Determine $6\mathbb{Z}\cap 10\mathbb{Z}$ Determine $6\mathbb{Z}\cap 10\mathbb{Z}$:
I believe the correct answer is $30\mathbb{Z}$
Can this be thought of as least common multiple?
Also what would $6\mathbb{Z}\cup 10\mathbb{Z}$ be? 
$\mathbb{Z}$?
Thank you for your time.
 A: In general, $\;n\Bbb Z\cap m\Bbb Z=\text{l.c.m.}\,(n,m)\Bbb Z\;$ , so it is true that $\;6\Bbb Z\cap10\Bbb Z=30\Bbb Z\;$ . 
In general, union of subgroups is not a subgroup.
A: Let $\mathbb{S}=\{ s_1, s_2,s_3,...,s_k \}$ be any kind of set, in this case, the set $\mathbb{S}$ is discrete and has a finite cardinality, but it could as well be a continuous set like $\mathbb{R}$ or discrete with infinite cardinality like $\mathbb{Z}$. Then you say that the product with a scalar $k$ just multiplies each element, which would then give $$k\mathbb{S}=\{ ks_1,ks_2,ks_3,...,ks_4 \}.$$
Then  $6 \mathbb{Z}$ is only the set of all multiples of $6$ and that $10 \mathbb{Z}$ is the set of all multiples of $10$. This is because the scalars $6$ and $10$ multiply, in the set, all of the possible whole numbers, giving the multiples of each. 
That means that the intersection of the two sets $\mathbb{6Z} \cap \mathbb{10Z}$ is the set of all multiples of both $6$ and $10$, so of all the multiples of the largest common multiple of both $6$ and $10$, in this case $30$ as you said.
The answer is then really $\mathbb{30Z}$. You can then maybe see why the union of those two sets can't be expressed as simply as the intersections, usually they are no more subgroups.
