A group $G$ may be generated by two elements $a$ and $b$ of coprime order and yet not be cyclic.
The simplest family of examples is that of the dihedral groups $D_n$ with $n$ odd. The group $D_n$ is defined to be the group of plane isometries sending a regular $n$-gon to itself and it is generated by the rotation of $2\pi/n$ radians and any axial symmetry. These two isometries have orders $n$ and $2$ respectively, yet they don't commute.
Instead, if $G$ is abelian and generated by two elements of coprime order, then $G$ is cyclic. This can be done in two steps:
(1) prove that if $G=\langle a,b\rangle$ with ${\rm ord}(a)=m$, ${\rm ord}(b)=n$ and ${\rm GCD}(m,n)=1$, then $G\simeq\Bbb Z/m\Bbb Z\times\Bbb Z/n\Bbb Z$;
(2) prove that $\Bbb Z/m\Bbb Z\times\Bbb Z/n\Bbb Z\simeq\Bbb Z/mn\Bbb Z$ if ${\rm GCD}(m,n)=1$.
Maybe it's worth reminding the structure theorem for finite abelian groups: A finite abelian group $G$ with $|G|=n$ is always isomorphic to a product $C_1\times C_2\times\cdots\times C_k$of cyclic groups with $|C_i|=e_i$,such that $e_1\mid e_2\mid\cdots\mid e_k$ and $\prod_{i=1}^ke_i=n$.
Best Answer
Is false. Take $G=\mathbb{Z}_2\oplus\mathbb{Z}_2$. Or more explicitely $G=\{e, a, b, c\}$ with addition
$$e\mbox{ is neutral}$$ $$a+a=e$$ $$b+b=e$$ $$c+c=e$$ $$a+b=c$$ $$a+c=b$$ $$b+c=a$$
Also known as the Klien four-group. This group is abelian. It is not cyclic because it is of order $4$ while every (nontrivial) element is of order $2$.
Finite (or even finitely generated) abelian groups are all known. If $G$ is a finite abelian group then
$$G\simeq\mathbb{Z}_{p_1^{a_1}}\oplus\cdots\oplus \mathbb{Z}_{p_m^{a_m}}$$
for some (not necessarily distinct) primes $p_1,\ldots, p_m$ and naturals $a_1,\ldots,a_m$.
For the finitely generated case you have to add $\mathbb{Z}^k$ term.
Abelian groups have this neat property: every subgroup is normal. Now if $G$ is of order $n$ then by Cauchy's theorem it has a subgroup $H$ of prime order $p | n$. This subgroup is nontrivial. It is proper if $n$ is not prime. If additionally $G$ is abelian then $H$ is normal, hence $G$ is not simple.
So the only possibility for an abelian group to be simple is when it is of prime order. Again Cauchy's theorem implies that such group has to be cyclic and indeed every $\mathbb{Z}_p$ is simple.