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Let G be the set of ordered pairs $(a,b)$ where a and b are rational numbers and $a\not= 0$. Let $(a,b)*(c,d)=(ac, ad+b)$.

  1. Show that $G$ forms a group under $*$
  2. Show $A = \{(a,b) \in G: a=1\}$ forms a subgroup.
  3. Does $B=\{(a,b) \in G: b=1\}$ form a group? Explain your answer.
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In view of your earlier question it really would be nice to see some indication that you know what conditions have to be checked to determine that a set and operation form a group, and that a subset of a group is actually a subgroup. –  Brian M. Scott Dec 7 '12 at 23:21
I think (a,b)(c,d) should be (ac,ad+bc) –  Amr Dec 7 '12 at 23:23
Thank you for your help. I have worked out part i) on my own. What about the other two parts? –  bbr4in Dec 7 '12 at 23:28
1) Show that this group law is associative. Guess a neutral element and prove that it is indeed neutral. Guess inverse elemenst and prove that they are indeed inverse. 2) Use a subgroup criterion 3) What about $(1,1)*(1,1)$? –  Hagen von Eitzen Dec 7 '12 at 23:30
Thank you very much for your help, Hagen von Eitzen. –  bbr4in Dec 7 '12 at 23:42
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