# Calculating the number of elements of some order of a direct product of groups

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.

For $C_6$, observe that $\gamma_C(1)=1$, $\gamma_C(2)=1$, $\gamma_C(3)=2$ and $\gamma_C(6)=2$. The order of an element $(a,b)$ with $a,b\in C_6$ is the lcm of their respective orders and hence is again a divisor of $6$. Therefore we have $\gamma_G(1)=\gamma_C(1)^2=1$ as $(a,b)$ has order $1$ iff both $a$ and $b$ have order $1$. For bigger orders $m$ it seems easier to compute the number of pairs $(a,b)$ that have order that is a divisor of $m$ (instead of exactly $m$). For example, $(a,b)$ has order dividing $2$ (i.e. equal to $1$ or $2$) iff both $a$ and $b$ have order dividing $2$ (i.e. equal to $1$ or $2$). Thus we conclude \begin{align} \gamma_G(1)&=&\gamma_C(1)^2&=&1\\ \gamma_G(2)+\gamma_G(1)&=&(\gamma_C(2)+\gamma_C(1))^2&=&4\\ \gamma_G(3)+\gamma_G(1)&=&(\gamma_C(3)+\gamma_C(1))^2&=&9\\ \gamma_G(6)+\gamma_G(3)+\gamma_G(2)+\gamma_G(1)&=&(\gamma_C(6)+\gamma_C(3)+\gamma_C(2)+\gamma_C(1))^2&=&36&.\end{align} Solving these equations, we obtain therefore $\gamma_G(1)=1$, $\gamma_G(2)=3$, $\gamma_G(3)=8$, $\gamma_G(6)=24$.
• Kind of but still not 100%. Do you do, for example $\gamma_G(2) + \gamma_G(1)$ because these are all the elements that will have order 2? And so For order three you just have 3 and 1 but for order 6 you have all of them? And I still don't understand that squared bit. Are you using that lcm "formula" I put up in my OP? – Kaish Dec 28 '12 at 20:37
• I think I get the squared bit. It's because of the direct product right? Like if we are doing the number of elements of order 2, then in the first $C_6$ we will have $\gamma_C(2) + \gamma_C(1)$ and in the other $C_6$ we will have the same and as its a direct product, we multiply these two together to get $(\gamma_C(2) + \gamma_C(1))^2$, but I still don't get why the LHS would be (in this case) $\gamma_G(2) + \gamma_G(1)$. – Kaish Dec 28 '12 at 20:50
• Then can you think about it like this? After thinking about it my way to get the $(\gamma_C(2) + \gamma_C(1))^2$, from here you then have to minus all the elements of $\gamma_G(1)$ as they automatically are counted when doing $\gamma_C(2)$ twice and so subtracting them gives us just one set of $\gamma_G(1)$ elements in the answer. – Kaish Dec 28 '12 at 20:53