The question is as follows in a past exam paper:
Suppose that $G$ is a group. An isomorphism from $G$ to itself is called an automorphism.
Prove that the set $Aut(G)$ of all automorphisms of $G$ is a group under the operation of composition of functions.
For each $x \in G$, prove that $\theta_x:G \to G$ given by $\theta_x(g)=x^{-1}gx$ is an automorphism of $G$, and let $Inn(G)=\{ \theta_x \mid x \in G \}$ denote the set of all such automorphisms (known as inner automorphisms).
Prove that $Inn(G) \leqslant Aut(G)$
My Answer
for closure; let $f_1,f_2 \in Aut(G$), so I need to show that $f_1 \circ f_2$ is an automorphism. I know that $f_1,f_2 \in Aut(G)$ are bijective, so once I show that $f_1 \circ f_2$ is bijective then I have shown closure holds for composition of functions.
So, let $x,y \in G$ s.t. $f_1 \circ f_2(x)=f_1 \circ f_2(y) \Rightarrow f_1(f_2(x))=f_1(f_2(y)) \Rightarrow f_2(x)=f_2(y)$, since $f_1$ is injective, and $x=y$, since $f_2$ is injective. Hence
$f_1 \circ f_2$ is injective.
for surjectivity, since $f_1 \in Aut(G)$ is surjective, then $\exists x \in G$ s.t. $f_1(x)=y$ for some $y \in G$, and $f_2 \in Aut(G)$ is surjective, then $\exists z \in G$ s.t. $f_2(z)=x$, therefore $f_1 \circ f_2(z)=y$ since $f_1(f_2(z))=y \Rightarrow f_1(x)=y$ for any $y \in G$
It folows that closure holds for $f_1 \circ f_2$.
For associativity, given that composition of functions is associative, we take it that composition in G is associative.
For the identity; I know that the identity function, $I_f$, always returns the same value that was used as its argument, is an element of $Aut(G)$. So, let $f \in Aut(G)$ s.t. $f \circ I_f(x) =f(x)=I_f \circ f(x)$. It is clear that the identity exists.
For the inverse; since $f \in Aut(G)$ and is bijective, then $\exists f^{-1} \in Aut(G)$ s.t. $f \circ f^{-1}=I_f =f^{-1} \circ f$
It follows that $Aut(G)$ under the operation of composition of functions is a group.
For the second part of the question. I need to show, firstly, that $\theta_x$ is an automorphism, that is bijectivity, and secondly, homomorphism.
So, $\forall g,h \in G$ let $$\theta_x(g)=\theta_x(h) \\ x^{-1}gx=x^{-1}hx \\ xx^{-1}gx=xx^{-1}hx \\ egxx^{-1}=ehxx^{-1} \\ ge=he\\g=h$$ Hence $\theta_x$ is injective
For surjectivity, let $\theta_x(xgx^{-1})=x^{-1}(xgx^{-1})x=(x^{-1}x)g(xx^{-1})=ege=g$, so it follows that $\theta_x$ is bijective.
To show that $\theta_x$ is a homomorphism let $g,h \in G$ s.t. $\theta_x(gh)=x^{-1}ghx=x^{-1}gxx^{-1}hx = (x^{-1}gx)(x^{-1}hx)= \theta(g) \theta(h)$.
Since $\theta_x$ is a homorphism and is bijective, it folows that it is an automorphism.
For the final part, I need to show closure and that an inverse exists to prove $Inn(G) \leqslant Aut(G)$.
I need a bit of help with this. Closure is obvious since $$\theta_x \circ \theta_y(g)=\theta_x(\theta_y(g))=\theta_x(y^{-1}gy)=x^{-1}(y^{-1}gy)x=(x^{-1}y^{-1})g(yx)=\theta_{xy}(g)$$
I have to find a $\theta_x$ that satisfies the inverse property, any help would be appreciated.
Best Answer
Need improvement as commented above. Here is how to satisfy inverse property?
$$\theta_x \circ \theta_{x^{-1}}(g)=\theta_x(\theta_{x^{-1}}(g))=\theta_x(xgx^{-1})=x^{-1}(xgx^{-1})x=g=\theta_e(g)$$
Similarly, $$\theta_{x^{-1}} \circ \theta_{x}(g)=\theta_e(g)$$