= Solution
Differentiate the defining relation at $\Lambda=I+tX+O(t^2)$. The linear term gives the <Lie algebra>
$$
\mathfrak g=\{X\in M_3(\mathbb R):X^T\eta+\eta X=0\}.
$$
Solving this equation between <matrices> yields
$$
X=\begin{pmatrix}0&a&b\\a&0&-c\\b&c&0\end{pmatrix}
=aK_1+bK_2+cJ,
$$
where a convenient <basis> is
$$
K_1=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix},\qquad
K_2=\begin{pmatrix}0&0&1\\0&0&0\\1&0&0\end{pmatrix},\qquad
J=\begin{pmatrix}0&0&0\\0&0&-1\\0&1&0\end{pmatrix}.
$$
The first two generate <Lorentz boosts> and the last a spatial <rotation>. Direct <commutators> give
$$
\boxed{[J,K_1]=K_2,\qquad[J,K_2]=-K_1,\qquad[K_1,K_2]=-J.}
$$
There are three independent real coefficients, so $\dim_{\mathbb R}\mathfrak g=3$. A <Cartan subalgebra> is $\mathbb RJ$, and the <rank of a semisimple Lie algebra> is one. Explicitly, if $X=aK_1+bK_2+cJ$ and $Y=a'K_1+b'K_2+c'J$, their <commutator> vanishes exactly when
$$
bc'-cb'=ca'-ac'=ba'-ab'=0.
$$
For nonzero $X$, these equations force $Y$ to be proportional to $X$, so there is no two-dimensional commuting <subalgebra>. The split real rank is also one, realized by the semisimple <Lorentz boost> generator $K_1$. Thus
$$
\boxed{\mathfrak g=\mathfrak{so}(1,2),\qquad\dim\mathfrak g=3,\qquad\operatorname{rank}\mathfrak g=1.}
$$
For this <Lorentz algebra in three spacetime dimensions>, disconnected components of the full <Lorentz group> do not change its tangent <Lie algebra> at the identity.
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