Whilst studying Riemanninan charts, I tried to prove the statement that the composition of two Riemanninan isometries is still a Riemanninan isometry. Let me elaborate:
Say we have a Riemanninan chart $(P,g)$ together with two Riemannian isometries $A$ and $B$, both from $(P,g) \to (P,g)$. Then I would like to prove that $A \circ B$ is also a Riemannian isometry from $(P,g) \to (P,g)$.
So if I'm understanding this correctly, I have $A: (P,g) \to (P,g)$ and $B: (P,g) \to (P,g)$. It seems to be obvious that their composition also maps from $(P,g) \to (P,g)$? Could someone maybe help me how to elaborate on this?
Using the hints I got in the comments, I tried to complete the proof:
We have our Riemannian chart $(P,g)$ with two isometries $A,B$ from $(P,g) \to (P,g)$. Since $A$ and $B$ are Riemannian isometries, the metric will be preserved. Hence, $A^*g=g$ and $B^*g=g$.
$(A \circ B)^* = B^* \circ A^*$, and therefore $(A \circ B)^*g=g$, making the composition a Riemanninan isometry.