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 $G = C_{6} \times C_6$. List all possible orders for G, and for each $m \geq 1$ of them calculate the value of $\gamma_G(m)$.
From my other question, I thought that I would do this:
lcm(6,6) = 6
Integers dividing 6 = 1, 2, 3, 6. I therefore use the Euler function on these numbers to get the number of elements in each order, but thats wrong. The correct answer is:
$$\begin{matrix} m: & 1 & 2 & 3 & 6 \\ \varphi(m): & 1 & 3 & 8 & 24 \end{matrix} $$
Why is this?
EDIT: I found a theorem: G and H are groups. For any $g \in G$, ord(g) = m. $h \in H$, ord(h) = k. Then
$$ ord(g, h) = lcm(m,k) = \frac{m \cdot k}{(m , k)} $$
What does the comma bit mean at the bottom of the fraction?
Is this theorem useful?
