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Group Theory: Is the following a valid Group?

  1. Oct 8, 2015 #1
    1. The problem statement, all variables and given/known data

    Is the following a valid group?

    The values contained in the set of all real numbers ℝ, under an operation ◊ such that x◊y = x+y-1

    2. Relevant equations

    Axioms of group theory:
    Closure
    Associativity
    There must exist one identity element 'e' such that ex=x for all x
    There exists an inverse 'y' for every element such that xy = yx = e

    3. The attempt at a solution

    Testing of each axiom:

    Closure: All x◊y will give a value existing in ℝ. Closure holds.

    Associativity: Test: x◊y◊z
    = x◊(y◊z) = x◊(y+z-1) = x+y+z-2
    = (x◊y)◊z = (x+y-1)◊z = x+y+z-2
    Associativity holds.

    Identity: e◊x = x
    Group definition says e◊x = e+x-1.
    Hence, e◊x = e+x-1 = x
    This is true for all x when e = 1
    There is an identity

    Inverses: x◊y = y◊x = e = 1
    Meaning, x+y-1 = y+x-1
    x,y∈ℝ so the additive property defined by operation ◊ means that all elements will be able to have an inverse in ℝ such that x◊y=e
    Inverses exist

    Final answer, YES THIS IS A GROUP.

    So, that's my attempt at an answer. However I don't know if I've really gone about it the right way, or even if I'm correct. Thoughts?

    Thanks.
     
  2. jcsd
  3. Oct 8, 2015 #2

    Mark44

    Staff: Mentor

    I don't see anything wrong with what you did, although I have to admit it's been many years since I did this stuff.
     
  4. Oct 8, 2015 #3
    Well, that's a positive enough response for now! Thanks :)
     
  5. Oct 9, 2015 #4

    Fredrik

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    Staff Emeritus
    Science Advisor
    Gold Member

    The part about inverses can be improved a bit. You didn't make it clear how you know that each x has an inverse. The best way is to find a formula for the inverse of an arbitrary x.
     
  6. Oct 9, 2015 #5
    Thanks, I've gone over it and written:

    x◊y = x+y-1 = e = y+x-1 = y◊x
    Meaning
    y = e+1-x
    which is true and different for all x because it only uses addition to create the inverse
     
  7. Oct 9, 2015 #6

    HallsofIvy

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    Didn't you just show that e= 1? So you are saying that if y= 1+ 1- x= (1+ 1)- x= 2- x then x◊ y= x+ y- 1= x+ (2- x)- 1= 1.
     
  8. Oct 10, 2015 #7
    I wasn't sure if I was 'allowed' to do a proof with e=1, or if I was supposed to keep it as 'e' so it's more general.

    As I'm primarily a physics student, I haven't really done much on proofs before.

    If the 'e=1' method is fine then I'll certainly use it! :biggrin:
     
  9. Oct 10, 2015 #8

    Fredrik

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    You can and should use it. The equality ##e=1## means that the symbols ##e## and ##1## represent the same element of the set. If they represent the same element, then the expression ##e+1-x## must represent the same element as the expression ##2-x##. In other words, since ##e=1##, we have ##e+1-x=2-x##.

    Since you have completed the solution, I'll show you mine. It's essentially the same as yours.

    For all ##x,y,z\in\mathbb R##, we have
    $$(x\diamond y)\diamond z= (x+y-1)+z-1= x+(y+z-1)-1=x\diamond(y\diamond z).$$ So ##\diamond## is associative. For all ##x\in\mathbb R##, we have
    $$x\diamond 1=x+1-1=x$$ and
    $$1\diamond x=1+x-1=x.$$ So ##1## is an identity of ##\diamond##. Let ##x\in\mathbb R##. We have
    $$x\diamond (2-x)=x+(2-x)-1=1$$ and
    $$(2-x)\diamond x=(2-x)+x-1=1.$$ Since ##1## is an identity of ##\diamond##, this means that ##2-x## is an inverse of ##x##. Since ##x## is an arbitrary element of ##\mathbb R##, this means that every element of ##\mathbb R## has an inverse.
     
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