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I am working from these lecture notes. For this example,

Example. List the elements of the cyclic subgroup of $\mathbb{Z}_8\times \mathbb{Z}_{15}$ generated by $(6,10)$.

$$ \begin{eqnarray*} \langle (6,10) \rangle &=& \left\{ (0,0), (6,10), (4,5),(2,0),(0,10), (6,5),\right.\\ & &\left.(4,0),(2,10),(0,5),(6,0),(4,10),(2,5)\right\}. \end{eqnarray*}$$

do I just write down several multiples of 6 and 15 and look at what those multiples are $\mod 8$ or is there a more efficient way to do this?

Also, for this example,

Example. Explain why no element of $\mathbb{Z}_8\times \mathbb{Z}_{10}$ can generate the group.

In fact, the largest order of an element of $\mathbb{Z}_8\times \mathbb{Z}_{10}$ is $(8,10)=40$. The element $(1,1)$ has order $40$. And if $(x,y)\in \mathbb{Z}_8\times \mathbb{Z}_{10}$, then $40\cdot (x,y)=(40x,40y)=(0,0)$. So the order of $(x,y)$ is no greater than $40$.

why does the fact that the largest possible order is $40$, tell us that no element can generate the group $\mathbb{Z}_8 \times \mathbb{Z}_{10}$?

Thanks in advance.

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2  
Would you mind copying the relevant questions in the pdf to your question? If you aren't sure of the necessary latex commands, someone can edit it for you. –  Michael Albanese Jan 19 '13 at 13:26
    
And please: do NOT number the examples in your question if they are not numbered in the PDF document! Better, mention the example as "third example in page 3" or something like that... –  DonAntonio Jan 19 '13 at 13:52

2 Answers 2

up vote 4 down vote accepted

generated by g means the set (which turns out to be a cyclic subgroup) {g, g+g, g+g+g, ...} until you get back to where you started.

so {(6,10), (12, 20), (18, 30), (24, 40), ...} and written reduced that's {(6,10), (4, 5), (2, 0), (0, 10), ...}.

A slightly easier way to do this is write out the group by 6 in Z_8 then by 10 in Z_15 and piece them together:

  • 6, 4, 2, 0
  • 10, 5, 0

since they have different periods it'll be like this:

 6  4  2  0  6  4  2  0  6  4  2  0
10  5  0 10  5  0 10  5  0 10  5  0

which is exactly what's written there.


The group Z_8 x Z_10 has 80 elements (there are 80 pairs (x,y) where x is < 8 and y < 10). An element can only generate the whole group if it has the same order as the whole group.

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In example 2, the subgroup is generated by $1$ element. You start by writing $(6, 15)$ and its multiples (mod 8 and 15 of course) until you get $(0, 0)$. That seems efficient enough to me.

In example 10, the group $\mathbb Z_8 \times \mathbb Z_{10}$ has $80$ elements. A group generated by 1 generator will have the same size as the order of the generator. Knowing that all elements have order not greater than $40$ means you cannot generate a group of order greater than $40$ with 1 element.

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