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As usual, this is probably a lot easier than I'm making it out to be. I just don't know... where to start. I know that if I have one generator, $g$, of a cyclic group of order $m$, then all the other generators are $g^k$, where $(k,n)=1$.

Is the order of $\mathbb{Z}_9 \times \mathbb{Z}_{10}$... 90? Does $(1,1)$ generate the group?

Am I stumbling in the right direction here?

Jyrki Lahtonen
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Indigo
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  • Yes, the order is 90. And Yes.$g=(1,1)$ generates. But as a teacher I would be more interested in whether you fully understand why $g$ generates the whole group. But before we head there, the group operation in $\Bbb{Z}9\times\Bbb{Z}{10}$ is based on addition, so instead of "powers" of $g$ you should look at ________? – Jyrki Lahtonen Nov 04 '15 at 05:30
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    $(a,b)$ generates $\mathbf{Z}9\times\mathbf{Z}{10}$ if and only if $a$ generates $\mathbf{Z}9$ and $b$ generates $\mathbf{Z}{10}$. (Prove this.) – fkraiem Nov 04 '15 at 05:34
  • @JyrkiLahtonen multiples? – Indigo Nov 04 '15 at 05:38
  • Correct. In proving that $g$ generates the whole group it may (?) be illuminating to go via the route that $g_1=(1,0)$ and $g_2=(0,1)$ together surely generate all of it. Why? Therefore it suffices to show that $g_1$ and $g_2$ are both multiples of $g$. Have you done examples that allow you to see why that happens? – Jyrki Lahtonen Nov 04 '15 at 05:44
  • @JyrkiLahtonen I think I... Might have. Honestly the textbook leaves something to be desired in its explanation of this. – Indigo Nov 04 '15 at 05:53
  • Ok. We get $(a,b)=a g_1+ b g_2$, so we can write everything using $g_1$ and $g_2$ (and the group operation) alone. And to get both $g_1$ and $g_2$ in terms of $g$, you can check that $9g$ and $10g$ are (almost) what you need. The answerers have taken a route using a bit more results. The tools they use are hopefully familiar. The route I implied went back to a bit lower level. The goal here is that you form a coherent picture of what's going on. Good luck! – Jyrki Lahtonen Nov 04 '15 at 06:05

2 Answers2

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The natural map $\mathbb{Z}_{90}\rightarrow \mathbb{Z}_9\times\mathbb{Z}_{10}$ given by sending elements of $\mathbb{Z}_{90}$ to their classes mod 9 and 10 is an isomorphism (this is the Chinese remainder theorem). Thus, obviously 1 is a generator of $\mathbb{Z}_{90}$, so its image in $\mathbb{Z}_9\times\mathbb{Z}_{10}$ is $(1,1)$, so yes $(1,1)$ generates the group.

There are not many other generators. Note $\phi(90) = \phi(9)\phi(2)\phi(5) = 6\cdot 1\cdot 4 = 24$. The generators are precisely the image of elements of $\mathbb{Z}_{90}^\times$ in $\mathbb{Z}_9\times\mathbb{Z}_{10}$.

oxeimon
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OK, let's just check hands on! Let $(n,m)$ be an element of $\mathbb{Z}_9\times \mathbb{Z}_{10}$. We define the orbit of this element as $O(n,m)=\{(kn, km)|k\in \mathbb{Z}\}$, more intuitively just $(n,m), (2n,2m), (3m,3m), \cdots$. The orbit of $(n,m)$ is the whole group if and only if $(n,m)$ is a generator.

So let's check: If $n=0,3,6,$ then the orbit cannot be the whole thing. For $0$ becuase $k.0 = 0$, for $3$ becuase $3.3=9\mod{9}=0$ and for $6$ because $3\times 6=18\mod{3}=0$.

For $m = 0,2, 4, 5, 6, 8$ you have the same problem in $\mathbb{Z}_{10}$. Hence the only viable candidates for generating have $m=1, 3,7,9$ and $n=1,2,4,5,7,8$. Define $$S=\{(n,m)|n=1,2,4,5,7,8, m=1,3,7,9\}$$

Next thing is to check which of these are actually generators. We need to check how long does it take $(n,m)$ to come back to itself. Or what is the first $k$ such that $(n,m) = (kn,km)$ in $\mathbb{Z}_9\times \mathbb{Z}_{10}$. Now if $(k-1)n\mod{9}=0$ and as we saw none of the candidates have any common factor with $9$, so $k-1=9\ell$. Also similar we must have $(k-1)=10\ell'$. Meaning $9\ell=10\ell'$. Again since $9$ and $10$ have no common factor, the smallest $\ell=10$ and $\ell'=9$. $k-1=90$, meaning $k=91$. Since the order of the group is $90$, this means all elements in $S$ are generators. We have $6\times 4=24$ generators.

Now let me ask you this, with this in mind, can you generalize this for $\mathbb{Z}_q\times\mathbb{Z}_{r}$ with $(q,r)=1$ ($q,r$ have no common factors)? Can you generalize this even further?

Hamed
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