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Proving a group is abelian

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data
    (From chapter three of Pinter's A book of Abstract Algebra)
    Prove that each of the following sets, with the indicated operation, is an abelian group:

    a) [itex]x * y = x + y + k [/itex] (k is some constant), for the set of real numbers
    b) [itex]x * y = \frac{xy}{2}[/itex] on the set [itex] \{ x | x \in R; x \neq 0 \}[/itex]
    c) [itex]x * y = x + y + xy[/itex] on the set [itex]\{ x | x \in R; x \neq -1 \}[/itex]
    d) [itex]x * y = \frac{x + y}{xy + 1}[/itex] on the set [itex]\{ x | x \in R; -1 < x < 1 \}[/itex]

    2. The attempt at a solution

    a) This was my over-complicated proof, which I have two questions about:

    Suppose * is an operation on the set of real numbers. Suppose a, b belong to the set of real numbers. The operation * is defined as the binary operation x * y = x + y + k, where k is a constant and x, y belong to the set of real numbers. Say q = a * b = a + b + k, and p = b * a = b + a + k. Due to the commutative property of addition, p = q.

    My first question is, is it bad to use, say, the commutative property of addition in a proof?

    My next is how could I make this proof simpler? It looks overly complex to me... but this almost looks too skimpy:

    Suppose a group <R, *> where R is the set of real numbers and * is the operation x * y = x + y + k, where k is some constant and x, y [itex]\in[/itex] R. Given that a, b [itex]\in[/itex] R, a * b = a + b + k. Also, b * a = b + a + k. Therefore, because a + b + k = b + a + k by the commutative property of addition, a * b = b * a and the group <R, *> is abelian.

    b) Suppose a group <R, *> where R is the set of [itex]\{ x | x \in R; x \neq 0 \}[/itex] and * is the operation [itex]x * y = \frac{xy}{2}[/itex], where x, y [itex]\in[/itex] R. Given that a, b [itex]\in[/itex] R, suppose [itex]a * b = \frac{ab}{2} = b * a = \frac{be}{2}[/itex]. To check, multiply the equation by 2 to yield [itex]ab = ba[/itex], which remains true due to the commutative property of multiplication. Thus, the group <R, *> is abelian.

    c) Suppose a group <R, *> where R is the set of [itex]\{ x | x \in R; x \neq -1 \}[/itex] and * is the operation [itex]x * y = x + y + xy[/itex], where x, y [itex]\in[/itex] R. Given that a, b [itex]\in[/itex] R, suppose [itex]a * b = a + b + ab[/itex] and [itex]b * a = b + a + ba[/itex]. Suppose [itex]a + b + ab = b + a + ba[/itex], implying that a * b = b * a, or that the operation * is commutative. The equation, simplified by subtraction, yields [itex]ab = ba[/itex], which are equivalent by the commutative property of multiplication. Thus, the group <R, *> is abelian.

    d) Suppose a group <R, *> where R is the set of [itex]\{ x | x \in R; -1 < x < 1 \}[/itex] and * is the operation [itex]x * y = \frac{x + y}{xy + 1}[/itex], where x, y [itex]\in[/itex] R. Given that a, b [itex]\in[/itex] R, suppose [itex]a * b = \frac{a + b}{ab + 1}[/itex] and [itex]b * a = \frac{b + a}{ba + 1}[/itex] and [itex]\frac{a + b}{ab + 1} = \frac{b + a}{ba + 1}[/itex]. Simplifying yields [itex]b + a = a + b[/itex], due to the fact [itex]ab + a = ba + 1[/itex]. Because [itex]a * b = b * a[/itex], equivalent by the commutative property of multiplication, the group <R, *> is abelian.

    Thanks for any ... review? Sort of a meaty post, and thanks for making it this far haha.
    Also, I know they don't really need formal proofs, but I like the practice.
     
    Last edited: Aug 5, 2011
  2. jcsd
  3. Aug 5, 2011 #2

    Mark44

    Staff: Mentor

    Comment on your proof of part a.
    Yes, this is overcomplicated. What you're trying to prove for the commutivity part is that x*y = y*x for all elements x and y of your group.

    Let x and y be any real numbers, and let k be a real constant.
    x * y = x + y + k = y + x + k (because addition in the reals is commutative)
    = y * x




     
  4. Aug 5, 2011 #3
    And don't forget associativity, inverses, and an identity! We're tasked with proving its an abelian GROUP. :smile:
     
  5. Aug 5, 2011 #4
    Thanks, I figured - still used to looking at really long proofs haha. Also, thanks for clarifying a point I actually didn't bring up - whether or not you have to specify k as a real or not.

    And @springy: I thought abelian was a synonym for commutative; so proving identity, associativity, etc was separate?

    Thanks
     
  6. Aug 5, 2011 #5

    Mark44

    Staff: Mentor

    An abelian group is one for which the operation is commutative. Since you have to prove that each is an abelian group, you need to show that each is a group, and then that for each, the operation is commutative.
     
  7. Aug 5, 2011 #6
    So you don't have to prove that there exists an inverse,associativity, or anything else for it to be an abelian group, correct?

    Sorry for all the simple questions - just wanna make sure I have the basics down pat before moving on
     
  8. Aug 5, 2011 #7
    "Abelian" is an adjective that applies to the noun "group."

    Here's a situation that is analogous to your proofs: If I told you to prove to me that that (pointing to an object) is a blue book, then all you just did is prove that the object is blue! You still need to show me it's a book. Makes sense, right?

    So if I ask you to prove to me that some set along with an operation is an abelian group, you have to show me that it's a group also. EDIT: And what does that involve? Dealing with associativity, inverses, and an identity.

    Mark44 gave the definition of an abelian group.
     
  9. Aug 5, 2011 #8
    I now understand your post - I thought you were speaking of proving a group to be abelian. Now I realize what you're talking about... sorry for wasting your time - been a long day. Thanks so much for your help.
     
  10. Aug 5, 2011 #9
    Oh, OK. I see where the misunderstanding was. Just to be absolutely crystal clear, once you've established that something is, in fact, a group, then all you need to do to show it's an abelian group is prove commutativity. The issue I pointed out was that you hadn't established that those sets and operations formed groups before tackling commutativity.

    And no, you are not wasting anybody's time. We're all here to help answer questions. Feel free to ask away!
     
  11. Aug 5, 2011 #10
    Thanks, I sort of hijacked the language the book was using and assumed it considered them groups! So it's completely a misunderstanding on my end, thanks for your replies and help :)
     
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