# Inner automorphisms form a normal subgroup of $\operatorname{Aut}(G)$

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 $(\operatorname{Aut}(G),\circ)$.

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 $f,f^{-1} \in Aut(G), p_a \in Inn(G):$ $(f^{-1} \circ p_a \circ f)(x) = f^{-1}(p_a(f(x))) = f^{-1}(a^{-1} \cdot f(x) \cdot a) = f^{-1}(a^{-1}) \cdot f(x) \cdot f^{-1}(a) = b^{-1} \cdot f(x) \cdot b.$ $f(x)$ is in $G$, so are $b$ and $b^{-1}$, and $b^{-1} \cdot f(x) \cdot b$ defines another $p_b \in Inn(G)$. The proof is complete.

• You showed only that $p_a$ is bijective, not that it is an isomorphism. – Hagen von Eitzen Aug 12 '14 at 16:52
• Can you explain me why $f^{-1}(a^{-1}\cdot f(x) \cdot a)= f^{-1}(a^{-1})\cdot f(x) \cdot f^{-1}(a)$ and the $f^{-1}$ doesn't affect $f(x)$, please? – MonsieurGalois May 13 '16 at 7:01

$p_{a}$ is bijective because it has the two-sided inverse $p_{a^{-1}}$.

Then you can verify that is a homomorphism: $$p_{a}(x) p_{a}(y) = a^{-1} x a a^{-1} y a = a^{-1} x y a = p_{a}(xy).$$

To prove that $Inn(G)$ is a normal subgroup of $Aut(G)$, show that $$p_{a^{-1}} \circ p_{b} = p_{b a^{-1}},$$ and $$\alpha p_{a} \alpha^{-1} = p_{\alpha(a)},$$ if $\alpha \in Aut(G)$.

Show that $$Inn(G)$$ is a normal subgroup: for $$f,f^{-1} \in Aut(G), p_a \in Inn(G):$$ $$(f \circ p_a \circ f^{-1})(x) = f(p_a(f^{-1}(x))) = f(a^{-1} \cdot f^{-1}(x) \cdot a) = f(a^{-1}) \cdot f(f^{-1}(x)) \cdot f(a) = (f(a))^{-1} \cdot x \cdot f(a)=p_{f(a)}(x)$$ and $$f(a)\in G$$. So $$p_{f(a)} \in Inn(G)$$.

we proved that for each $$f \in Aut(G),p_a \in Inn(G)$$

$$f \circ p_a \circ f^{-1}=p_{f(a)} \in Inn(G)$$ $$\hspace{2em}$$ ie $$\hspace{2em}$$ $$f \circ Inn(G) \circ f^{-1}\subseteq Inn(G)$$ that prove that $$Inn(G)$$ is a normal subgroup in $$Aut(G)$$