# Group Theory: Show that $G/Z(G) \cong \operatorname{Inn}(G)$?

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."

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?

• Use the first isomorphism theorem.
– user121880
Dec 5, 2015 at 1:25
• +1 for a detailed, well thought out question that allowed responders to quickly identify what the stumbling block is. Dec 5, 2015 at 1:28

Given the other answers, question 4 follows directly from the isomorphism theorem: $$G/\ker\phi \cong \phi(G)$$
• $f: G \rightarrow H$ is a homomorphism then $G/Ker(f) \cong Im(f)$? I'm actually stuck on the first isomorphism theorem on another question of mine. I've just hit a wall. Dec 5, 2015 at 1:36