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 $\gamma_G(1)=1$, $\gamma_G(2)=3$, $\gamma_G(3)=8$, $\gamma_G(6)=24$.

| cite | improve this answer | |
  • $\begingroup$ Could you please explain how you get number of elements a bit more please. I don't understand why you do what you did. $\endgroup$ – Kaish Dec 28 '12 at 16:56
  • $\begingroup$ I hope it's better now. $\endgroup$ – Hagen von Eitzen Dec 28 '12 at 18:48
  • $\begingroup$ 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? $\endgroup$ – Kaish Dec 28 '12 at 20:37
  • $\begingroup$ 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) $\gamma_G(2) + \gamma_G(1)$. $\endgroup$ – Kaish Dec 28 '12 at 20:50
  • $\begingroup$ 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. $\endgroup$ – Kaish Dec 28 '12 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.