I have quite a long exercise that I have to solve but I do have the solutions for it. I'm having trouble understanding some steps, however.
Let $G$ be a group. We denote by $x*y$ the product of two elements $x, y \in G$, and by $1$ the neutral element of $G$. Let (Bij$(G), \circ)$ be the group of bijective maps from $G$ to itself. We denote by Aut$(G)$ the subset of Bij$(G)$ consisting of automorphisms of $G$.
The exercise is as follows: For $g \in G$, we define a map $\rho_g : G \to G$ by setting $\rho_g(x) = g * x * g^{-1}$. Show that $\rho_g$ is an automorphism of $G$.
Well first I have to show that: $\rho_g(x*y)=\rho_g(x)*\rho_g(y)$.
$\rho_g(x*y)= g * (x * y) * g^{-1} = g * x * g^{-1}*g * y * g^{-1} = \rho_g(x)*\rho_g(y)$
Now my first question is at this step. Can I always expand $g * (x * y) * g^{-1}$ to $g * x * g^{-1}*g * y * g^{-1}$?
Or why does this work exactly?
Also then showing bijectivity. For injectivity, I checked some other exercises and found that If the kernel is trivial, then it is injective. The solutions to this exercise are as follows:
$\ker(\rho_g) = \{x \in G , g*x*g^{-1}=1\} = \{x \in G , g*1*g^{-1}=1\} = \{1\}$.
Why do I solve this for $1$? Does the kernel not have to be trivial for a map to be injective?
For surjectivity, the proof seems almost too simple. They argue that $\rho_g(x) = y$ has the solution given by $x = g * y * g^{-1}$.
I guess this does make sense although I'm not sure if I can always prove it this way. Can I just always solve for $x$ to prove surjectivity?