List all the possible orders for G and calculate $\gamma_G(m)$ Consider a finite group G. For any integer $m \geq 1$ set $\gamma(m) = \gamma_G(m)$ to be 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 ord(g) = m.
Consider the cyclic group $G = C_{36} = \{1, a, ..., a^{35} \}$. List all possible orders for G, and for each $m \geq 1$ of them calculate the value of $\gamma_G(m)$.
I understand that the order of g is the smallest integer $m$ such that $g^m = 1_G$, but how do I find out this order number? Do I go through each element in $C_{36}$ and see if I can raise it to some power ($\geq 1$) to give me the identity element?
 A: Hint: Use Lagrange's Theorem.
If you haven't covered Lagrange's Theorem, consider the factors of $36$, the order of $G = C_{36}$.


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*Which integers divide $36$? This will give you the orders of all the subgroups of $C_{36}$ (all the $m$s about which you need to be concerned).Then consider which elements of each subgroup generate the subgroup; the order of any such generator will equal the order of the subgroup it generates (subgroups of cyclic groups are cyclic).

*Find all the generators of $G = C_{36}$. Apply what do you know about elements that are coprime to $36$. Count them: this will be $\gamma(36)$.
Finally, after finding the orders $m_i$ of each element $g_i\in G = C_{36},\;$ count the number of elements with the same order $m$. This amounts to computing $\;\gamma(m)$ for each distinct $m$ .
A: I would try to discover the general principle rather than going through each element in turn.
Can you answer this question if you change 36 into smaller numbers, like 3,4,5,6? These cases are small enough for you to work out completely, and you should see a pattern emerging.
