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?

  • 7
    $\begingroup$ Use the first isomorphism theorem. $\endgroup$
    – user121880
    Dec 5, 2015 at 1:25
  • 2
    $\begingroup$ +1 for a detailed, well thought out question that allowed responders to quickly identify what the stumbling block is. $\endgroup$ Dec 5, 2015 at 1:28

1 Answer 1


Given the other answers, question 4 follows directly from the isomorphism theorem: $$G/\ker\phi \cong \phi(G)$$

  • $\begingroup$ How would I go about showing this? I apologise for my stupidity. I feel the clarity has disappeared after spending so much time on these questions. $\endgroup$
    – Geometry
    Dec 5, 2015 at 1:31
  • $\begingroup$ @Geometry, have you seen the first isomorphism theorem? $\endgroup$
    – lhf
    Dec 5, 2015 at 1:32
  • $\begingroup$ $ 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. $\endgroup$
    – Geometry
    Dec 5, 2015 at 1:36
  • $\begingroup$ Wow. I just realised how simple it is. I'm incredibly tired and immune to coffee right now. I apologise. $\endgroup$
    – Geometry
    Dec 5, 2015 at 1:43

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