Transformation of spinors due to Lorentz group Assume we have a Dirac spinor $\psi(x)$ which satisfies the Dirac equation:
$$(i\gamma^{\mu}\partial_{\mu} - m)\psi(x) = 0.$$
If we boost our spacetime coordinates to a new system with a Lorentz transformation we get
$$x^{\mu} \longrightarrow x'^{\mu} = \Lambda^{\mu}_{\,\,\nu}x^{\nu},$$
$$\psi(x) \longrightarrow \psi'(x') = S(\Lambda) \psi(x) \equiv S \psi(x)$$
where $S$ is a matrix.
If we want $\psi'(x')$ to satisfy the Dirac equation we need to impose some conditions on the way the spinor transforms, i.e. on the matrix $S$:
$$(i\gamma^{\mu}\partial'_{\mu} - m)\psi'(x') = 0 \Leftrightarrow(i\gamma^{\mu}\partial_{\rho}(\Lambda^{-1})^{\rho}_{\,\,\mu} - m)S\psi(x)
= 0, $$
If the following holds true:
$$S^{-1} \gamma^{\mu} (\Lambda^{-1})^\rho_{\,\,\mu} S = \gamma^{\rho}.$$
The text im trying to follow concludes that thus:
$$S^{-1} \gamma^{\mu} S = \Lambda^{\mu}_{\,\,\rho} \gamma^{\rho}.$$
But I can't seem to get the $\Lambda$ matrix onto the right side, since I see no reason that it should commute with $S$. Where'd I go wrong?
 A: Start from your expression:
$$ \tag{1} S^{-1} \gamma^\mu \Lambda_\mu^{\,\,\rho} S
= S^{-1} \gamma^\mu (\Lambda^{-1})_{\,\,\mu}^{\rho} S
= \gamma^\rho.$$
Expliciting the spinor indices and using the orthogonality of $\Lambda$ it reads
$$ \tag{2}
S^{-1}_{\alpha\beta} \gamma^\mu_{\beta \gamma} \Lambda^{\,\,\rho}_{\mu} S_{\gamma \delta}
= \gamma^\rho_{\alpha \delta}, $$
where it is important to note that this are all numbers, hence every object commutes with every other.
Then write it as
$$ \tag{3} S^{-1}_{\alpha\beta} \gamma^\mu_{\beta \gamma} S_{\gamma \delta} \Lambda^{\,\,\rho}_{\mu}
= \gamma^\rho_{\alpha \delta}, $$
now multiplying by $\Lambda^\mu_{\,\,\rho}$ and using the orthogonality relation
$$ \Lambda^\mu_{\,\,\nu} \Lambda_\mu^{\,\,\sigma} = \delta_\nu^{\,\,\sigma}, $$
you have your result.
A: Lambda matrices have Lorentz indices, on the other hand $S$ has spinor indices. These matrices belong to the whole different kinds of linear operators, acting on different spaces.
In your situation, $\gamma^{\mu}\Lambda^{-1}$ is just a shorthand for $\gamma^{\nu} {\Lambda^{-1}}_{\nu}^{\mu}$.
Sure they commute. If you expand all Greek and spinor (Latine) indices, it will be obvious (note that gamma matrices have actually three indices: $(\gamma^{\mu})^{a}_b$).
