Show that nZ intersection mZ= lZ

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SUMMARY

The discussion centers on two mathematical questions regarding group theory. The first question demonstrates that the intersection of nZ and mZ equals lZ, where l is the least common multiple of m and n, confirming the properties of additive groups. The second question explores the conditions under which the union of two subgroups H and K of a group G forms a subgroup, concluding that this occurs if and only if one subgroup is a subset of the other. The discussion emphasizes the importance of precise definitions and logical reasoning in proving these concepts.

PREREQUISITES
  • Understanding of additive groups, specifically (Z,+)
  • Knowledge of least common multiples (LCM)
  • Familiarity with subgroup properties and definitions
  • Ability to apply logical reasoning in mathematical proofs
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  • Study the properties of additive groups in detail
  • Learn how to calculate least common multiples (LCM) effectively
  • Research subgroup criteria and their implications in group theory
  • Explore logical proof techniques in mathematics
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Mathematicians, students of abstract algebra, and anyone interested in group theory and its applications will benefit from this discussion.

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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.
 
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