$G$ is set $\{x \in \mathbb{R} : x \neq -1\}$ with group operation $x\circ y=x+y+xy$.
Show that $f(x)=x-1$ is an isomorphism from the multiplicative group $\mathbb{R}^*$ to $G$.
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Does $R$ here mean the real numbers? I'll format the question assuming this is the case. – hardmath Oct 11 '15 at 21:12
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The answer of lhf shows that $g(x)=x+1$ and its inverse $f(x)=x-1$ are group isomorphisms. – Dietrich Burde Oct 11 '15 at 21:19
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in that question is set f(x)=x+1. but in here is x-1. but they got same answer, i still dont know how to solve it, sorry. – Helen Oct 11 '15 at 21:25
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To show $f$ is a homomorphism, you must show that:
$f(xy) = f(x) \circ f(y)$, that is:
$xy - 1 = (x - 1) \circ (y - 1) = (x - 1) + (y - 1) + (x -1)(y-1)$
$= x - 1 + y - 1 + xy - x - y + 1 = xy + (x - x) + (y - y) + 1 - 1 - 1$
but this is clearly true.
Now $f$ is a bijection between $\Bbb R$ and itself, and $f$ maps $0$ to $-1$, showing $f$ is thus also a bijection from $\Bbb R^{\ast} = \Bbb R\setminus\{0\}$ to $\Bbb R\setminus \{-1\}$, which is the underlying set of $G$.
David Wheeler
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