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Groups and subgroups

  1. Jul 23, 2011 #1
    Well, apparently, I'm not too clear on a few things.

    For #4, what is <[8]> in the group Z18? What does the <> mean around the congruence class? Is my work correct?

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110723_161645.jpg?t=1311456158 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110723_161721.jpg?t=1311456173 [Broken]


    For #8, I understand making a closed subset of Z. It appears H = {[0], [1], [3]} is closed under addition in Z4. What exactly is the additive group Z?

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110723_161701.jpg?t=1311456186 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110723_161733.jpg?t=1311456197 [Broken]
     
    Last edited by a moderator: May 5, 2017
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  3. Jul 23, 2011 #2

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    An additive group is one where the function is addition (a group is a set and a function, remember).
    Z is the standard notation for the integers.

    I recall that we used <> to denote the generated set. For instance <[8]> in Z_18 is the set of all numbers, modulo 18, that can be generated by x*8.

    For instance, [8] is a member, as is [16], as is [6] (16+8=24=[6]).
     
  4. Jul 23, 2011 #3
    Did you take a look at the problems and my work?
     
  5. Jul 23, 2011 #4

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    Yeah, I sure did. Was there something that made you think I didn't?

    I wasn't wanting to go way in depth.

    To be honest, I couldn't really follow the work for the first question.

    As for question 2, it asked about Z, the integers. You seemed to instead work in Z_4, which is a different set. I was trying to hint that this was not correct.
     
  6. Jul 23, 2011 #5
    For #4, I said <[8]> = <[2]> in Z18. Is that correct?

    Then, I generated a set in Z18 from <[2]>: {[16], [14], [12], [10], [8], [6], [4], [2], [0]}.

    The order of this set is 9.

    For #8, I know what a closed subset of Z is. A subset H of a group G has to be nonempty and if a, b are contained in H, then ab-1 is in H. They're asking for a subset of Z. Do they want a set of congruence classes in Z?
     
  7. Jul 23, 2011 #6

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    I am not sure if <[8]>=<[2]> in Z_18 or not, I don't recall that being a rule, though.

    I believe that for question 8 they are asking for a set that is closed under addition but is not a subset of Z. They are not asking for a modulo.

    (Z,+) is a group, and is also a sub-group of (R,+).

    There are more axioms then the two you stated.
    if G=(S,*) is a group, then
    S contains the identity I under *
    for all x in S there exists x^-1 s.t. x*x^-1=I
    for all x,y in S x*y is in S.
    a*(b*c)=(a*b)*c

    So I guess you gotta find a set that is closed under addition but doesn't meet one of those criteria.
     
  8. Jul 23, 2011 #7
    Those two iff conditions are a later theorem. The other properties were listed earlier. I suppose the earlier properties would be easier to produce a counterexample with a particular set.

    The additive group is (Z,+). Now, I just need a closed subset that is not a group of the additive group (Z,+).
     
  9. Jul 24, 2011 #8

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    "The additive group is (Z,+). Now, I just need a closed subset that is not a group of the additive group (Z,+). "
    Yep, do any come to mind?
    With this kind of thing I always find it easier to decide which axiom I am going to not meet and then decide on the set.
     
  10. Jul 24, 2011 #9
    Well, the first thing that comes to mind is the set of positive even integers. There's no identity or inverse.
     
  11. Jul 24, 2011 #10

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    Perfect. Though, the positive integers alone would also have worked. So, did the thread clear up what you were wondering about?
     
  12. Jul 24, 2011 #11
    Yeah, that's true. The identity exists, but an inverse doesn't exist.

    Does my work for #4 make sense?
     
  13. Jul 24, 2011 #12

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    I couldn't really follow your working for question 4.
    What made it so you can say <[2]>=<[8]>? I didn't catch that part.
     
  14. Jul 24, 2011 #13

    Hi, I looked at your solution. Once you have <[8]> = <[2]>, your enumeration of the elements of <[2]> is fine.

    However, you didn't say how you know that <[8]> = <[2]>. It's true, but how do you know it's true? You either have to calculate it out by enumerating the elements of <[8]>; or else you have to have some other argument.
     
  15. Jul 24, 2011 #14
    No, they asked for a subset, not a congruence class. You're overthinking this. A subset is just a subset. What's a subset of Z? How could a subset be closed under addition but not be a subgroup of Z when Z is regarded as an additive group?
     
  16. Jul 24, 2011 #15
    Well, in Z18, 13[2] = [26] = [8]. Is that how I correctly show they're equivalent?
     
    Last edited: Jul 24, 2011
  17. Jul 24, 2011 #16
    Didn't we start out in Z18?
     
  18. Jul 24, 2011 #17
    Oops. Yes.
     
  19. Jul 25, 2011 #18
    Is it because (8,18) = 2? Is that how I found a similar subgroup?
     
  20. Jul 26, 2011 #19

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    I don't understand what you're asking in that last post.

    As for the other thing,
    You said
    "Well, in Z18, 13[2] = [26] = [8]. Is that how I correctly show they're equivalent? "

    But isn't 26[1]=[26]=[8]? But <[1]>\=<[8]> Unless I am forgetting some theorem here, I don't think that your method is a full proof that <[2]>=<[8]>. I may be wrong, though.
    Personally, to be safe, I would just find <[8]> directly.
     
  21. Jul 26, 2011 #20
    gcd(8, 18) = 2. That's why I said they're equivalent.
     
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