# Let $H<(\mathbb{Z},+)$ and that $H$ contains $12,30,54$. What are the possibilities for $H$?

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?

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).