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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 $\operatorname{ord}(g) = m$. 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 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:

$\operatorname{lcm}(6,6) = 6$

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. $h \in H$, $\operatorname{ord}(h) = k$. Then

$ \operatorname{ord}(g, h) = \operatorname{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?

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    You have to let $g$ and $h$ run through $C_6$, then the computations go like this: $ord(0,0)=lcm(ord(0),ord(0))=lcm(1,1)=1$ or $ord(2,3)=lcm(ord(2),ord(3))=lcm(3,2)=6$. A 6-by-6 table is doable, but you can also handle this$a$bit smarter by making a 4-by-4 table (for each of the orders) while keeping track how many elements there are for each order. Or you could write $C_6 \times C_6$ as $C_4 \times C_9$...2012-12-12

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