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
2
$\begingroup$
abstract-algebra
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2Sounds like a typical "check your understanding" exercise. Have you tried to figure out something on your own? – 2011-10-13
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0Yes, I do not know where to start. Any suggestions? – 2011-10-13
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0Consider 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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0m is divisible by n? I'm not sure what things you're referring to. – 2011-10-13
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0Close. 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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0Ah. 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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0If $H \subseteq K$, 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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0Ah. 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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0If 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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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