My question is:
Consider a finite group $G$. For any integer $m \geq 1$ set $\gamma(m) = \gamma_G(m)$ the number of elements $g \in G$ such that ord($g$) = $m$. We say that $m$ is a possible order for $G$ if $\gamma(m) \geq 1$, that is, if there is at least one element $g \in G$ such that $\operatorname{ord}(g) = m$. Consider the group $G = C_{6} \times C_6$. List all possible orders for $G$, and for each $m \geq 1$ of them calculate the value of $\gamma_G(m)$.
I know that if we have two groups $G, H$ then for some elements in these groups, $\text{ord}(g) = k$ and $\text{ord}(h) = l$. Then, we can say that
$ \text{ord}(g,h) = \text{lcm}(k,l) = \frac{k \cdot l}{\text{gcd}(k,l)}$
The possible orders will be all the numbers that divide the $\text{lcm}(6,6) = 1, 2, 3$ and 6. But I'm not sure how I would go about using the formula to calculate the number of elements in each of these orders. Can someone help please.