Question

If $\mathit{G}$ is a group, prove that $\mathit{G}\mathbf{/}\mathit{Z}\mathbf{\left(}\mathbf{G}\mathbf{\right)}$ is isomorphic to the group Inn $\mathit{G}$ of all inner automorphisms of $\mathit{G}$ (see Exercise 37 in Section 7.4).

Short answer

It is proved that $G/Z\left(G\right)$ is isomorphic to the group $G$.

Step-by-step solution

Define the mapping

Define the mapping $f:G/Z\left(G\right)\to \text{Inn}\left(G\right)$ as $f\left(\left[g\right]\right)=I_g$, where $I_g\left(h\right)=gh{g}^{-1},h\in G$.

Prove that $f$ is an isomorphism.

First show that $f$is well defined as:

Assume that $\left[a\right],\left[b\right]\in G/Z\left(G\right)$, such that $I_a=I_b$. Then, by the definition of inner automorphism, we have:

$\begin{array}{l}ag{a}^{-1}=bg{b}^{-1},\forall g\in G\\ \Rightarrow b^{-1}ag=g{b}^{-1}a,\forall g\in G\end{array}$

This implies that $b^{-1}a\in Z\left(G\right)$, that is, $\left[a\right]=\left[b\right]$ in $G/Z\left(G\right)$.

Thus, role="math" localid="1654531327626" $f$ is well defined.

Prove that f is a group isomorphism

Firstly, show that f is group homomorphism as:

Assume that $\left[x\right],\left[y\right]\in G/Z\left(G\right)$, then:

$$\begin{gathered} f\left(\left[x\right]{\left[y\right]}^{-1}\right)=f\left(\left[x{y}^{-1}\right]\right) \\ \begin{array}{c}=I_{x{y}^{-1}}\\ =I_x\circ I_{y^{-1}}\\ =I_x\circ {I_y}^{-1}\\ =f\left(x\right)\circ f{\left(y\right)}^{-1}\end{array} \end{gathered}$$

This implies that f is a group homomorphism.

Now, show that f is one-to-one and onto as:

Assume that $\left[x\right],\left[y\right]\in G/Z\left(G\right)$, such that:

$\begin{array}{c}f\left(\left[x\right]\right)=f\left(\left[y\right]\right)\\ I_x=I_y\\ xg{x}^{-1}=yg{y}^{-1}\end{array}$

This implies that $y^{-1}xg=g{y}^{-1}x,\forall g\in G$.

That is,

$\begin{array}{l}{y}^{-1}x\in Z\left(G\right)\\ \left[x\right]=\left[y\right]\end{array}$

Hence, f is one-to-one.

Assume that $I_c\in \text{Inn}\left(G\right)$, then:

role="math" localid="1654531602595" $f\left(\left[c\right]\right)=I_c$

This implies that f is onto.

Hence, it is proved that f is group isomorphism and $G/Z\left(G\right)$is isomorphic to $\text{Inn}\left(G\right)$.