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1999 Paper 3 Q7
D: 1680.5 B: 1516.0
groups binary operation associativity group axioms subgroup identity element inverse element

Let \(a\) be a non-zero real number and define a binary operation on the set of real numbers by $$ x*y = x+y+axy \,. $$ Show that the operation \(*\) is associative. Show that \((G,*)\) is a group, where \(G\) is the set of all real numbers except for one number which you should identify. Find a subgroup of \((G,*)\) which has exactly 2 elements.

Solution: Claim: \(*\) is associative. Proof: Then \(x*(y*z) = x*(y+z+ayz) = x + (y+z+ayz) + ax(y+z+ayz) = x + y + z + a(yz + xy + zx) + a^2xyz\) and \((x*y)*z = (x+y+axy)*z = (x+y+axy) + z+ a(x+y+axy)z = x + y + z + a(yz + xy + zx) + a^2xyz\) so \(x*(y*z) = (x*y)*z\) and we are done. Let \(G = \mathbb{R} \setminus \{-\frac1{a} \}\) In order to show that \((G, *)\) is a group we need to check:

  1. closure \(x*y = x + y + axy\) is a real number
  2. associativity we have already checked
  3. identity \(x*0 = x + 0 + 0 = x = 0*x\), so \(0\) is an identity
  4. inverses \(x*\left ( \frac{-x}{1+ax} \right ) = x - \frac{x}{1+ax} + ax \frac{-x}{1+ax} = \frac{x +ax^2 - x - ax^2}{1+ax} = 0\) so \(x\) has an identity, assuming \(1+ax \neq 0\) which is true for everything in our set
Consider the set \(\{0, \frac{-2}{a} \}\). Then \(\frac{-2}{a} * \frac{-2}{a} = \frac{-4}{a} + a \frac{4}{a^2} = 0\), so this is a group of order \(2\)

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