Suppose $G$ is an Abelian group then possible orders of $ab$ where $a,b\in G$ and are disjoint order elements. Kindly vet my attempt.
First of all, group is Abelian, so can be, at best a subgroup of $S_n, n\in \mathbb{N}$, say a cyclic group $\mathbb{Z_n}$ or product of disjoint cycles of such type.
Let $G$ be cyclic group of $n$ elements, and that also implies additive operation.Hence, $C_n\cong \mathbb {Z_n}.$
Next, let the two elements of disjoint orders be $|a|=x, |b|=y.$
An element $a$ with $|a|= x$ will have disjoint cycles as of lengths:
Say, $a= \mathbb{Z}_{x}, b= \mathbb{Z}_{y}.$
The abelian group forbids dihedral groups too, but can be a cyclic subgroup of $D_n$, for suitable natural $n$.
Wlog, $x\gt y.$
So, $|G|\ge x$, i.e. number of elements in the group is at least $x$.
Need ascertain the order of the set formed by product of two elements.
As order of one element is lesser than the other, so at least one of them is not a generator of the group $G$.
Need find the criteria to find order of product of cycles in the two elements.
Such example based on product of disjoint or single cycle, can be expected in symmetric groups , say take some abelian subgroup of $n=6, |S_6|= 6.5.4.3.2.1= 30.24= 720$.
Second question concerns how to interpret elements of group product, represented as tuples in Cartesian product.
Should such tuples(set of elements, one from each group in product)  be taken as pair of generators, if all elements in a tuple are generators?
Else, if not generators (either /both of them), then how to interpret?
 A: Your $a$ and $b$ as groups is wrong; instead, they are elements of $G$.
Here is a hint:
Since $G$ is abelian, we have for all $x,y\in G$, that
$$
(xy)^m=x^my^m$$
for any $m\in \Bbb Z$.
A: Theorem Suppose $G$ is an abelian group, $x,y \in G$, and the orders of $x$ and $y$ are finite. Then $|xy|$ is also finite and is a divisor $\operatorname{lcm}(|x|,|y|)$.
Proof Let $m = \operatorname{lcm}(|x|,|y|)$. Then $|x|$ and $|y|$ are both divisors of $m$, so $x^m = e$ and $y^m = e$. Since $G$ is abelian, $(xy)^m = x^my^m = ee = e$, so $|xy|$ must be a divisor of $m$.
The obvious way to apply this theorem to the problem is to let $x = a$ and $y = b$, and deduce that $|ab|$ is a divisor of $\operatorname{lcm}(60,36) = 180$.
However, we could also take $x = a^{-1}$ and $y = ab$, so that $xy = b$, and the theorem tells us that $36$ is a divisor $\operatorname{lcm}(60, |ab|)$. An integer is divisible by $36$ if, and only if, it is divisible by both $4$ and $9$. Since $60$ is divisible by $4$ but not $9$, we must have that $|ab|$ is divisible by $9$.
Similarly, we could choose $x = ab$ and $y = b^{-1}$, so that $xy = a$, and find that $60$ is a divisor $\operatorname{lcm}(|ab|,36)$. An integer is divisible by $60$ if, and only if, it is divisible by both $12$ and $5$. Since $36$ is divisible by $12$ but not $5$, we must have that $|ab|$ is divisible by $5$.
Therefore we can deduce that $|ab|$ is a divisor of $180$ and is divisible by $45$, so can only be $45$, $90$, or $180$.
In order to show these three orders are possible, in $C_{180}$ (a cyclic group of order $180$) choose three elements, $s$, $t$ and $u$, such that $|s| = 4$, $|t| = 15$ and $|u| = 9$.
If $a = s^{-1}t$ and $b = su$, then $|a| = 60$, $|b| = 36$, and $|ab| = |tu| = 45$.
If $a = st$ and $b = su$, then $|a| = 60$, $|b| = 36$, and $|ab| = |s^2tu| = 90$.
There is no example where the product has order $180$ in $C_{180}$, so let $C_4$ be a cyclic group of order $4$ generated by $v$, and in $C_{180} \times C_4$ let $a = (t,v)$ and $b = (su,1)$. Then $|a| = 60$, $|b| = 36$ and $|ab| = |(stu,v)| = 180$.
A: Two tools:

*

*First, since $a,b$ commute, we have $$(ab)^n=\underbrace{ab\cdot ab\cdots ab}_{\text {n-times}}=a^nb^n$$
So $(ab)^{180}=a^{180}b^{180}=1$.

*

*Next whenever $g^n=e$, we have $\mid g\mid \mid n$.

Proof:  Let $l=\mid g\mid $.  Write $n=ql+r$, with $0\le r\lt l$.  Then $e=g^n=g^{ql+r}=g^{ql}g^r=eg^r=g^r\implies r=0$, because of minimality of the $l\square $
So $\lvert ab\rvert \mid180$.
