Listing elements of subgroup generated by $\{12,42\}$ in the integers with addition

Question

The subgroup generated by these elements should contain both $12\mathbb{Z}$ and $42\mathbb{Z}$ but also ideals of the form $$ (12k+42j)\mathbb{Z},\;j,k\in\mathbb{Z} $$

Is this the best answer I can give? If so, is this the best way to denote this?

edit/answer: extrapolating from what was given below, we can rewrite what I wrote as $$ (12k+42j)\mathbb{Z}=6\mathbb{Z} $$ Since Bezout's identity guarantees the existence of some integers $j,k$ with $$ 42j+12k=gcd(42,12)=6 $$ And since this is the smallest divisor of both numbers, it divides all other divisors (euclidean algorithm) and will include any other ideal generated by common divisors of the two numbers.

Answer 1

The only subgroups of $\mathbb Z$ are those of form $n\mathbb Z$.

In this case we need to find $n$ so that $12,42\in n\mathbb Z$.

So $n$ is a common divisor of $12$ and $42$, hence $n\in \{\pm1,\pm2,\pm3,\pm6\}$

Clearly, the smallest of all of these options for $n\mathbb Z$ is when $n=\pm6$.

So the subgroup you want is the subgroup of multiple of $6$.

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In general one can prove that the subgroup generated by $a_1,a_2\dots a_n$ is $m \mathbb Z$ with $m=gcd(a_1,a_2\dots a_n)$

Answer 2

Hint: $\Bbb Z$ is cyclic, so all of its subgroups are cyclic, i.e., each subgroup is generated by only one element. What is the generator of your subgroup?