Here is the question that I want to answer part(d) in it:
Let $R$ be a ring and $I \subset R$ a two-sided ideal, with quotient homomorphism $\pi : R \rightarrow R/I.$ Let $\operatorname{End_{I}(R)}$ be the set of $\varphi \in \operatorname{End(R)}$ such that $\varphi(I) \subset I,$ and let $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}.$
$(a)$ Given $\varphi \in \operatorname{End_{I}(R)},$ show that there exists $\bar{\varphi} \in \operatorname{End(R/I)}$ such that $\bar{\varphi} \pi = \pi \varphi.$
$(b)$ Given $\varphi, \psi \in \operatorname{End_{I}(R)},$ show that $\overline{\varphi \psi} = \bar{\varphi} \bar{\psi}.$
$(c)$ Given $\varphi \in \operatorname{Aut_{I}(R)},$ show that $\varphi(I) = I$ and $\bar{\varphi} \in \operatorname{Aut(R/I)}.$\ Hint: Use part $(b).$
$(d)$ Let $G$ be a group which acts on $R$ by ring automorphisms and restricts to $I.$ Show that the $G-$action on $R$ induces a $G-$action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr).$
My question is:
If I managed to prove parts (a), (b) and (c), could anyone give me a hint on how to prove (d)?
It seems to me that after a) b) and c) have been checked part d) is just an obvious rephrasing of them.
Just convince yourself that to have a group $G$ acting on $R$ by automorphisms amounts to have a group homomorphism $G\rightarrow{\rm Aut}(R)$ and then that given that the image lands in ${\rm Aut}_I(R)$ the induced map $$ G\rightarrow{\rm Aut}(R/I),\qquad g\mapsto\bar\phi_g $$ is again an homomorphism.