Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.