Lattice of subgroups

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1. The problem statement, all variables and given/known data

How do the atoms and coatoms of the lattice of all subgroups of the group [itex]\mathbb{Z}(+,-,0)[/itex] look like?

2. Relevant equations

Let [itex](L,\le)[/itex] be a lattice and [itex]e, f \in L[/itex] is the minimum (maximum) elements of L. Then we say that [itex]a, b \in L[/itex] is the atom (coatom) of L if a covers e (b covers f).

3. The attempt at a solution

I guess that all subgroups of the given group are of form [itex]H = k\mathbb{Z} = \left\{ k.x | x \in \mathbb{Z}\right\}[/itex] and that the ordering on the lattice will be set inclusion.

But I don't know how its Hasse diagram will look like (I think I need it to solve the problem).

Thank you for any hint!
 
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matt grime

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When is kZ contained in mZ?
 
When is kZ contained in mZ?
If and only if k divides m, I guess. But I can't imagine how the Hasse diagram will behave behave as I will approach to its top, ie. the whole group [itex]\mathbb{Z}[/itex] - what subgroups will be residing under the top, ie. how will the coatoms look like?
 

matt grime

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What makes you think you can picture it? Or draw it? The lattice has generators for each prime, i.e. infinitely many of them. So the lattice is the lattice on countably infinitely many generators, plus a vertex greater than all (the infinitely many) other vertices (and there is no reason in a poset why any vertex has to have a predecessor).
 
What makes you think you can picture it? Or draw it? The lattice has generators for each prime, i.e. infinitely many of them. So the lattice is the lattice on countably infinitely many generators, plus a vertex greater than all (the infinitely many) other vertices (and there is no reason in a poset why any vertex has to have a predecessor).
I knew I couldn't draw the complete Hasse diagram, just the general situation on the bottom (atoms) and below the top (coatom). I guess the atoms (elements covering [itex]0Z = \left\{0\right\}[/itex]) would be [itex]pZ[/itex], where p is a prime, but I really don't know what will the coatoms look like...

Btw we weren't told what is a generator of a lattice...
 

matt grime

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If I were you I'd go get my definition of atom and coatom sorted. You say 'the atom', implying there is one, then say there is one for all primes. And why isn't 4 an atom? I.e. what does 'cover' mean. It must have someting to do with x covers y if y is a subgroup of x, possibly maximal. But this can't make sense for your definition of coatom which as stated says b is a/the coatom if b covers f where you define f to be maximal. Since nothing can cover properly a maximal element there is something wrong. A coatom is surely something that is covered by the maximal element. And I will take 'covered' to mean x covers y if y<x and y is maximal. What makes you think Z has coatoms?

I think we have things the wrong way round, at least when i try to make 'common sense' interpretations of the words, anyway. pZ is a very large group it contains mZ for all m divisible by p. It is in fact a maximal subgroup of Z. Atoms ought to be small So presumably pZ is a coatom. There are no minimal non-trivial subgroups of Z - any subgroup H has a subgroup 2H if H is not trivial. Hence there are no atoms.

So we have a hasse diagram with Z at the top, and a line to each of theinfinitely many subgroups pZ, p a prime, then a line from pZ and qZ to pqZ, etc - this is a lattice with infinitely many generators - plus e at the bottom on its own.
 
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Thank you matt, that helped me a lot for understanding the problem!
 

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