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.

  • $\begingroup$ Examine your proof of surjectivity to identify $(\theta_x)^{-1}$. $\endgroup$ Jun 1, 2020 at 20:05
  • 2
    $\begingroup$ Your first proof is wrong. You are missing the homomorphism property several times, e.g. you have to show that $f_1 \circ f_2$ is a homomorphism as well. $\endgroup$
    – Jan
    Jun 1, 2020 at 20:06
  • $\begingroup$ I assumed that given the question is "Prove that the set $Aut(G)$ of all automorphisms of G is a group under the operation of composition of functions.", the set of automorphisms are homophisms. It is the set of automorphisms under the operation of composition of functions that needs to be shown to be a group. $\endgroup$
    – alortimor
    Jun 1, 2020 at 20:20

1 Answer 1


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, $$\theta_{x^{-1}} \circ \theta_{x}(g)=\theta_e(g)$$


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