Showing $12\Bbb Z<\Bbb Z$ as groups.

Question

I'm having a bit of trouble answering the following question:

The subset $m\Bbb Z = \{mn\in \Bbb Z\mid n\in\Bbb Z \}$ is a subset of the group $\Bbb Z$. By writing down the elements of the subgroups $4\Bbb Z$ AND $6\Bbb Z$, identify the intersection, $4\Bbb Z \cap 6\Bbb Z$, and show that this is a subgroup of $\Bbb Z$.

I have worked out that

$$\Bbb Z = \{\dots , -3,-2,-1,0,1,2,3,4,5,6\dots \},$$

$$4 \Bbb Z= \{\dots, -12,-8,-4,0,4,8,12,16,20,24, \dots \},$$

$$6\Bbb Z = \{\dots, -18,-12,-6,0,6,12,18,24,30,36, \dots\}.$$

So $4\Bbb Z \cap 6\Bbb Z = 12\Bbb Z$.

but I am unsure about how to approach the second part of the question and show that $12\Bbb Z$ is subgroup of $\Bbb Z$.

Any help will be greatly appreciated, thanks in advance!

Answer 1

Since $0=12\times 0$, we have $0\in 12\Bbb Z$. So $12\Bbb Z$ is a non-empty subset of $\Bbb Z$.

Let $a, b\in 12\Bbb Z$. Then $a=12x, b=12y$ for some $x, y\in \Bbb Z$. Consider $a-b$. If $a-b\in 12\Bbb Z$, then, by the one-step subgroup lemma, $12\Bbb Z$ is a subgroup of $\Bbb Z$. But $a-b=12(x-y)\in 12\Bbb Z$.

Hence $12\Bbb Z$ is a subgroup of $\Bbb Z$.

There's nothing special about the number $12$ here.

Answer 2

Shaun's answer shows directly that $12\mathbb{Z}$ is a subgroup of $\mathbb{Z}$. You can also argue that the intersection of two subgroups is always also a subgroup.