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A question about groups

  1. Dec 29, 2012 #1
    1. The problem statement, all variables and given/known data

    What exactly does G={ f: R -> R : f(x)=ax+b, where a is not equal to zero} is a group under composition, mean? So what are the elements of G? Are they (for example) f(x)=ax+b and g(x)=a'x+b'? Or are they f(x)=ax+b and f(y)=ay+b?

    Thanks in advance

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 29, 2012 #2

    Dick

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    The first, of course, they are all linear polynomials in x.
     
  4. Dec 29, 2012 #3
    Thanks...
     
  5. Dec 29, 2012 #4

    Dick

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    I read the problem statement, the other interpretation doesn't make much sense. f(x)=ax+b and g(y)=ay+b define the same function. Some example of elements of G are 2x+1, x-1, -2x+4, etc etc.
     
  6. Dec 29, 2012 #5
    Thanks. Yes, I figured that out...it doesn't make any sense...I don't know why I was confused. :)
     
  7. Dec 30, 2012 #6

    HallsofIvy

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    On the other hand, although the group is defined as {f(x)= ax+ b}, it is not a good idea to use "f" to represent two different members. Rather, say that f(x)= ax+ b and g(x)= a'x+ b'. The group operation, "composition" would give fg= a(a'x+ b')+ b= aa'x+ (ab'+ b) and gf= a'(ax+ b)+ b'= aa'x+ (a'b+ b'). Since, in general, [itex]ab'+ b\ne a'b+ b'[/itex] this group is not commutative.

    Of course, the function f(x)= 1x+ 0= x is the group identity: if g(x)= ax+ b then fg= 1(ax+ b)+ 0= ax+ b and gf= a(x)+ b= ax+b. What is the inverse of f(x)= ax+ b?
     
  8. Dec 30, 2012 #7
    The inverse is: [tex]f^{-1}(x) = a^{-1}x - a^{-1}b[/tex], right? Because...

    [tex]a(a^{-1}x - a^{-1}b) + b = (x-b)+b = x[/tex]
     
  9. Dec 30, 2012 #8

    Dick

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    Yes, that's exactly right.
     
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