Listing the orders of the elements and calculating the number of elements in the order

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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Well, what made you use Euler's totient function for this? It is true that there is a specific case where that gives the answer, but this is not such a case. – Tobias Kildetoft Dec 12 '12 at 18:39
I thought that was how you calculate the number of elements in the order. How else do you do it? – Kaish Dec 12 '12 at 18:46
Euler's totient function gives you the number of elements of the given order in a cyclic group. This is not quite a cyclic group, but we can still use something like this, since we can see what the order of an element $(a,b)$ will be given the orders of $a$ and $b$. – Tobias Kildetoft Dec 12 '12 at 18:48
So how would I calculate those orders then? – Kaish Dec 12 '12 at 19:05
Yes, the theorem you found is very useful for this. The comma bit at the denominator should have been $(m,k)$ which means $\rm{gcd}(m,k)$, ie their greatest common divisor. – Tobias Kildetoft Dec 12 '12 at 19:19