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Am I verifying that this is a group correctly?

  1. Dec 3, 2015 #1
    (a,b)*(c,d)=(ac,bc+d)on the set {(x,y)∈ℝ*ℝ:x≠0}
    1.(b,a)*(d,c)=(bd,ad+c) so not commutative
    2.[(a,b)*(c,d)]*(e,f)=(ace,(bc+d)e+f)=(ace,bce+de+f)
    (a,b)*[(c,d)*(e,f)]=(ace,bce+de+f) so associative

    Is that correct so far? What do I do next?
     
  2. jcsd
  3. Dec 3, 2015 #2

    fresh_42

    Staff: Mentor

    The multiplication is associative as you showed in 2. However, your multiplication in 1. is wrong although the conclusion is right. The commutative property would mean (a,b)*(c,d) = (c,d)*(a,b) which is not true. But it does not mean to change the components. Still missing are: 3. multiplication is defined within your set, 4. there exist an element 1=(1,0) with 1*(a,b) = (a,b) and 5. an inverse (c,d) for each (a,b) such that (c,d)*(a,b) = 1.

    EDIT: The usual order is: 3, 2, 4, 5, 1.
     
  4. Dec 3, 2015 #3
    ok so:
    3.(a,b)*(e1,e2)=(a,b)
    (ae1,be1+e2)=(a,b)
    ae1=a e1=a/a=1
    be1+e2=b e2=b-b(1)=0
    (a,b)*(1,0)=(a(1),b(1)+0)=a,b
    (1,0)*(a,b)=(1(a),0(a)+b)=a,b so it has an identity
    4. (a,b)*(a',b')=(1,0)
    (aa',ba'+b')=1,0
    aa'=1 a'=1/a
    ba'+b'=0 b'=-ba'=-b/a
    (a,b)*(1/a,-b/a)=(a/a,b/a+(-b/a))=1,0
    (1/a,-b/a)*(a,b)=(a/a,-ba/a+b)=1,0 so it has an inverse

    so it is a group. Did i do it right?
     
  5. Dec 3, 2015 #4

    fresh_42

    Staff: Mentor

    Yes, that is correct. It's probably easier to write the elements as column vectors:
    ##\begin{pmatrix} a \\ b \end{pmatrix} ∈ \{ \begin{pmatrix} x \\ y \end{pmatrix} ∈ ℝ^2 ## | ## x ≠ 0 \}##

    Just to be complete: You should at least mention that your multiplication takes place within the set of elements. That is quite obvious here because for
    ##\begin{pmatrix} a \\ b \end{pmatrix} , \begin{pmatrix} c \\ d \end{pmatrix} ∈ \{ \begin{pmatrix} x \\ y \end{pmatrix} ∈ ℝ^2 ## | ## x ≠ 0 \}## with ##\begin{pmatrix} a \\ b \end{pmatrix} * \begin{pmatrix} c \\ d \end{pmatrix} = \begin{pmatrix} ac \\ bc + d \end{pmatrix}##, ##ac ≠ 0.##

    As I said, it's obvious here. But that property is often forgotten and not always obvious at first sight. Actually it has to be placed at the beginning to ensure the binary operation doesn't leave the set, assumed group.

    EDIT: You showed that the left identity is also the right identity. This is not really necessary since it follows from the associative property. (Try to show how, if you like.) And you did not show, that the left inverse is also the right inverse. Why? Ok, you don't have to, since it can be followed from the other rules, as well. (Same exercise, if you like to do.) I'm just curious why you didn't whereas you did with the identity.
     
    Last edited: Dec 3, 2015
  6. Dec 3, 2015 #5
    What do you mean?
     
  7. Dec 3, 2015 #6
    i'm not sure what you mean. I did show the left and right inverses.
     
  8. Dec 3, 2015 #7

    fresh_42

    Staff: Mentor

    Sorry for that, I overlooked it. (Nevertheless it's a good exercise in group theory to show that lefts and rights are equal by using the associative property.)
     
  9. Dec 3, 2015 #8
    its alright. Thanks anyways
     
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