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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?2011-10-13
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    Yes, I do not know where to start. Any suggestions?2011-10-13
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    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?2011-10-13
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    m is divisible by n? I'm not sure what things you're referring to.2011-10-13
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    Close. Make it concrete. Suppose $m=3n$ then $n$ divides $m$. Notice the divisibility and containment relations are reversed.2011-10-13
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    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?2011-10-13
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    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.2011-10-13
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    Ah. For some reason, the code in the very first sentence isn't showing up. Do you mind writing it in words? Thanks!2011-10-13
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    If H is a subset of K, then the union of H and K is just K (the "bigger" set).2011-10-13
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    @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

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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 $m$.