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Let $H<(\mathbb{Z},+)$ where ($\mathbb{Z},+$) is the abelian group of integers under addition. If the numbers $12$, $30$, and $54$ are contained in $H$, what are the possibilities for $H$?

To me, I immediately assume since they're all even numbers and you can't 'reach' an odd number through addition or subtraction of even numbers that $H=\langle 2\rangle$.

Although I can see how this solution could be correct also:

$H=\langle\gcd(12,30,54)\rangle=\langle 6\rangle$

Which one is correct?

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Note that any subgroup of $\mathbb Z$ is of the form $n\mathbb Z$ for some $n\in\mathbb N$. Therefore, $H=n\mathbb Z$ for some $n$. Now, since $H$ contains only the multiples of $n$ and it contains $12,30,54$, hence, $n$ can be any common factor of $12,30,54$, i.e. possible values of $n$ are $6,3,2$ (we are excluding $n=1$ because then $H$ will be $\mathbb Z$, not a proper subgroup).

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  • $\begingroup$ Seems enough for the OP. $\endgroup$ – mrs Mar 20 at 13:41
  • $\begingroup$ Thanks! This is perfect. $\endgroup$ – upanddownintegrate Mar 20 at 14:11

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