I came across with the following theorem (in Cellular Automata and Groups):
"Let G be a finitely generated residually finite group. Then the automorphism group Aut(G) is also residually finite".
Let $a_0 \in Aut(G), a_0 \ne id_G$. Then, there exists a $g_0 \in G$ such that $a_0(g_0) \ne g_0$. Since G is residually finite, there exists a finite group F and a homomorphism $φ: G \rightarrow F$ such that $φ(a_0(g_0)) \ne φ(g_0)$.
Consider the set H to be the intersection of all $Ker(ψ)$ where $ψ \in Hom(G,F)$.
Then H is a normal subgroup of G. Observe that for every $a \in Aut(G)$ we have $a(H)= a(\cap kerψ)= \cap a(kerψ)= \cap ker(ψ \cdot a^{-1})= \cap kerψ$.
I don't get the last "=".
Can anyone help?
Thanks in advance.
Since $a\in\operatorname{Aut}(G),$ as $ψ$ runs through all elements of $\operatorname{Hom}(G,F),$ so does $ψ\circ a^{-1}.$ Hence $$\bigcap_{\psi\in\operatorname{Hom}(G,F)}\ker(ψ\circ a^{-1})= \bigcap_{\varphi\in\operatorname{Hom}(G,F)}\ker\varphi.$$