Group Theory in General Relativity

In Special Relativity, the Lorentz Group is the set of matrices that preserve the metric, i.e. $\Lambda \eta \Lambda^T=\eta$.

Is there any equivalent in General Relativity, like: $\Lambda g \Lambda^T=g$?

(We could at least take locally $g\approx\eta$, so we recover the Lorentz group, but I don't know whether we could extend this property globally.)

Why does Group Theory have much less importance in General Relativity than in QFT and particle physics?

Answer

As you point out, the Minkowski metric $\eta = \mathrm{diag}(-1,+1, \dots, +1)$ in $d+1$ dimensions possesses a global Lorentz symmetry. A highbrow way of saying this is that the (global) isometry group of the metric is the Lorentz group. Well, translations are also isometries of Minkowski, so the full isometry group is the Poincare group.

The general notion of isometry that applies to arbitrary spacetimes is defined as follows. Let $(M,g)$ be a semi-Riemannian manifold, then any diffeomorphism $f:M\to M$ (coordinate transformation essentially) that leaves the metric invariant is called an isometry of this manifold.

A closely related notion that is often useful in relation to isometries is that of Killing vectors. Intuitively a killing vector of a metric generates an "infinitesimal" isometry of a given metric. Intuitively this means that they change very little under the action of the transformations generated by the Killing vectors.

Isometries and Killing vectors are a big reason for which group theory is relevant in GR. Killing vectors often satisfy vector field commutator relations that form a Lie algebra of some Lie Group.

Addendum (May 28, 2013). Remarks on symmetric spaces and physics.

One can show that in $D$ dimensions, a metric can posses at most $D(D+1)/2$ independent killing vectors. Any metric that has this maximum number of killing vectors is said to be maximally symmetric.

Example. Consider $4$-dimensional Minkowski space $\mathbb R^{3,1}$. The isometry group of this space, the Poincare group, has dimension $10$ since there are $4$ translations, $3$ rotations, and $3$ boosts. On the other hand, in this case we have $D=4$ so that the maximum number of independent killing vectors is $4(4+1)/2 = 10$. It turns out, in fact, that each rotation, translation, and boost gives rise to an independent Killing vector field, so that Minkowski is maximally symmetric.

One can in fact show that there are (up to isometry) precisely three distinct maximally symmetric spacetimes: $\mathrm{AdS}_{d+1}, \mathbb R^{d,1}, \mathrm{dS}_{d+1}$ called anti de-Sitter space, Minkowski space, and de-Sitter space respectively, and that these spacetimes all have constant negative, zero and positive curvature respectively. The isometry groups of these spacetimes are well-studied, and these spacetimes form the backbone of a lot of physics. In particular, the whole edifice of $\mathrm{AdS}/\mathrm{CFT}$ relies on the fact that $\mathrm{AdS}$ has a special isometry group that is related to the conformal group of Minkowski space.