Factor Group over Center Isomorphic to Inner Automorphism Group

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.

• There is a pretty natural map $G \to \operatorname{Inn}(G)$. Verify that is a homomorphism, and determine its kernel. – 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.

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

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