Show that nZ intersection mZ= lZ

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In summary, the conversation discusses two questions related to group theory. The first question asks to prove that the intersection of two additive groups is equal to the least common multiple of the two groups. The second question asks to show that the union of two subgroups is also a subgroup if and only if one subgroup is a subset of the other. The conversation also includes a suggestion to write down the definitions and where to seek help if needed. An attempted solution to the second question is also provided, but it is pointed out that it is not entirely correct.
  • #1
uob_student
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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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  • #2
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.
 
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  • #3
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
 
  • #4
http://umm.biz/index.php?act=Attach&type=post&id=754324

Is this solution true for question two ??
 
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  • #5
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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Related to Show that nZ intersection mZ= lZ

1. What is nZ and mZ in the equation nZ intersection mZ= lZ?

In this equation, nZ and mZ represent sets of numbers that are multiples of n and m respectively.

2. What does the intersection symbol (∩) mean in the equation nZ intersection mZ= lZ?

The intersection symbol (∩) represents the common elements between two sets. In this equation, it represents the set of numbers that are multiples of both n and m.

3. How do you prove that nZ intersection mZ= lZ?

To prove that nZ intersection mZ= lZ, you must show that all elements in nZ intersection mZ are also in lZ, and vice versa. This can be done by showing that any element in nZ intersection mZ is a multiple of both n and m, and therefore a multiple of l. Similarly, any element in lZ is a multiple of l and therefore also a multiple of both n and m, making it a part of nZ intersection mZ.

4. What are some applications of the equation nZ intersection mZ= lZ?

This equation has applications in number theory, specifically in understanding the relationship between multiples of two numbers. It can also be used in algebraic equations and in various mathematical proofs.

5. Can nZ intersection mZ be an empty set?

Yes, it is possible for nZ intersection mZ to be an empty set. This occurs when there are no common multiples between n and m. For example, if n = 3 and m = 5, there are no common multiples, so nZ intersection mZ would be an empty set.

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