$G$ is a finite abelian group, $G$ is cyclic if and only if order of $G$ is equal to exponent of $G$ Let $G$ be a finite abelian group.
How do I prove if exponent of $G$ is equal to order of $G$ , $G$ is cyclic?
 A: First prove that if there are elements of relatively prime orders $a,b$ then there is an element of order $ab$. 
A: For the first implication, assume that $G$ is cyclic so there is some $x \in G$ that generates $G$. Every element of $G$ can be written as $x^n$ for some $n \in \mathbb Z$ and those elements satisfy $$(x^n)^{|G|} = (x^{|G|})^n = 1^n = 1,$$
so $\exp(G)$ is at most $|G|$. On the other hand, the order of $x$ is $|G|$, so the exponent of $G$ cannot be smaller than $|G|$. This shows that $\exp(G) = |G|$.
Edit: My previous proof contained a mistake but this one should be correct:
Assume that the exponent of $G$ is $|G| = p_1^{e_1}\cdots p_r^{e_r}$. The exponent is defined as the $\operatorname{lcm}$ of the orders of all elements of $G$, so for each prime factor $p_i$ there must be an element $x_i \in G$ whose order is divisible by $p_i^{e_i}$, otherwise the exponent would not be divisible by $p_i^{e_i}$. If $x_i$ has order $p_i^{e_i} m_i$ then $y_i := x_i^{m_i}$ has order $p_i^{e_i}$. It is an easy exercise to show that $y := y_1 \cdots y_r$ has order $p_1^{e_1}\cdots p_r^{e_r} = |G|$, hence $G$ is generated by $y$.
