Lorentz transformation

Proposition

Given that the interval $x^{\mu }{x}_{\mu }$ is invariant under a Lorentz transformation, prove that the Lorentz transformation is orthogonal.

1.	$x{\text{'}}^{\mu }x{\text{'}}_{\mu }=x^{\alpha }{x}_{\alpha }$.	Given.
2.	$x{\text{'}}^{\mu }x{\text{'}}_{\mu }=g_{\mu \nu }x{\text{'}}^{\mu }x{\text{'}}^{\nu }$	metric tensor
3.	$x{\text{'}}^{\mu }={\mathrm{\Lambda }}^{\mu }{}_{\alpha }{x}^{\alpha }$	Lorentz transformation
4.	$x{\text{'}}^{\nu }={\mathrm{\Lambda }}^{\nu }{}_{\beta }{x}^{\beta }$	Lorentz transformation
5.	$x{\text{'}}^{\mu }x{\text{'}}_{\mu }=g_{\mu \nu }{\mathrm{\Lambda }}^{\mu }{}_{\alpha }{x}^{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }{x}^{\beta }$	Substitute 3 and 4 into 2.
6.	$x^{\alpha }{x}_{\alpha }=g_{\alpha \beta }{x}^{\alpha }{x}^{\beta }$	metric tensor
7.	$g_{\mu \nu }{\mathrm{\Lambda }}^{\mu }{}_{\alpha }{x}^{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }{x}^{\beta }=g_{\alpha \beta }{x}^{\alpha }{x}^{\beta }$	From 5, 1, and 6.
8.	$g_{\mu \nu }{\mathrm{\Lambda }}^{\mu }{}_{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }{x}^{\alpha }{x}^{\beta }=g_{\alpha \beta }{x}^{\alpha }{x}^{\beta }$	Rearrange 7.
9.	∴ $g_{\mu \nu }{\mathrm{\Lambda }}^{\mu }{}_{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }=g_{\alpha \beta }$	From 8, since $x^{\alpha }$ and $x^{\beta }$ may be arbitrary.
10.	$g_{\alpha \beta }{g}^{\alpha \gamma }={\delta }^{\gamma }{}_{\beta }$	Kronecker delta
11.	$g_{\mu \nu }{\mathrm{\Lambda }}^{\mu }{}_{\alpha }{\mathrm{\Lambda }}^{\nu }{}_{\beta }{g}^{\alpha \gamma }={\delta }^{\gamma }{}_{\beta }$	Multiply both sides of 9 by $g^{\alpha \gamma }$, then apply 10.
12.	$g_{\mu \nu }{\mathrm{\Lambda }}^{\mu \gamma }{\mathrm{\Lambda }}^{\nu }{}_{\beta }={\delta }^{\gamma }{}_{\beta }$	Contracting the indices $\alpha$ in 11.
13.	${\mathrm{\Lambda }}_{\nu }{}^{\gamma }{\mathrm{\Lambda }}^{\nu }{}_{\beta }={\delta }^{\gamma }{}_{\beta }$	Contracting the indices $\mu$ in 12.
14.	$({\mathrm{\Lambda }}^T{)}^{\gamma }{}_{\nu }{\mathrm{\Lambda }}^{\nu }{}_{\beta }={\delta }^{\gamma }{}_{\beta }$	Swap the order of indices in order to transpose the first $\mathrm{\Lambda }$ of 13.
15.	∴ ${\mathrm{\Lambda }}^T={\mathrm{\Lambda }}^{-1}$ $◻$	14 may be paraphrased as ${\mathrm{\Lambda }}^T\mathrm{\Lambda }=I$.

Reference: http://www.physics.gla.ac.uk/~dmiller/lectures/RQM_2008.pdf, page 9.