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 $(\mathbb{Z}, +)$?
Conditions of being a subgroup
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abstract-algebra
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0@Bill, when the discussion converges, I'd encourage you to write it up as an answer, rather than leave it in the comments. – 2011-10-14
1 Answers
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$m\mathbb{Z} \subseteq n\mathbb{Z}$ $\Longleftrightarrow$ $m \in n\mathbb{Z}$ $\Longleftrightarrow$ $m$ 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 $m$.