G/Z(G) is Isomorphic to Inn(G)

# G/Z(G) is Isomorphic to Inn(G)

Proposition 1: Let $G$ be a group. Then $G/Z(G)$ is isomorphic to $\mathrm{Inn}(G)$. |

*Recall that $Z(G) = C_G(G)$ is the center of $G$, i.e., all elements of $g$ that commute with every element of $G$. Also recall that $\mathrm{Inn}(G)$ is the set of all inner automorphisms of $G$.*

**Proof:**Let $\phi : G \to \mathrm{Inn}(G)$ be defined for all $a \in G$ by:

\begin{align} \quad \phi(a) = i_a \end{align}

- Then $\phi$ is a homomorphism since for all $a, b \in G$ we have that for all $g \in G$:

\begin{align} \quad [\phi(ab)](g) = [i_{ab}](g) = (ab)g(ab)^{-1} = abgb^{-1}a^{-1} =ai_b(g)a^{-1} = [i_a \circ i_b](g) = [\phi(a) \circ \phi(b)](g) \end{align}

- So $\phi(ab) = \phi(a) \circ \phi(b)$. Now observe that:

\begin{align} \quad \ker (\phi) &= \{ a \in G : i_a = i_e \} \end{align}

- Note that $i_a = i_e$ if and only if $i_a(g) = i_e(g)$ for all $g \in G$, i.e., if and only if $aga^{-1} = g$ for all $g \in G$, i.e., if and only if $ag = ga$ for all $g \in G$. So $i_a = i_e$ if and only if $a \in Z(G)$. Thus:

\begin{align} \quad \ker (\phi) = Z(G) \end{align}

- By The First Group Isomorphism Theorem we have that:

\begin{align} \quad G/Z(G) = G/\ker(\phi) \cong \phi(G) \end{align}

- Furthermore, observe that $\phi$ is surjective since for all $i_a \in \mathrm{Inn}(G)$ we have that $a \in G$ is such that $\phi(a) = i_a$. So $\phi(G) = \mathrm{Inn}(G)$. From above we conclude that:

\begin{align} \quad G/Z(G) \cong \mathrm{Inn}(G) \quad \blacksquare \end{align}

Corollary 2: Let $G$ be a group. If $\mathrm{Inn}(G)$ is cyclic then $G$ is an abelian group. |

**Proof:**From proposition 1 we know that $G/Z(G)$ is isomorphic to $\mathrm{Inn}(G)$. So if $\mathrm{Inn}(G)$ is abelian then $G/Z(G)$ is abelian.

- But by the result on the If G/Z(G) is Cyclic then G is Abelian page, we have that $G/Z(G)$ being abelian implies that $G$ is abelian. $\blacksquare$