Showing a map is homomorphism of groups and computing the kernel I just started learning about homomorphisms and was assigned this problem:

Where $$\psi_G(h)  = ghg^{-1}$$
My solution: To show it is a homomorphism
$$\psi_g(ab) = gabg^{-1} = gag^{-1}gbg^{-1} = gag^{-1}gbg^{-1} = \psi_g(a)\psi_g(b)$$
So it is a homomorphism by definition.
To find the kernel, the kernel is $$g \in G | \psi_G(g) = 1_H$$
Then $$\psi_g(1_G * g) = \psi_g(1_G)\psi_g(g) = \psi_g(g)$$
Therefore the kernel is $1_G$.
This is the outline of my proof. I am not sure if I need anything else, does 'homomorphisms of groups' have a more specific meaning that I am missing? Thank you!
 A: Several issues.

*

*What you first proved was that $\psi_g$ is a group homomorphism. What you proved later is that $\psi_g$ is one-to-one. That is not quite what you want to prove. You want to prove that the map $g\longmapsto \psi_g$ is a group homomorphism. That said, it is a good thing that you are checking that the function is well-defined: that is, that $\psi_g$ is actually an element of the set you need it to be in, $\mathrm{Aut}(G)$. You proved it is indeed a homomorphism and that it is one-to-one, but unless you also know $G$ is finite you aren't quite done proving tha $\psi_g$ is an automorphism. You also need to show it is surjective, or that it has an inverse that is a homomorphism.


*Once you do that, you will only have proven that $\psi_g\in \mathrm{Aut}(G)$. Which is a good thing to prove if you did not know it already. You still have to show that map $G\to \mathrm{Aut}(G)$ is a group homomorphism, and find its kernel.


*So what you need to show is that the image of $gh$ is the product of the image of $g$ and the image of $h$. That is, that $\gamma(gh) = \gamma(g)\gamma(h)$; not that $\psi_g(xy)=\psi_g(x)\psi_g(y)$. So, to that end... What is the group operation of $\mathrm{Aut}(G)$? It's composition
of functions. So what you need to show is that
$$\psi_{(gh)} = \psi_g\circ\psi_h.$$
If this holds, then you have a homomorphism.


*To find the kernel... what is the trivial element of $\mathrm{Aut}(G)$? It's the identity map. So you are trying to figure out what are the $g\in G$ for which $\gamma(g)=e_{\mathrm{Aut}(G)} = \mathrm{id}_G$; that is, what are the $g\in G$ such that $\psi_g=\mathrm{id}_G$; equivalently, what are the $g\in G$ such that $\psi_g(x)=x$ for all $x\in G$.
