We can almost reach the Lie algebra of any isometry group in a single go. For every $X$ in the Lie algebra we must have:
because $e_{tX}$ has to be in the isometry group by definition as shown at Section "Lie algebra of a matrix Lie group".

$∀v,w∈V,t∈R(e_{tX}v∣e_{tX}w)=(v∣w)$

Then:
so we reach:
With this relation, we can easily determine the Lie algebra of common isometries:

$dtd(e_{tX}v∣e_{tX}w) ∣∣∣∣∣ _{0}=0⟹(Xe_{tX}v∣e_{tX}w)+(e_{tX}v∣Xe_{tX}w)∣∣∣∣∣ _{0}=0⟹(Xv∣w)+(v∣Xw)=0$

$∀v,w∈V(Xv∣w)=−(v∣Xw)$

Bibliography:

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