Show that nZ intersection mZ= lZ

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Homework Help Overview

The original poster presents two questions related to group theory and additive groups. The first question involves showing that the intersection of nZ and mZ equals lZ, where l is the least common multiple of m and n. The second question concerns the conditions under which the union of two subgroups H and K of a group G is itself a subgroup.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants inquire about the original poster's attempts and suggest starting with definitions to clarify the problem. There are discussions about the validity of a proposed solution for the second question and the need for precise mathematical communication.

Discussion Status

The discussion is ongoing, with participants encouraging the original poster to share their work and clarify their difficulties. Some guidance has been offered regarding the logical structure of the arguments, particularly for the second question, but no consensus has been reached on the correctness of the proposed solution.

Contextual Notes

There is a noted emphasis on the importance of clear mathematical communication, as well as the distinction between subgroup and subset notation. Participants are also exploring the implications of the assumptions made in the problems presented.

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hello

i have two questions and i need answers for them

first one:

in the additive group (Z,+)
show that nZ intersection mZ= lZ

, where l is the least common multiple of m and n.



The second question is :

Given H and K two subgroups of a group G , show the following:

(H union K) subgroup of G if and only if H subset of K or K subset of H

:confused: :smile:
 
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What have you done and what do you think you need to do? Remember, the first thing to do (if you can't do the question straight away) is to write down the definitions of everything so you know what you have to do.
 
Last edited:
You're not going to get answers! Show us what you have done and, as precisely as you can, where you have difficulty and you will get help
 
http://umm.biz/index.php?act=Attach&type=post&id=754324

Is this solution true for question two ??
 
Last edited by a moderator:
You need to learn to communicate by posting the maths here not posting a link to a zipped file with two jpgs in it.

It has the seed of a correct solutin, but no, I think it is not correct. You say:

let x be in H and Y be in K and then show if xy is in H y is in H and then conclude that K C H, and dually that if xy is in K then x isin K and H C K. But that doesn't follow. If for ALL x and y in H and K respectively then either xy is always in H or xy is always in K and it would follow, but there is no reason to suppose this at all.

Suppose that neither H C K nor K C H, then there are elements x in H but not in K and y in K but not in H. Where is xy? (I.E. apply your argument now to get a contradiction.)

Perhaps better is to do: if H C K we're done, if not then (and apply your argument to well chosen x in H).

You also should take care between using < and C, subgroup and subset.
 
Last edited:

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