I am using the definition of group action of $G$ on $X$ which says that any group homomorphism from $G$ to $\operatorname{Aut(X)}$ is a group action.
And I am trying to prove $(d)$ in the following question (after proving $(a), (b), (c)$)
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 trial:
I was given hint here Show that the $G$-action on $R$ induces a $G$-action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr).$ to prove that the given action ( $G-$action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr).$) is a homomorphism which I do not know exactly how to do as the hint mentioned $\bar{\varphi_{g}}.$ Also, I do not know if I must show that this action is a group homomorphism or ring homomorphism, could anyone clarify this for me please?
Also, I was told that the meaning of the question is:
It means that G is a subgroup of the group Aut(R) and g(I)=I for each g in I. Elements of Aut(R) are bijective ring homomorphisms f:R-->R, which are called automorphisms.
Here is my trial to show that it is a ring homomorphism:
(a) Showing that it preserves addition:
${}^{(g_{1} + g_{2})} \pi(r) = \pi({}^{(g_{1} + g_{2})}r) = {}^{(g_{1} + g_{2})}r + I = {}^{g_{1}}r + {}^{g_{2}}r + I = ({}^{g_{1}}r + I) + ({}^{g_{2}}r + I) = {}^{g_{1}} \pi(r) + {}^{g_{2}} \pi(r)$
I am not sure from my third equality here, specifically the first $+$ sign in it, if it is correct or not, could anyone check it for me please?
(b) Showing that it preserves multiplication:
${}^{(g_{1}g_{2})} \pi(r) = \pi({}^{(g_{1}g_{2})}r) = {}^{(g_{1}g_{2})}r + I = {}^{g_{1}}(^{g_{2}}r) + I = {}^{g_{1}}(^{g_{2}}r + I) = {}^{g_{1}} \pi({}^{g_{2}}r)$
Where the fourth equality because I know that $g(I) = I \forall g \in G.$ And after the last step I do not know how to complete, could anyone show me the rest of steps or the correct ones please?
(c) for showing that the identity of $G$ is mapped to the identity of $\operatorname{Aut(R/I)}$ I do not know how should I prove that, any help will be appreciated in that proof.
Also I have a question: what is the relation between $G-$action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr)$ and the function $\bar{\varphi_{g}}.$ Is it just a name for the homomorphism between $G$ and $\operatorname{Aut(R/I)}$? vould anyone explain that to me please?
For a) and b): Why are you using two different operations ($g_1+g_2$ and $g_1g_2$) for elements of G? For c): The action (denoted by ".") of the group's identity, say e, on elements $\pi(r)$ of $R/I$ is given by $e.\pi(r)=\pi(e.r)=\pi(r)$. Finally, the action $\cdot:G\times R/I\rightarrow R/I$, given by $(g,\pi(r))\mapsto g.\pi(r)$ corresponds with a homomorphism $\overline{\varphi}:G\rightarrow Aut(R/I)$ given by $g\mapsto \overline{\varphi}_g:R/I\rightarrow R/I$, where $\overline{\varphi}_g(\pi(r)):=\pi(\varphi_g(r))$, for the homomorphism $\varphi:G\rightarrow Aut(R)$ which induces $\overline{\varphi}.$