I'm having some trouble proving that for a group $G$, $G/Z(G)\cong\text{Inn}(G)$, where $Z(G)$ is the center of the group defined as $Z(G)=\{z\in G:gz=zg\forall g\in G\}$ and $\text{Inn}(G)$ is the inner automorphism group.

I think we need to find some kind of homomorphism, but I'm not really sure how to. Thanks in advance.

  • 4
    $\begingroup$ There is a pretty natural map $G \to \operatorname{Inn}(G)$. Verify that is a homomorphism, and determine its kernel. $\endgroup$ – Daniel Fischer Oct 13 '13 at 20:53

Consider the map $\phi:G\to\text{Aut}(G)$ defined as $\phi(g)=\varphi_g$, where $\varphi_g$ is the automorphism of $G$ defined by $\varphi_g(h)=ghg^{-1}$.

Lemma 1: $\phi$ is a homomorphism.

Proof: We have $\phi(g_1g_2)=\varphi_{g_1g_2}$, and $$ \varphi_{g_1g_2}(h) =(g_1g_2)h(g_1g_2)^{-1}=g_1(g_2hg_2^{-1})g_1^{-1}=\varphi_{g_1}(\varphi_{g_2}(h)).$$

Lemma 2: $\text{ker}(\phi)=Z(G)$.

Proof: We have $$\begin{align*}\text{ker}(\phi) &= \{g:\phi(g)=e\} \\ &= \{g:\varphi_g=e\}\\ &= \{g:\varphi_g(h)=h\} \\ &= \{g:gh = hg\} \\ &= Z(G).\end{align*}$$ Finally, by the first isomorphism theorem, we have $G/\text{ker}(\phi) = G/Z(G)\cong im(\phi) = \text{Inn}(G)$, as desired.

  • 1
    $\begingroup$ You didn't show it's an epimorphism. $\endgroup$ – badatmath Feb 23 '18 at 12:44

You need the homomorphism $G\mapsto c_g$, where $c_g$ is the inner homomorphism given by $h\mapsto ghg^{-1}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.