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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 $(\mathbb{Z}, +)$?

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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

1 Answer 1

$m\mathbb{Z} \subseteq n\mathbb{Z}$ $\Longleftrightarrow$ $m \in n\mathbb{Z}$ $\Longleftrightarrow$ $m$ 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 $m$.

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