Here is the question. It's rather long so I apologise. "Let $G$ be a group. Let $\operatorname{Aut}(G)$ be the set of all isomorphisms of $G$. This is a group under composition, known as the group of automorphisms of $G$. If $g \in G$ then we know that the map $\theta_{g}:G \rightarrow G$ defined by $\theta_g(a)=g^{-1}ag$ is an isomorphism."
Questions I have answered.
1) Show that if $g,h \in G$ then $\theta_{gh} = \theta_{h} \circ \theta_{g}$.
Answer: Let $ a \in G$ observe that $\theta_{gh}(a)= (gh)^{-1}a(gh) = h^{-1}g^{-1}agh = h^{-1}\theta_g(a)h=\theta_h(\theta_g(a))=(\theta_h \circ \theta_g)(a)$. This is true $\forall a \in G \space \space \therefore\space \space \theta_{gh} = \theta_{h} \circ \theta_{g}$.
2) Define a map $\phi: G \rightarrow \operatorname{Aut}(G)$ by $\phi(g)= \theta_{g^-1}$. Show that $\phi$ is a homomorphism.
Answer: We need to show that $\theta_{gh^{-1}} = \theta_{g^{-1}} \circ \theta_{h^{-1}} $. However we know that $\theta_{gh} = \theta_h \circ \theta_g$ and therefore we can conclude that $\theta_{gh^{-1}} = \theta_{h^{-1}g^{-1}} = \theta_{g^{-1}} \circ \theta_{h^{-1}} $.
3) Find $\ker(\phi)$.
Answer: $\theta_g(a) = g^{-1}ag \space \therefore \space \theta_{g^{-1}}(a) = gag^{-1}$. To find $\ker(\phi)$ find $\theta_{g^{-1}}(a)=a$ so we have $gag^{-1}=a$ and then $gag^{-1}g = ag$ so we have $ga=ag$ and so $\ker(\phi)= Z(G)$ where $Z(G)$ is the centre, all the elements in the group that commute with all other elements in the group.
And finally, question 4: The image of $\phi$ is called $ \operatorname{Inn}(G)$, the group of inner automorphisms. Show that $G/Z(G) \cong \operatorname{Inn}(G)$.
I can't work out the answer to this final question. I apologise for the length, I just wanted to explain in as much detail as possible. Can anyone help me?
