# Given certain elements of a proper subgroup $H$ of $\mathbb{Z}$, determine $H$

Question: Suppose that $H$ is a proper subgroup of $\mathbb{Z}$ under addition and that $H$ contains $18$, $30$, and $40$. Determine $H$.

Can we determine $H$ without knowing the order of $\mathbb{Z}$? Please don't tell the answer; I just need a hint.

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I am slightly confused by the first sentence in the second paragraph. We already know the order the $\mathbb{Z}$ (it is infinite). Do you perhaps mean the index of $H$ in $\mathbb{Z}$? –  user1729 Mar 27 '13 at 10:54
The wording of the question isn't very good. We can't actually determine $H$ just by knowing that it contains certain elements. I presume what is meant is that $H$ contains only those elements and the elements that it has to contain as a consequence of containing those ones (i.e. those elements generate $H$), but it should say that. –  Tara B Mar 27 '13 at 12:05
@TaraB: In this question, $H$ is a proper subgroup containing the element $2$, and so it can be determined. But in general, yes, you are right. –  user1729 Mar 27 '13 at 12:47
@user1729: Ah, you are right. I had missed that the word 'proper' gives the necessary extra information. –  Tara B Mar 27 '13 at 12:49

If $H\leq\mathbb Z$ so for some $n\neq 1,0$, $~~H=n\mathbb Z$. Now think of this: $$18=nk,~ 30=nk',~ 40=nk'',~~ k,k',k''\in\mathbb Z$$

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what does $H=nZ$ means???Order of $Z$ is divisible by order of $H$ –  TLE Mar 27 '13 at 13:08
$n\mathbb{Z}$ is shorthand for the set $\{nk; k\in\mathbb{Z}\}$, which is the subgroup of $\mathbb{Z}$ generated by the element $n$ (why?). Note that all subgroups of a cyclic group are themselves cyclic, so $H=n\mathbb{Z}=\langle n\rangle$ for some $1\neq n\in\mathbb{Z}$. –  user1729 Mar 27 '13 at 13:22
$\ddot\smile{}{}{}{}{}{}{}$ –  amWhy Mar 27 '13 at 14:16
@amWhy: Morning Friend. ;-) –  Babak S. Mar 27 '13 at 14:18

Hint $1$: The greatest common devisor of $18$, $30$ and $40$ is which number and why is this relevant?

Hint $2$: To find $H$, think about what happens if $\operatorname{gcd}(18, 30, 40)$ is prime and if it is composite.

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Thanks for the comment. I was not on here to help the OP. +1 –  Babak S. Mar 27 '13 at 13:47

As an example of the sort of reasoning you might use, you might think that since $18 \in H$ and $30\in H$, we know that $18 + 18 - 30 = 6 \in H$. Using reasoning like this, you can arrive at a unique answer.

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