A set $G$ has an operation $\ast$ and is closed and associative under that operation.
(a) there exists $e$ in $G$ such that $a \ast e=a$
Prove $e$ is the identity by showing $e\ast a=a$
(b) For all $a$ in $G$ there exists $b$ such that $a \ast b=e$
Prove $b$ is an inverse by showing $b \ast a=e$
For (a) I tried $e \ast e=e$, $e \ast a \ast e= a \ast e$, and $e \ast a = a$. But I'm not sure if that's allowed since I'm very new to abstract still.
For (b) I'm completely stumped. My professor gave the hint to start with $b \ast a$ and develop $e$ but I keep thinking to do it via commutative ways which I know isn't guaranteed to happen in a group. Is there a way to manipulate associativity and closure to get there?
Any help is appreciated.