2

Let $m\mathbb{Z}$ and $n\mathbb{Z}$ be subgroups of $(\mathbb{Z}, +)$. What condition on $m$ and $n$ is equivalent to $m\mathbb{Z}\subseteq n\mathbb{Z}$? What condition on $m$ and $n$ is equivalent to $m\mathbb{Z}\cup n\mathbb{Z}$ being a subgroup of $(\mathbb{Z}, +)$?

Zev Chonoles
  • 129,973
Lily
  • 301
  • 2
    Sounds like a typical "check your understanding" exercise. Have you tried to figure out something on your own? – Hans Lundmark Oct 13 '11 at 19:32
  • Yes, I do not know where to start. Any suggestions? – Lily Oct 13 '11 at 19:33
  • Consider if $m\mathbb{Z} \subseteq n\mathbb{Z}$, then $m = m \cdot 1 \in m\mathbb{Z} \subseteq n \mathbb{Z}$. Elements of $n\mathbb{Z}$ are thing like $-2n,-n,0,n,2n,3n$ and $m$ is one of these. What are such things called? – Bill Cook Oct 13 '11 at 19:37
  • m is divisible by n? I'm not sure what things you're referring to. – Lily Oct 13 '11 at 19:46
  • Close. Make it concrete. Suppose $m=3n$ then $n$ divides $m$. Notice the divisibility and containment relations are reversed. – Bill Cook Oct 13 '11 at 19:49
  • Ah. So the condition on m and n is equivalent to m contained in n is that n divides m? What about the union of the two? – Lily Oct 13 '11 at 19:55
  • If $H \subseteq K$, then $H \cup K = K$ (the "bigger" one). In general the union of 2 subgroups is only a subgroup when one is contained in the other. This is true for most every algebraic system. – Bill Cook Oct 13 '11 at 20:08
  • Ah. For some reason, the code in the very first sentence isn't showing up. Do you mind writing it in words? Thanks! – Lily Oct 13 '11 at 20:40
  • If H is a subset of K, then the union of H and K is just K (the "bigger" set). – Bill Cook Oct 13 '11 at 21:13
  • @Bill, when the discussion converges, I'd encourage you to write it up as an answer, rather than leave it in the comments. – Gerry Myerson Oct 14 '11 at 01:17

1 Answers1

2

$m\mathbb{Z} \subseteq n\mathbb{Z}$ $\Longleftrightarrow$ $m \in n\mathbb{Z}$ $\Longleftrightarrow$ $m$ is a multiple of $n$ (or equivalently $n$ divides $m$).

So the "divides" relation on integers is the same as "$\supseteq$" on the corresponding subgroups.

Next, the union of two subgroups is a subgroup if and only if one is contained in the other. So $m\mathbb{Z} \cup n\mathbb{Z}$ is a subgroup if and only if $m\mathbb{Z} \subseteq n\mathbb{Z}$ or $n\mathbb{Z} \subseteq m\mathbb{Z}$. So the union is a subgroup if and only if either $m$ divides $n$ or $n$ divides $m$.

Bill Cook
  • 29,244