The Galilean transformations form a group of bijections of the coordinate space

[Injectivity] Assume $g_{(b,a,v,R)}=g_{(b^{\mathrm{\prime }},a^{\mathrm{\prime }},v^{\mathrm{\prime }},R^{\mathrm{\prime }})}$. Evaluating at $(0,0)$ and comparing first coordinates gives $b=b^{\mathrm{\prime }}$. Evaluating at $(0,x)$ then gives $Rx+a=R^{\mathrm{\prime }}x+a^{\mathrm{\prime }}$ for every $x\in {\mathbb{R}}^3$; at $x=0$ this gives $a=a^{\mathrm{\prime }}$, and after subtracting, $(R-R^{\mathrm{\prime }})x=0$ for every $x$, which for $x=e_1,e_2,e_3$ compares the three columns of the matrix $R-R^{\mathrm{\prime }}$ and gives $R=R^{\mathrm{\prime }}$. Evaluating at $(t,0)$ then gives $vt+a=v^{\mathrm{\prime }}t+a^{\mathrm{\prime }}$ for every real $t$; subtracting the known $a=a^{\mathrm{\prime }}$ and taking $t=1$ gives $v=v^{\mathrm{\prime }}$. Hence $\theta ={\theta }^{\mathrm{\prime }}$.

[Composition] For every $(t,x)\in C$, applying first $g_{(b_1,a_1,v_1,R_1)}$ and then $g_{(b_2,a_2,v_2,R_2)}$ gives $(t+b_1+b_2,R_2(R_{1}x+v_{1}t+a_1)+v_2(t+b_1)+a_2)$. By the linearity and associativity identities of [L3] this equals $(t+(b_1+b_2),(R_2{R}_1)x+(R_2{v}_1+v_2)t+(R_2{a}_1+v_2{b}_1+a_2))$, which is $g_{{\theta }_3}(t,x)$ for ${\theta }_3:=(b_1+b_2,R_2{a}_1+v_2{b}_1+a_2,R_2{v}_1+v_2,R_2{R}_1)$; and $R_2{R}_1\in SO(3)$ by [L2], so ${\theta }_3\in \mathrm{\Theta }$ and the composite is Galilean with the displayed parameters. Taking the second quadruple to be $(0,0,0,I_3)$ makes the composite equal $g_{{\theta }_1}$ by [L2], and taking the first to be $(0,0,0,I_3)$ makes it equal $g_{{\theta }_2}$; hence $g_{(0,0,0,I_3)}$ is the identity map of $C$, and the composite lies in $G$.

[Inverse] Let $\theta =(b,a,v,R)$ and $\eta :=(-b,-R^{\mathsf{T}}(a-vb),-R^{\mathsf{T}}v,R^{\mathsf{T}})$; by [L2] $R^{\mathsf{T}}\in SO(3)$, so $\eta \in \mathrm{\Theta }$. For $(t^{\mathrm{\prime }},x^{\mathrm{\prime }})\in C$ put $t:=t^{\mathrm{\prime }}-b$ and $x:=R^{\mathsf{T}}(x^{\mathrm{\prime }}-v(t^{\mathrm{\prime }}-b)-a)$. Then $t+b=t^{\mathrm{\prime }}$, and $Rx+vt+a=R{R}^{\mathsf{T}}(x^{\mathrm{\prime }}-v(t^{\mathrm{\prime }}-b)-a)+v(t^{\mathrm{\prime }}-b)+a=x^{\mathrm{\prime }}$ by $R{R}^{\mathsf{T}}=I_3$ of [L2]; hence $g_{\theta }(t,x)=(t^{\mathrm{\prime }},x^{\mathrm{\prime }})$ and $g_{\theta }$ is surjective. Expanding $x$ gives $x=R^{\mathsf{T}}{x}^{\mathrm{\prime }}+(-R^{\mathsf{T}}v)t^{\mathrm{\prime }}+(-R^{\mathsf{T}}(a-vb))=g_{\eta }(t^{\mathrm{\prime }},x^{\mathrm{\prime }})$, so $g_{\theta }\circ g_{\eta }$ is the identity of $C$. If $g_{\theta }(t_1,x_1)=g_{\theta }(t_2,x_2)$ then $t_1=t_2$ from the first coordinate, and $R(x_1-x_2)=0$ from the second, giving $x_1=x_2$ because $R$ is invertible by [L2]; so $g_{\theta }$ is injective, and with surjectivity it is a bijection.

[Two-sided inverse] By step 1.2 the composite $g_{\theta }\circ g_{\eta }$ has parameters $(-b+b,R{a}_{\eta }+v{b}_{\eta }+a,R{v}_{\eta }+v,R{R}^{\mathsf{T}})$ with $(b_{\eta },a_{\eta },v_{\eta },R_{\eta })=\eta$, so its components are $0$, $R(-R^{\mathsf{T}}(a-vb))+v(-b)+a=-(a-vb)-vb+a=0$, $R(-R^{\mathsf{T}}v)+v=-v+v=0$ and $R{R}^{\mathsf{T}}=I_3$; hence $g_{\theta }\circ g_{\eta }=g_{(0,0,0,I_3)}$, the identity of $C$ by step 1.2. This proves clause 2 with the displayed inverse.

[Subgroup] By steps 1.2 and 2.1 the set $G$ contains the identity of the bijection group of $C$ and is closed under composition and inversion; by step 1.1 the parameter map is injective, and every $g_{\theta }$ is a bijection by steps 1.3 and 2.1. Therefore $G$ is a subgroup of the group of all bijections of $C$ under composition, and clauses 1-4 hold.