Isometry group on a coset manifold In ''Einstein Gravity in a Nutshell'' Zee says 

''On a coset manifold $G/H$, the isometry group is evidently just $G$''

when discussing the relation between the Killing vector fields and Lie algebras (Ch IX 6 - App 3). Is there some intuition and/or physicist-level-of-rigor-proof for this ''evidently''? E.g. can we somehow know that Killing vectors for $S^{d}$ form Lie algebra $so(d+1)$ without calculating the commutators of Killing vectors?
 A: You might be possibly swamped by a surfeit of unfamiliar terms and symbols, but look at a beachball, or our earth, for d=2,
$$ x^2+y^2+z^2=R^2$$
All three rigid rotations: $L_x,L_y, L_z$, so SO(3), preserve the above constraint and map the earth onto itself preserving all distances (isometry).
If you eliminate the redundant coordinate z in favor of the other two, $z=\pm\sqrt{R^2-x^2-y^2}$, you singled out $L_z$, manifestly an earth-spin isometry, leaving z invariant. (In physics, this is a "linearly realized", unbroken symmetry.) 
But it is not the only one transiting on circles on the globe. Spinning around  a spot in the Pacific near Genovesa Island (Galapagos), you might arc your way from Greenwich to the North pole, etc... The motions $L_x,L_y$ do not leave z invariant (in physicsese, they are nonlinearly realized, "spontaneously broken" symmetries).
The symmetries of the earth are still the three $L_x,L_y, L_z$. You might like to illustrate this in your formalism.
A: In the generality as stated in your question this doesn't seem correct. I am assuming $G$ is a Lie group with a left-invariant metric.
There is an obvious mapping $G\to Isom(G/H)$ where $g\in G$ acts on $G/H$ by left multiplication. This does neither have to be injective not surjective however.
Consider the extreme case that $G=H$. In that case every element becomes trivial. The other extreme case, $G=e$ would require that all isometries are obtained as left-multiplications by a fixed element. This is not true.
As a concrete example, the sphere $S^2$ can be written as $SO(3)/SO(2)$, but its isometry group is $O(3)$. In fact, it can also be written $O(3)/O(2)$, and in general a representation of the form $G/H$ will not be unique, so that in this form the statement cannot hold. 
