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Let $\mathbb R$ be the set of all real numbers and $H = \mathbb R\times \mathbb R$. Define a binary operation $\circ$ on $H$ as follows $(p,q) \circ (r,s) = (pr,ps+q)$, where $(p,q),(r,s) \in H$? Is $(H, \circ )$ a group?

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What do you mean "how to define"? In what respect does what you write doffer from what can be called a definition? –  Hagen von Eitzen Dec 23 '13 at 16:32
    
What have you tried? Where are you having problems? –  Igor Rivin Dec 23 '13 at 16:36
    
I just want to prove that $(H, \circ )$ a group. –  Rokai Dec 23 '13 at 16:37
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@Rokai Do you know what axioms a group satisfies? –  tylerc0816 Dec 23 '13 at 16:40
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To prove this is a group, you need to check all the axioms of a group. You need an identity, so can you find an element $(x,y)$ such that $(p,q)\circ (x,y)=(p,q)=(x,y)\circ(p,q)$. If you expand out the $\circ$'s it should be easy to figure out what $x$ and $y$ need to be. Then look for the inverse of a general element $(p,q)$. Then see if associativity is satisfied.

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Actually i'm having difficulties around expanding $(p,q)\circ (x,y)=(p,q)=(x,y)\circ(p,q)$. Can you please help me? I'm new to this kind of problems. –  Rokai Dec 23 '13 at 16:57
    
You plug it into the definition of $\circ$. So $(p,q)\circ(x,y)=(px,py+q)$. Now to have this equal to $(p,q)$ you need $p=px, py+q=q$ What does that tell you about $x,$ and $y?$ Does it work on the other side? –  Ross Millikan Dec 23 '13 at 17:08
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Hint: Find the identity of the group and show not all elements have inverse in the group.

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