TITLE: Set of all Inner Automorphisms is a subgroup of the set of all Automorphisms of a group $G$
QUESTION [7 upvotes]: Show that the set of all inner automorphisms of $G$, we will denote as $\mathrm{Inn}(G)$, is a subgroup of the set of all automorphisms of $G$, we will denote $\mathrm{Aut}(G)$, where we take $\mathrm{Aut}(G)$ to have the operation of function composition.

I have shown that $\mathrm{Inn}(G)$ is non-empty and that it is closed under composition of functions. I am now trying to show that the inverses of inner automorphisms are also inner automorphisms.
This is what I have so far in this regard:
$f:G \rightarrow G$ is an inner automorphism with $f(x)=c^{-1}xc$ for some fixed $c \in G$. If $a \in G$ then since $f$ is an isomorphism, it is also a bijection and so there is some $x,y \in G$ where $f(x)=a$ and $f(y)=c$. It follows $x=f^{-1}(a)$ and $y=f^{-1}(c).$ Hence
$$f^{-1}(a)=f^{-1}(f(x))=f^{-1}(c^{-1}xc)=\left[ f^{-1}(c)\right]^{-1}f^{-1}(x)f^{-1}(c)=y^{-1}f^{-1}(x)y.$$
At this point I want $f^{-1}(x)=a$ but then $x=f(a)$ which seems suspicious as then $x=c^{-1}ac$ so then $f(x)=f(c^{-1}ac)=(c^{2})^{-1}ac^2=a$ so $(c^{2})^{-1}a=a(c^{2})^{-1}$ but we aren't assuming $G$ is abelian. Any tips?

REPLY [3 votes]: I would aim at proving something more by considering the map
$$
\varphi : G \to \operatorname{Aut}(G), \qquad g \mapsto (x \mapsto g x g^{-1})
$$
that maps $g \in G$ to the inner automorphism $f_{g} : x \mapsto g x g^{-1}$.
Show that $\varphi$ is a homomorphism of groups. It will follow that its image $\operatorname{Inn}(G)$ is a subgroup of $\operatorname{Aut}(G)$. Moreover, the first isomorphism theorem will tell you that there is an isomorphism
$$
\frac{G}{\operatorname{ker}(\varphi)} \cong \operatorname{Inn}(G),
$$
and one can verify that $\operatorname{ker}(\varphi) = Z(G)$ is the centre of $G$.