = 1 \}$ and $B=\{ g \in G : g^b = 1 \}$. Then by induction you have elements $\alpha \in A$ of order $a$ and $\beta \in B$ of order $b$. Then $\alpha \beta$ has order $ab=e$.
A: let $f$ be an element of maximal order $m$ in the abelian group $G$ and let $e$ be the exponent of $G$. Then $m|e$. If $m$ is strictly less than $e$ there must be another element $g$ of order $n$ with LCM$(m,n) \gt m$ but now the order of $fg$ is LCM$(m,n)$ contradicting the assumption that $f$ has maximal order.
