Definition: Inner Automorphism
Let \( (G,\cdot) \) be a group. The set of all automorphisms \( \theta_{a}:G\to G \) for all \( a\in G \), defined by
\[
\theta_{a}(x)=a\cdot x\cdot a^{-1} \quad \forall~x\in G,
\]
is said to be the Inner Automorphism of \( G \) and is denoted by \( \text{Inn}(G) \).
i.e.,
\[
\text{Inn}(G)=\{\theta_{a} \mid \theta_{a}:G \to G \text{ defined by } \theta_{a}(x)=a\cdot x\cdot a^{-1} \quad \forall~x\in G \text{ and } a\in G \}.
\]
Theorem
For any group \((G,\cdot)\), prove that \(\operatorname{Inn}(G)\) is a normal subgroup of \(\operatorname{Aut}(G)\).
Proof
Step 1:
To Prove \(\operatorname{Inn}(G)\) is a Subgroup of \(\operatorname{Aut}(G)\)
For each \(g \in G\), define the mapping
\[
\theta_g : G \to G \quad \text{by} \quad \theta_g(x) = g x g^{-1}.
\]
The set
\[
\operatorname{Inn}(G) = \{ \theta_g \mid g \in G \}
\]
Closure: Let \(\theta_g, \theta_h \in \operatorname{Inn}(G)\). Then for any \(x \in G\):
\[
\begin{align}
(\theta_g \circ \theta_h)(x) &= \theta_g\big(\theta_h(x)\big) \nonumber\\[5mm]
&= \theta_g\big(h x h^{-1}\big) \nonumber\\
&= g \big(h x h^{-1}\big) g^{-1} \nonumber\\
&= (gh) \, x \, (gh)^{-1} \nonumber\\
&= \theta_{gh}(x)\nonumber
\end{align}
\]
Since \(gh \in G\), the mapping \(\theta_{gh}\) is an element of \(\operatorname{Inn}(G)\). Thus, \(\operatorname{Inn}(G)\) is closed under composition.
Identity: For the identity element \(e \in G\),
\[
\theta_e(x) = e x e^{-1} = x,
\]
so \(\theta_e\) is the identity automorphism.
Inverses: For any \(g \in G\), the inverse of \(\theta_g\) is given by
\[
\theta_g^{-1} = \theta_{g^{-1}},
\]
since for any \(x \in G\):
\[
\theta_g \circ \theta_{g^{-1}}(x) = \theta_{gg^{-1}}(x) = \theta_e(x) = x.
\]
Therefore, \(\operatorname{Inn}(G)\) is a subgroup of \(\operatorname{Aut}(G)\).
Step 2:
To Prove \(\operatorname{Inn}(G)\) is Normal in \(\operatorname{Aut}(G)\)
To show normality, we need to prove that for every \(\varphi \in \operatorname{Aut}(G)\) and every \(\theta_g \in \operatorname{Inn}(G)\), the conjugate
\(\varphi \circ \theta_g \circ \varphi^{-1}\) is in \(\operatorname{Inn}(G)\).
Let \(\varphi \in \operatorname{Aut}(G)\) and \(g \in G\). For any \(x \in G\):
\[
\begin{align}
(\varphi \circ \theta_g \circ \varphi^{-1})(x) &= \varphi\Big(\theta_g\big(\varphi^{-1}(x)\big)\Big) \nonumber\\[5mm]
&= \varphi\Big(g\, \varphi^{-1}(x)\, g^{-1}\Big) \nonumber\\[5mm]
&= \varphi(g) \, \varphi\big(\varphi^{-1}(x)\big) \, \varphi(g)^{-1} \nonumber\\[5mm]
&= \varphi(g) \, x \, \varphi(g)^{-1}
\end{align}
\]
Equation (1) shows that
\[
\varphi \circ \theta_g \circ \varphi^{-1} = \theta_{\varphi(g)},
\]
which is an element of \(\operatorname{Inn}(G)\). Thus, \(\operatorname{Inn}(G)\) is normal in \(\operatorname{Aut}(G)\).
