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How do I find it?

For example, an easy one: $G\:=\:\left(\mathbb{Z},+\right)$

H and K are sub-groups: $n\mathbb{Z}, m\mathbb{Z}$ for different $n$ and $m$. And we know that $n$ and $m$ are generators of their sub-groups.

So I want to find the generators of $n\mathbb{Z}\cap m\mathbb{Z}$

It seems to be: $\dfrac{m\cdot n}{\left(m,n\right)}$ by intuition, am I wrong ? And if it is, how do I show that formally?

Ilya.K.
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2 Answers2

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If $x\in n\mathbb{Z}\cap m\mathbb{Z}$, then $x$ is an integer multiple of both $m$ and $n$, and therefore an integer multiple of $\text{lcm}(n,m)$. That is, if both $n$ and $m$ divide $x$, then $\text{lcm}(n,m)$ divides $x$. Many proofs of this are given here. This implies that every $x\in n\mathbb{Z}\cap m\mathbb{Z}$ is a multiple of $\text{lcm}(n,m)$, thus: $$n\mathbb{Z}\cap m\mathbb{Z} = \langle \text{lcm}(n,m)\rangle.$$ So, you were not wrong, since by basic number theory: $$nm = \gcd(n,m)\text{lcm}(n,m).$$ In general, the intersection of two finitely generated groups is not necessarily finitely generated.

Edit:

We want to show that every $x\in n\mathbb{Z}\cap m\mathbb{Z}$ is an integer multiple of $\text{lcm}(n,m)$ (think of how every $y\in n\mathbb{Z}$ is an integer multiple of $n$, and $n$ generates $n\mathbb{Z}$). If $x\in n\mathbb{Z}\cap m\mathbb{Z}$ then for some integers $k_1,k_2$, $$x=k_1n\quad\text{and}\quad x=k_2m.$$ By definition, this means that both $n$ and $m$ divide $x$. By the theorem linked above, this implies that $\text{lcm}(n,m)$ divides $x$, which by definition means that there exists a $k\in\mathbb{Z}$ such that $$x=k\cdot\text{lcm}(n,m).$$ Since $x$ was arbitrary, we can conclude that $$n\mathbb{Z}\cap m\mathbb{Z} = \text{lcm}(n,m)\mathbb{Z}.$$

To see why it has to be the least multiple, consider $3\mathbb{Z}\cap 5\mathbb{Z}$. Notice that $15\in 3\mathbb{Z}\cap 5\mathbb{Z}$, but $15\notin 30\mathbb{Z}$. So if we attempt to make a larger multiple a generator, some elements of $3\mathbb{Z}\cap 5\mathbb{Z}$ will be left out.

Community
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  • I think you right, but we need to add another word about why this is a legitimate concluson. So x is an integer multiple of both m and n, therefore a generator of this group that every element in it is an integer multiple of m and n, therefore to find it's generator, by a defenition of a generator, it is the smallest element, who gives us the whole group, and because we talking about Z and +, we need to find the smallest multiple of m and n, therefore it is exactly the definition of l.c.m. Correct me if I wrong, but this is the detailed explanation for this conclusion. Thanks for the link. – Ilya.K. Aug 13 '15 at 17:49
  • A generator is not necessarily the smallest element that produces the entire group. In fact, plenty of groups don't have any natural way to order the elements like $GL_n(\mathbb{R})$. If you consider the group $\mathbb{Z}/5\mathbb{Z}$, it is generated by $1,2,3$ and $4$. And if you haven't seen it already I edited the above. –  Aug 13 '15 at 17:53
  • Ok, but still I dont see enough explanation why if the lcm also devides x, then the generator is the lcm. I mean, so what? So the lcm devides it, why can we say that it creates the whole group of multiplies of m and n ? If it doesnt realte to "smallest or biggest" I mean intuitively, of course it does. But I see a hole in the proof. Why is it the lcm and not randomly choosen another multiple of m and n, for example ? – Ilya.K. Aug 13 '15 at 17:57
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    I'll elaborate above a little more. –  Aug 13 '15 at 18:01
  • Ok, so in the last paragraph, it seemed to be a proof by contradiction: Let the generator be bigger then the lcm. And now we need to show contradiction in general (and for this lcm is just a bound of letters, I think we should express it as the multiple of m and n devided by the gcd), I think would be the most correct way. I mean, ok, it is understood, but we took concrete numbers. We should do the same in general example in the proof by contradiction. Formally. I wiil think about it now. – Ilya.K. Aug 13 '15 at 18:52
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    That last part wasn't a proof, only an example of what happens when we try to make the generator larger than the least common multiple. The proof is outlined above. –  Aug 13 '15 at 18:56
  • Yes, but what if teoretically there are such magic number, that that example doesnt implies the same conclusion? We must show in generally that the lcm is creating the whole group of multiplies. Because eather you can write $x=k⋅3\cdot :lcm\left(n,m\right)$ or $x=k⋅4\cdot :lcm\left(n,m\right)$ or any natural number there, but the lcm generates them all. It stil bothers me, because the lcm isnt a simple operator. It is a simplify of algorithm to find the least common multiple. I understand that you conclude it by analogy of how m creates mZ, but it bother me that m is a number. – Ilya.K. Aug 13 '15 at 19:10
  • and the lcm is some number that we get by multipling m and n and dividing by the gcd. I mean, that is a whole monster there. And we claim, that that thing is generates the whole group only because it devides every element in it, and an example that shows and explains us why it cant be bigger then it. I dont know, but I feel something is missing and it doesnt satisfies me. – Ilya.K. Aug 13 '15 at 19:13
  • "We must show in general that the lcm is creating the whole group of multiples" -

    The proof above shows exactly that. I think you are making this way harder than it needs to be. An element $\ell\in n\mathbb{Z}\cap m\mathbb{Z}$ generates $n\mathbb{Z}\cap m\mathbb{Z}$ if every $x\in n\mathbb{Z}\cap m\mathbb{Z}$ can be written in the form $x=k\ell$ for some $k\in\mathbb{Z}$. We showed above that every $x\in n\mathbb{Z}\cap m\mathbb{Z}$ is of the form $x=k\cdot\text{lcm}(n,m)$ for some $k$, so we must have that $\text{lcm}(n,m)$ generates $n\mathbb{Z}\cap m\mathbb{Z}$.

     –  Aug 13 '15 at 19:22
  • Maybe you right, dont let it (and me) too much bother you, ill look at it tommorow morning. All that bothers me, that we can the same written form for a bit adjusted formes of multipliing the lcm or m and n. But maybe it is the hour )) – Ilya.K. Aug 13 '15 at 19:25
  • Don't worry, you don't bother me at all. Get some sleep and look it over tomorrow. –  Aug 13 '15 at 19:28
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The generator of $ mZ \cap nZ $ is the l.c.m of m and n. Intuitively take $3Z \cap 2Z$ it means that you are looking for the smallest common multiple of 2 and 3 which is 6

Salvatore
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