Proving the set of 2D Lorenz functions are a group. Observer in reference two moving at speed $v$ relative to observer one. Given $(x', t') = f_{v}(x,t) = \dfrac{1}{\sqrt{1-v^2}}(x -vt, t - vx)$ The 2-dimensional Lorenz group is the set of functions: $L = \\{f_v \hspace{2mm} | \hspace{2mm} -1 < v < 1\\}$ Prove that $L$ is a group under composition. * * * I am initially stuck on proving whether the function is a binary operation, I keep ending up with an ungodly $(x', t')$ upon applying the function composition. EDIT: As a demonstration of my attempt: $f_v(f_v(x,t)) = \left( \dfrac{\dfrac{x-vt}{\sqrt{1-v^2}} - v \left(\dfrac{t-vx}{\sqrt{1-v^2}} \right)}{\sqrt{1-v^2}}, \dfrac{\dfrac{t-vx}{\sqrt{1-v^2}} - v\left( \dfrac{x-vt}{\sqrt{1-v^2}} \right)}{\sqrt{1-v^2}} \right )$ Which can be slightly simplified albeit I can't see a nice expression surfacing.

Probably you are seeking for answer in Einstein's special theory of relativity with scaling speed of light to $1$. Now let $(x', t') = f_{v}(x,t) = \dfrac{1}{\sqrt{1-v^2}}(x -vt, t - vx)$ therefore $$f_{-v}(x',t')=\dfrac{1}{\sqrt{1-v^2}}(x'+vt',t'+vx')=\dfrac{1}{\sqrt{1-v^2}}\left(\dfrac{x-vt}{\sqrt{1-v^2}}+\dfrac{vt-v^2x}{\sqrt{1-v^2}},\dfrac{t-vx}{\sqrt{1-v^2}}+\dfrac{vx-v^2t}{\sqrt{1-v^2}}\right)=\dfrac{1}{\sqrt{1-v^2}}\left(\dfrac{x(1-v^2)}{\sqrt{1-v^2}},\dfrac{t(1-v^2)}{\sqrt{1-v^2}}\right)=(x,t)$$or $$f_{-v}f_v(x,t)=(x,t)$$this is because if frame 1 moves with speed $v$ respect to frame 2 then frame 2 moves with speed $-v$ respect to frame 1.