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Groups Axioms

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data

    For each of the following definitions for a * b and a given set, determine which of the axioms
    G0, G1, G2, G3 are satisfied by a * b. In which cases do we obtain a group?

    1) a - b on the set Z
    2)a + b - ab on the set R | {1}
    3)ab on {2^n | n in Z}

    2. Relevant equations

    3. The attempt at a solution

    The four axioms for group are:

    G0 For all a, b in G, a*b in G.
    G1. Associativity. For all a, b, c in G, (a * b) * c = a * (b * c).
    G2. Identity. a * e = a = e * a.
    G3. Inverses. a * b = e = b * a.

    How am i going to check these axioms for example 1 which says a - b on the set Z ?
  2. jcsd
  3. Feb 20, 2012 #2

    If you subtract two integers, do you get another integer?

    Does it matter the order you subtract integers?
    Example is (2-3)-5 = 2-(3-5)???

    Is there integer such any number minus that integer is the the original integer?

    Is there an integer when subtracted to another integer you obtain the value(from above)?

    Can you provide a counter example to any of the above? If so, you don't have a group. If not, prove them all which is somewhat trivial.
  4. Feb 20, 2012 #3
    Ok but how are you going to prove these axioms with just these equations and letters without giving any number. I dont understand what is happening with the letters
  5. Feb 21, 2012 #4


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    Associative: is (a- b)- c= a- (b- c)?

    Identity: does there exist an integer so that, x, such that a- x= x- a= a for all a?

    Inverse: if there does exist such an a, given x, does there exist a y such that x- y= a?
  6. Feb 21, 2012 #5


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    if that is so, you have an uphill battle in learning group theory. i will try to explain, but maybe i won't do that so well.

    a) to prove G0, where "*" is subtraction (that is, by a*b, we mean a-b), we need to show that for any two integers a and b, that a-b is in Z.

    now, you might think, "which two integers"? it shouldn't matter. we use the letters a and b, because we aren't thinking of any two PARTICULAR numbers (like 3 and 5), but any two numbers at all.

    since, there are an infinite number of integers, we just don't have the time or the space to verify a rule for "every" integer, unless we can do it SYMBOLICALLY. while certain rules for integers can indeed be proved "from scratch" (using only logic, and a suitable defintion of "integer"), i'm sure your instructor will allow the following rules about integers to be taken "as fact":

    a+b is always an integer
    (a+b)+c = a+(b+c)
    a+b = b+a
    a+0 = 0+a = a
    a+(-a) = (-a)+a = 0
    ab is always an integer
    a(bc) = (ab)c
    ab = ba
    (a)(1) = (1)(a) = a
    a(b+c) = ab+ac

    these are all rules that you should have learned many years ago. note that these rules are only true for the integers, for other sets, they may, or may not be true.

    for your first example, you need to verify G0,G1, G2 and G3 for a*b = a-b. if at any point, one of the rules fails, you can stop there. i'll leave the verification of G0 to you. it may be helpful to remember that:

    a-b = a+(-b)

    for G1, we have to either show:

    1) a-(b-c) = (a-b)-c, for EVERY set of 3 integers {a,b,c} -OR-

    2) for AT LEAST ONE set of 3 integers, a-(b-c) ≠ (a-b)-c.

    suppose a and b were zero, can you think of a c which shows G1 cannot be true for all a,b,c?
  7. Feb 21, 2012 #6
    So for 1)

    we have:

    G0: For all a,b in Z , a-b in Z because a+(-b)
    G1: For all a,b,c in Z (a-b)-c=a-(b-c) which is not true (how are we going to show this?)

    Therefore it is not a group.
  8. Feb 21, 2012 #7
    By reading post 2.
  9. Feb 21, 2012 #8
    ok thanks.
    Now for 2) which says a + b - ab on the set R | {1}

    we have:

    G0: For all a,b in R| {1} , a+b-ab in RZ because a+b+(-ab)

    G1: For all a,b,c in R| {1} (a+b-ab)-c= a+b-(ab-c) ???
  10. Feb 21, 2012 #9


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    (a*b)*c is not what you've written out. It's
    a*b+c-(a*b)c = a+b-ab+c-(a+b-ab)c
  11. Feb 21, 2012 #10


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    G0: For all a,b in R-{1} , a+b-ab in R-{1} because a+b+(-ab).....?

    this sentence is unfinished.

    it should be clear that a+b-ab is indeed a real number. the question really is, is it in R - {1}?

    if not, then we do not have closure. and the sticking point here, is we need to be sure that if a,b are real numbers that aren't 1, then a+b-ab is a real number that isn't 1.

    ok, suppose a+b-ab = 1. then a(1-b) + b = 1, and a(1-b) = 1 - b.

    is it ok to divide by 1-b? why, or why not? what do we get after we do?
  12. Feb 21, 2012 #11

    a + b - ab on the set R | {1}

    G0: For all a,b in R| {1} , a+b-ab in R| {1}

    So we need to check that a and b are real number excluding 1 as well as the expression a+b-ab.

    Let a+b-ab = 1 , a(1-b) + b = 1 and a(1-b) = 1 - b

    if b is not 1 then we divide both sides by (1-b) as 1-b is not zero.

    which gives a=1 which is not acceptable.

    Therefore G0 does not hold.
    However even if G0 does not hold and it is not a group i have to check for G1,G2,G3.

    What about G1?
  13. Feb 21, 2012 #12


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    i don't think you understand what you just did.

    we first supposed that a ≠ 1, b ≠ 1.

    you then showed that IF a+b-ab = 1, then a = 1.

    that is, a = 1 is the ONLY value for a that makes a+b-ab = 1 true, so long as b is not 1, as well.

    but a = 1 is NEVER true, since a is in R - {1}. so a+b - ab is never 1, so a+b - ab is in R - {1}. this PROVES G0, not disproves it.

    for G1, you need to show that (a*b)*c = a*(b*c).

    (a*b)*c = (a+b-ab)*c. to evaluate this, we need to treat a+b-ab as a single number:

    a+b-ab = x (x is just temporary, we'll get rid of it later).

    x*c = x+c-xc...now we can "substitute a+b-ab back in for x":

    (a*b)*c = (a+b-ab)*c = x*c = x+c - xc = (a+b-ab) + c - (a+b-ab)c

    = a + b + c - ab - ac - bc + abc

    now you evaluate what a*(b*c) is.
  14. Feb 21, 2012 #13
    x*c = x+c-xc. is this a rule ?
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