For an arbitrary group $(G,\cdot)$ let $\operatorname{Aut}(G) = \{f: G \to G \mid f \text{ is an isomorphism}\}$ be the set of all automorphisms of the group $G$. We assume that $(\operatorname{Aut}(G),\circ)$ where $\circ$ is composition of mappings is a group.
1) Prove that for arbitrary $a \in G$ there is an automorphism $p_a: G \to G;\quad p_a(x) = a^{-1}xa$.
2) Prove, that $\operatorname{Inn}(G) = \{p_a \mid a \in G\}$ is a normal subgroup of $(\operatorname{Aut}(G),\circ)$.
Best Answer
In the beginning, we must show that mapping $p_a: G \to G: p_a(x) = a^{-1}xa$ exists. Then we will show, that it is injective and surjective. If so, then it is isomorphism and automorphism.
Because $(G,\cdot)$ is a group, for $a \in G $ and $x \in G$: $a^{-1}xa \in G$. Because of that, we can safely define $p_a: G \to G: p_a(x) = a^{-1}xa$, knowing that we begin and end in $(G,\cdot)$.
Let's assume that $p_a$ isn't injective. Then $\exists x_1,x_2 \in G, x_1 \ne x_2, $ such as $p_a(x_1) = p_a(x_2)$. Then $a^{-1}x_1a = a^{-1}x_2a$. $(G,\cdot)$ is a group, so we can "multiply" this expression by $a$ from the left side and by $a^{-1}$ from the right side: $aa^{-1}x_1aa^{-1} = aa^{-1}x_2aa^{-1} \Rightarrow x_1=x_2$, but $x_1 \ne x_2$. Therefore $p_a$ must be injective.
Let's assume that $p_a$ isn't surjective. Then $\exists y \in G: \forall x \in G:f(x) \ne y$. Then $a^{-1}xa \ne y$. We can transform it into $\exists y \in G: \forall x \in G: a^{-1}ya \ne x $. But that cannot be, because that means that result of $a^{-1}ya$ isn't in $(G,\cdot)$. Therefore $p_a$ must be surjective.
Now, for the second part, we should prove two things: that $Inn(G)$ is a subgroup of $Aut(G)$, and that it is a normal subgroup.
We must show that neutral element of $Aut(G)$ is in $Inn(G)$. We don't know how he looks like, but it isn't hard to guess that $id$ is a neutral element of $Aut(G)$. $id(x) = x$, so we must find $p_a$ with same properties. It isn't hard, too: $p_1(x) = 1 \cdot x \cdot 1 = x$. So $id = p_1, p_1 \in Inn(G)$.
Then, we must show that $\forall p_a,p_b \in Inn(G)$, $p_b \circ p_a \in Inn(G)$. $(p_b \circ p_a)(x) = p_b(p_a(x)) = p_b(a^{-1}xa) = b^{-1}a^{-1}xab = (ab)^{-1}x(ab) = p_{ab}(x) \in Inn(g)$.
Third part is existence of inverse element to $p_a$ in $Inn(G)$. This element would be $p_{a^{-1}}$: $(p_{a^{-1}} \circ p_a)(x) = (aa^{-1})^{-1}x(aa^{-1}) = 1 \cdot x \cdot 1 = x$.
Finally, we must show that $Inn(G)$ is a normal subgroup: for $f,f^{-1} \in Aut(G), p_a \in Inn(G):$ $(f^{-1} \circ p_a \circ f)(x) = f^{-1}(p_a(f(x))) = f^{-1}(a^{-1} \cdot f(x) \cdot a) = f^{-1}(a^{-1}) \cdot f(f^{-1}(x)) \cdot f^{-1}(a) = f^{-1}(a^{-1})\cdot x\cdot (f^{-1}(a^{-1}))^{-1}$. $f(x)$ is in $G$, so are $f^{-1}(a^{-1})$ and $(f^{-1}(a^{-1}))^{-1}$, and $f^{-1}(a^{-1})\cdot x\cdot (f^{-1}(a^{-1}))^{-1}$ defines another $p_b \in Inn(G)$. The proof is complete.