<s>[INST] How to prove that $(G,*)$ is a group? Let $G=\mathbb{R_0}\times\mathbb{R}$ where $\mathbb{R_0}=\mathbb{R}\setminus\{{0}\}$.  Define operation $*$ on $G$ by $(a,b)*(x,y)=(ax,a^2y+b)$.
I'd like to prove that $(G,*)$ is a group.  Immediately $(a,b), (x,y)\in G$ implies $(ax,a^2y+b)\in G$.
I need help to prove associativity of $*$.
Thanks for replies.
 [/INST] *

*First note that $*$ is well-defined. If $(a,b), (x,y) \in G$, then $a,x \in \Bbb R^*$, so $ax \in \Bbb R^*$, which means that $(ax, a^2 y + b) \in \Bbb R^* \times \Bbb R = G$.

*We show that $(1, 0) \in G$ is the neutral element of $G$: 
$$\begin{align*} (1,0) * (x,y) &= (1\cdot x, 1^2 y + 0) = (x,y) \\ (x,y) * (1,0) &= (x\cdot 1, x^2 \cdot 0 + y) = (x,y) \end{align*}$$

*Next we show, that for $(x,y) \in G$, $\left(\frac 1 x, -\frac{y}{x^2} \right) \in G$ is the inverse of $(x,y)$:
$$\begin{align*}(x,y) * \left( \frac{1}{x}, - \frac{y}{x^2} \right) &= \left( x \cdot \frac 1 x, x^2 \cdot \left(-\frac{y}{x^2} \right) + y \right) = (1,0) \\ \left( \frac 1 x, - \frac{y}{x^2} \right) * (x,y) &= \left( \frac{1}{x} \cdot x, \left( \frac{1}{x}\right)^2 \cdot y - \frac{y}{x^2} \right) = (1,0) \end{align*}$$

*It remains to show, that the operation $*$ is associative. Let $(a,b), (c,d), (e,f) \in G$. Then 
$$\begin{align*} \left( (a,b) * (c,d) \right) * (e,f) &= \left( ac, a^2d+b) *(e,f) \right) = \left( ace, a^2c^2f + a^2d + b\right) \; , \\ (a,b) * \left( (c,d) * (e,f) \right) &= (a,b) * \left( ce, c^2f + d\right) = \left(ace, a^2 c^2 f+a^2 d + b\right) \; . \end{align*}$$
So we have shown, that $G$ is a group.
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