$G$ 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?

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If this is homework, would you add the homework tag? –  Gerry Myerson Apr 22 '12 at 12:56
no, it's not homework. –  Mohan Apr 22 '12 at 12:58

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 $y$.
I'm using the following definition of exponent: It is the lcm of order of all the elements in $G$. Then why should $x$ has order $|G|$? –  Mohan Apr 22 '12 at 13:33
I don't follow this part: "by the definition of the exponent, $x$ has order $|G|$". My definition of exponent is the largest $k$ such that $g^k$ for all $g \in G$, but Wikipedia differs... –  lhf Apr 22 '12 at 13:43
First prove that if there are elements of relatively prime orders $a,b$ then there is an element of order $ab$.