Finite abelian group contains an element with order equal to the lcm of the orders of its elements

I will quote a question from my textbook, to prevent misinterpretation:

Let $G$ be a finite abelian group and let $m$ be the least common multiple of the orders of its elements. Prove that $G$ contains an element of order $m$.

I figured that, if $|G|=n$, then I should interpret the part with the least common multiple as $lcm(|x_1|,\dots,|x_n|)=m$, where $x_i\in G$ for $0\leq i\leq n$, thus, for all such $x_i$, $\exists a_i\in\mathbb{N}$ such that $m=|x_i|a_i$. I guess I should use the fact that $|x_i|$ divides $|G|$, so $\exists k\in \mathbb{N}$ such that $|G|=k|x_i|$ for all $x_i\in G$. I'm not really sure how to go from here, in particular how I should use the fact that $G$ is abelian.

Solution 1:

A finite abelian group can be written as a (finite) direct product of cyclic groups: $$ G=C_{m_1}\times C_{m_2}\times\dots\times C_{m_r} $$ where $C_n$ denotes a cyclic group of order $n$. Thus the order of any element in $G$ divides $\operatorname{lcm}(m_1,m_2,\dots,m_r)$. On the other hand, if $g_i$ is a generator of $C_{m_i}$, the element $$ g=(g_1,g_2,\dots,g_r) $$ has order precisely $\operatorname{lcm}(m_1,m_2,\dots,m_r)$.

Fill in the details.

Solution 2:

Hint : in an abeilan group, for any two elements $a,\, b$ of different order, there is an element in the group, whose order is lcm of order of $ a, \,b$