= Solution
Continue with metric signature $+---$. The Pauli matrices obey $\{\sigma^i,\sigma^j\}=2\delta^{ij}I_2$. Block multiplication in the <chiral gamma-matrix representation> gives
$$
(\gamma^0)^2=I_4,\qquad
\gamma^0\gamma^i=\begin{pmatrix}-\sigma^i&0\\0&\sigma^i\end{pmatrix},\qquad
\gamma^i\gamma^0=-\gamma^0\gamma^i,
$$
and
$$
\gamma^i\gamma^j=-\begin{pmatrix}\sigma^i\sigma^j&0\\0&\sigma^i\sigma^j\end{pmatrix}.
$$
Hence \b[$\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I_4$], the required <Clifford algebra>. Multiplying the <Dirac equation> by $i\gamma^\nu\partial_\nu+m$ yields
$$
(i\gamma^\nu\partial_\nu+m)(i\gamma^\mu\partial_\mu-m)\psi
=-(\Box+m^2)\psi=0.
$$
The antisymmetric part of $\gamma^\nu\gamma^\mu$ drops out because partial derivatives commute. Therefore every component of the <Dirac spinor> satisfies the <Klein-Gordon equation>. The converse is not true: four arbitrary scalar solutions need not obey the first-order <Dirac equation>.
The <Lorentz group> consists of real invertible matrices $\Lambda$ preserving the <Minkowski metric>: $\Lambda^Tg\Lambda=g$. Writing $\Lambda=I+\omega+O(\omega^2)$ gives $\omega_{\mu\nu}=-\omega_{\nu\mu}$, so \b[there are six continuous parameters: three spatial rotation angles and three boost rapidities]. The <proper orthochronous Lorentz group> is the connected component with determinant $+1$ that preserves time orientation. <Parity> and <time reversal> are additional discrete operations and are not encoded by the six real parameters in an exponential near the identity.
For the real-generator convention used here, vector generators can be written
$$
(M_{\rm vec}^{\rho\sigma})^\mu{}_{\nu}
=g^{\rho\mu}\delta^\sigma{}_{\nu}-g^{\sigma\mu}\delta^\rho{}_{\nu}.
$$
Take $\Omega_{\rho\sigma}=-\Omega_{\sigma\rho}$ and use the same parameters in the spinor representation. The <Lorentz-spinor generators from a Clifford algebra> are
$$
\boxed{M_{\rm sp}^{\rho\sigma}=\frac14[\gamma^\rho,\gamma^\sigma],\qquad
S(\Lambda)=\exp\left(\frac12\Omega_{\rho\sigma}M_{\rm sp}^{\rho\sigma}\right).}
$$
To verify the <Lorentz algebra>, first commute a generator with one <gamma matrix>:
$$
[M_{\rm sp}^{\rho\sigma},\gamma^\tau]
=g^{\sigma\tau}\gamma^\rho-g^{\rho\tau}\gamma^\sigma.
$$
For example, this follows by moving $\gamma^\tau$ through each product using the Clifford <anticommutator>. Applying the commutator derivation rule to $\frac14[\gamma^\tau,\gamma^\nu]$ then gives
$$
[M_{\rm sp}^{\rho\sigma},M_{\rm sp}^{\tau\nu}]
=g^{\sigma\tau}M_{\rm sp}^{\rho\nu}-g^{\rho\tau}M_{\rm sp}^{\sigma\nu}
+g^{\rho\nu}M_{\rm sp}^{\sigma\tau}-g^{\sigma\nu}M_{\rm sp}^{\rho\tau}.
$$
No extra factor of $i$ belongs in these generators with this commutator convention. Hermitian-generator conventions shift the factors of $i$ into the brackets and the exponential instead.
The same gamma commutator proves
$$
S^{-1}\gamma^\mu S=\Lambda^\mu{}_{\nu}\gamma^\nu.
$$
Indeed, to first order $S^{-1}\gamma^\mu S=\gamma^\mu+\frac12\Omega_{\rho\sigma}[\gamma^\mu,M_{\rm sp}^{\rho\sigma}]$, exactly the infinitesimal vector <action> defined above. Exponentiating proves the finite relation. The spinor field transforms as \b[$\psi'(x')=S(\Lambda)\psi(x)$ with $x'=\Lambda x$]; combining this relation with $\partial'_\mu=(\Lambda^{-1})^\nu{}_{\mu}\partial_\nu$ preserves the <Dirac equation>. A $2\pi$ rotation gives $S=-I_4$, so the spinor matrices give the double-cover <spin> representation, rather than a single-valued ordinary representation of the <Lorentz group> itself.
For a boost in direction $i$, $M_{\rm sp}^{0i}=\frac12\gamma^0\gamma^i$ is Hermitian, with eigenvalues $\pm\frac12$. A real <rapidity> $\eta$ therefore produces eigenvalues $e^{\pm\eta/2}$ in $S$, whose moduli are not one. \b[The finite-component spinor representation is not unitary for boosts], although the rotation matrices are unitary. The <Lorentz group> is noncompact; this does not forbid the unitary infinite-dimensional <action> on the physical Hilbert space of states.
Nevertheless, since $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$,
$$
(M_{\rm sp}^{\rho\sigma})^\dagger\gamma^0+\gamma^0M_{\rm sp}^{\rho\sigma}=0,
\qquad S^\dagger\gamma^0S=\gamma^0.
$$
This <Dirac spinor pseudo-unitarity> implies $\bar\psi'(x')=\bar\psi(x)S^{-1}$, and consequently
$$
\boxed{\bar\psi'(x')\psi'(x')=\bar\psi(x)\psi(x).}
$$
The <Dirac adjoint> is precisely the adjoint needed to form this <Lorentz scalar>; ordinary $\psi^\dagger\psi$ alone is not a scalar.
Let $\gamma^5=i\gamma^0\gamma^1\gamma^2\gamma^3=\operatorname{diag}(-I_2,I_2)$. It anticommutes with every <gamma matrix> and commutes with every $M_{\rm sp}^{\rho\sigma}$, so $S^{-1}\gamma^5S=\gamma^5$. Thus the <axial current> transforms as a <four-vector> under proper <Lorentz transformations>:
$$
j_5'{}^\mu(x')=\Lambda^\mu{}_{\nu}j_5^\nu(x).
$$
Under <parity> it is an <axial vector>, acquiring the additional <pseudovector> sign. Contracting two <axial currents> cancels that sign and uses the invariant metric. Therefore \b[$j_{5\mu}j_5^\mu$ is compatible with <Lorentz invariance>], including <parity> invariance of this contraction. <Lorentz invariance> of the interaction does not require the <axial current> to be conserved.
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