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Homework Help: Group Structure

  1. Aug 5, 2009 #1
    Group properties:
    1. [tex]\forall a, b, c \in G,
    (a * b) * c = a * (b * c).[/tex] (associativity)

    2. [tex]\exists e \in G[/tex] such that [tex]\forall x \in G,
    e * x = x * e = x.[/tex] (identity)

    3. [tex]\forall a \in G, \exists a' \in G[/tex] such that, [tex]
    a * a' = a' * a = e[/tex] (inverse)

    Determine whether the binary operation [tex]*[/tex] gives a group structure on the given set.

    Let [tex]*[/tex] be defined on Q by letting [tex]a * b = ab[/tex].

    Thought process:
    To begin, one has to understand the three properties of being a group- which is defined above. Can someone help me go through the process of testing the three properties from above to our specified problem?


    Last edited: Aug 5, 2009
  2. jcsd
  3. Aug 5, 2009 #2
    I'm going to assume your Q represents the field of rational numbers.

    The binary operation you have seems to be multiplication of rational numbers. So to test the 3 properties, you use arbitrary rational numbers and show that the operation is associative, there exists another element that is the inverse of another element, and that there exists an element that acts like the identity element. This shouldn't be too hard so I will stop with this.

    If you are still stuck, just let the arbitrary rational numbers be of the form [tex] \frac{a}{b} [/tex] where [tex] b \neq 0 [/tex] and [tex] a,b \in \bbmath{Z}[/tex].
  4. Aug 5, 2009 #3
    Question 1:
    Can I let "a, b, c" be rational numbers in the set G- as done in my proof below?

    Question 2:
    So as you said, this problem was fairly straightfoward. I did all three axiom test, and they all passed. But according to the solution, this problem fails at the third criterion. But according to my calculations,
    [tex]a * a' = aa' = e = a'a = a' * a,[/tex]
    which shows (along with the first two axiomatic criterions) that our given binary operation [tex]*[/tex] produces a group structure on our given set. Am I correct?

    Question 3:
    Also, to show a given binary operation is a group structure we have to prove the three axiomatic conditions (as stated above). However, the binary operation must fulfill closure. How can I show our given binary operation fulfills "closure"?

    Thanks again,

    Last edited: Aug 5, 2009
  5. Aug 5, 2009 #4
    Ah, my bad. I forgot that the set of rational numbers under multiplication is only a group if 0 isn't included. So Q isn't a group under multiplication but Q* is (Q* is convention for the rationals without 0). This is because 0 has no inverse in the rational numbers.

    Also to be a bit more complete, a group structure has to fulfill 4 axioms. 3 of them have already been mentioned in the first post and closure under the operation. Closure is just a way of saying that the performing the binary operation on 2 elements will not make another element not in the group.

    Let's use a concrete example such as Q*. Let [tex] \frac{a}{b}, \frac{c}{d} [/tex] be arbitrary elements in Q*. To prove closure, you want the operation to [tex] \frac{a}{b} \cdot \frac{c}{d} [/tex] still be an element of Q* where it should be of the form [tex] \frac{x}{y} [/tex].
  6. Aug 5, 2009 #5
    Is the logic in the quote above, along with the proof below reasonable?

    Proof (Closure):
    Let [tex]a = \frac{x}{y}, b = \frac{m}{n}[/tex] (such that [tex]y \neq 0, n \neq 0[/tex] since a, b are defined as rational elements).
    Therefore, [tex]a * b = \frac{x}{y} \frac{m}{n}[/tex]
    Since [tex]\frac{x}{y} \frac{m}{n}[/tex] is rational, it follows that our binary operation fulfills closure.


  7. Aug 5, 2009 #6
    Zero is a rational number. if a=0, what does a' equal?

    Question 3: First, can you write-out the symbolic statement for closure?
  8. Aug 5, 2009 #7
    [tex]a' = \frac{1}{0}[/tex] which is irrational, I'm guessing that means my proof is incorrect.

    What do mean, is the proof I wrote of closure insufficient?


    Last edited: Aug 5, 2009
  9. Aug 5, 2009 #8
    The thing is that Q under the operation of multiplication is not a group, so you won't be able to prove it is a group. But consider Q* (rationals without 0).

    Without showing fractions explicitly, I'm not sure how you can show rational numbers are closed under multiplication.
  10. Aug 6, 2009 #9
    Got it, since [tex]a' = \frac{1}{0},[/tex] then the third group axiom fails- which shows our particular binary operator does not fulfill the conditions of a group structure (even though the first two succeeds).

    Though our operation is not a group, a test of closure shows us our particular structure is closed on rationals so long as the denominator of the rational values are not zero (as shown in my proof).

    This can be shown in two cases:
    Case I:
    Assume that the numerator for either rational number is zero. Then if we multiply both rational numbers together, we get zero- which is a rational number.
    Case II:
    Assume neither values have a numerator of zero. Since the denominator for each variable is not zero (hence rational), then performing our binary operation [tex]*,[/tex] we get a real number. Since our result is a real number, it can be expressed as a rational.
    Thus, assuming our variables does not have a denominator of zero, we have shown that the binary operation always yields a real number, and all real numbers can be expressed as rational numbers.
  11. Aug 6, 2009 #10
    It looks good to me JL. I was responding to your previous post. The exception condition, that x and y are not equal to zero, is not necessary to prove closure of the rational numbers under multiplication--all rational numbers have a non-zero denominator by definition.

    Formally, the closure property look something like this.

    [tex](\forall a, b \in G) \wedge (\exists \; c \in G) \wedge (c = a * b) \ .[/tex]
    Last edited: Aug 6, 2009
  12. Aug 6, 2009 #11


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    You shouldn't use all these [tex]\wedge[/tex]'s. "[tex](\forall a,b\in G)[/tex]" is not a statement which can be true or false.

    Closure is just [tex](\forall a,b\in G)(a*b\in G)[/tex]
  13. Aug 6, 2009 #12
    But the statement [tex]\forall a,b\in G[/tex] can be assigned a truth value. However, I'm not familiar with your notation. What does one parenthetic expression followed directly with another parenthetic expression mean?
  14. Aug 7, 2009 #13


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    Science Advisor

    How? What would it mean for this to be true or false?

    Simple example: "for every integer, its square is also an integer" is a true statement. Notation: [tex]\forall z\in\mathbb{Z}:z^2\in\mathbb{Z}[/tex]
    When talking about mathematical induction, we often have a statement about natural numbers P(n), en we want to prove [tex]\forall n\in\mathbb{N}:P(n)[/tex].
    You can't just use a universal quantifier saying 'for all a and b in G', you have to specify which statement applies to all such a and b (namely, that their product is also in G).
    The parentheses just mean that the symbols inside them belong together. But in this case you can also just say [tex]\forall a,b\in G:a*b\in G[/tex].
    Last edited: Aug 7, 2009
  15. Aug 9, 2009 #14
    What function does : serve?
  16. Aug 9, 2009 #15
    Two comments:


    The following two logical sentences [itex](\forall x \in X)P(x) \text{ and }\forall x \in X:P(x)[/itex] both translate as

    "For all elements x in the set X, the proposition P is true for x." or more simply:

    "For all x in X, P(x) is true."

    When the quatifier is "there exists" the colon is typically read as "such that" or "where."


    The statement that all real numbers can be expressed as rational numbers is false. The proof that [itex]\sqrt{2}[/itex] is irrational (yet real) is a classic counter-example.

    I have not yet seen a proof that the rational numbers are closed under multiplication that doesn't involve letting a/b and c/d be two rational numbers (b, d =/= 0) and showing that the product ac/bd is also rational (i.e. that both ac and bd are integers and bd =/= 0).

    The critical principle here is the Zero Product Property of Real Numbers:

    If xy = 0 then either x = 0 or y = 0.

    The comment that Q is not a multiplicative group because 0 has not inverse is correct and this subtlety is the reason why that example appears in most elementary group theory exercises.

  17. Aug 9, 2009 #16
    Quick comment:

    The expression [itex]\frac{1}{0}[/itex] is not irrational. Unfortunatley, it isn't even a number. It is not defined. Moreover, it is not possible to give it numeric meaning without breaking some other (more important or useful) law of real numbers.

    I have had many calculus students who have claimed it equals infinity, which is equally false.

    I make this comment because it is an error (or similar to errors) that I see materialize too frequently and makes me suspect that some of my students have critical gaps in their pre-calculus background. I am not saying that's the case for jeff1levesque but I mention it for the public benefit.

  18. Aug 10, 2009 #17
    Elucidis, how do you know it is absolutely impossible to make division by 0 be defined?
  19. Aug 10, 2009 #18
    Firstly, the expression [itex]\frac{1}{0}[/itex] is the reciprocal of 0, i.e. that number which when multiplied by 0 gives the mulitplicative identity 1: that is [itex]\frac{1}{0}[/itex] * 0 = 1. But it is well established that 0 times any number is 0. From this we conclude that 0 = 1 and consequently every number = 0. This creates a trivial algebra, which is of little practical use (it still is an algebra though - just not interesting).

    If you assign it some numberic meaning, say k, then [itex]k^2=\frac{1}{0}\cdot\frac{1}{0} = \frac{1}{0} = k[/itex] indicating that k2 = k or equivalently k = 0 or 1. Both of these cases can be shown to be degenerate (i.e. concluding that 0 = 1) in a similar manner as above.

    Hence we cannot give any numberic meaning to the expression [itex]\frac{1}{0}[/itex] without collapsing the algebra to the trivial case.

  20. Aug 11, 2009 #19
    Unless I've missed something critical, "A such that B", or "A where B" could also be written "A in addtion to B", or simply "A and B" or even "B where A". Logically, they're all equivalent, though some are easier to read than others.
  21. Aug 12, 2009 #20
    This is true if A and B were both propositions. However [itex]\exists x \in X[/itex] is not a proposition - it's a quantifier, so

    [tex]\forall x \in \mathbb{R},\exists y \in \mathbb{R}:x+y=0[/tex]​

    is interpreted as "For all x in R, there exists y in R such that x + y = 0."

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