Set of all Inner Automorphisms is a subgroup of the set of all Automorphisms of a group $G$ 
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?
 A: You can do that via this fact that $Inn(G)$ is non empty (You have shown this) and the fact that for all $g,h\in G$ $$f_gf_h=f_{gh}$$ so $$f_gf_g^{-1}=f_e$$ where $e$ is the identity of the group (You can see that $f_e$ is the identity of $Aut(G)$). so $$(f_g)^{-1}=f_{g^{-1}}$$ and from this we have $$(f_g)^{-1}f_h=f_{g^{-1}}f_h=f_{hg^{-1}}\in Inn(G)$$ Here we satisfy the condition in which any subset of a group should have to be a subgroup.
A: You're overthinking it. How do you cancel out conjugation by $c$? Try conjugation by $c^{-1}$.
$$(c^{-1})^{-1}f(x)c^{-1}=cc^{-1}xc^{-1}c=x$$
A: Inner automorphism, as a subset of automorphism is of course bijective. So you do not need to prove that. All you need is to prove it is closed under map composition, which is straight forward, contains identity, which is also easy, and exists an inverse, for which simply take $(f_c)^{-1}$ to be $f_{c^{-1}}$.
A: 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 $G$.
A: The easiest way to show that a function has an inverse is to write down the inverse.  It's often the case that it's too hard to write down an inverse explicitly, but in this particular case it is easy.
What happens if you conjugate by $c$ and then conjugate by $c^{-1}$?
