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

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