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Let $G$ be a group. Say what it means for a map $\varphi: G \rightarrow G$ to be an automorphism. Show that the set-theoretic composition $\varphi \psi = \varphi \circ \psi$ of any two automorphisms $\varphi, \psi$ is an automorphism. Prove that the set $\mathrm{Aut}(G)$ of all automorphisms of the group $G$ with the operation of taking the composition is a group.

I have said: A map is an automorphism of a group $G$ if it is an isomorphism to itself.

a) For the next bit, I want to show if $\varphi, \psi$ is bijective, then $\varphi \circ \psi$ is bijective: For two elements $a, b \in G$ we have

$$\varphi \circ \psi (ab) = \varphi(\psi(ab)) = \varphi(\psi(a)\psi(b))$$

as $\psi$ is an isomorphism. Also, as $\varphi$ is an isomorphism, we have

$$\varphi(\psi(a) \psi(b)) = \varphi \circ \psi(a) \varphi \circ \psi(b)$$

Showing $\varphi \circ \psi$ is an isomorphism iff $\varphi, \psi$ are isomorphisms.

For the group bit, we want to prove the 3 group axioms.

1) Associativity: $\varphi \circ (\psi \circ \zeta) = (\varphi \circ \psi) \circ \zeta$. So for some $x \in G$, we get:

$$\varphi \circ (\psi \circ \zeta)(x) = (\varphi \circ \psi) \zeta(x) = \varphi(\psi(\zeta(x))) $$

2) Identity: If we let the identity automorphism, $e: G \rightarrow G$, be the map $e(x) = x$, then clearly we get that $e \circ \psi = \psi \circ e = \psi$.

3) Inverse: As the automorphisms are bijective (already proved) then we know that by definition of a bijection, there is well defined inverse such that $\psi^{-1}: G \rightarrow G$ exists.

(Second edit to correct proof for inverse): For any two elements $a,b \in G$, we want to see if $\psi^{-1}(ab) = \psi^{-1}(a)\psi^{-1}(b)$. Apply $\psi$ to both sides gives us

$$\psi \circ \psi^{-1}(ab) = \psi(\psi^{-1}(ab)) = ab$$

Doing the same on RHS gives us $ab$ and so we have proved the inverse exists and is unique.

Is this right and enough to prove this?

EDIT: Actually, can I just say that by definition of two bijective maps, the composition is also bijective and this is enough?

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You showed that $\phi\circ\psi$ is a homomorphism and not necessarily an isomorphism. For part (3) you showed that the function is invertible, but you do not say why the inverse is a homomorphism. –  peoplepower Dec 30 '12 at 15:08
    
@peoplepower What am I missing? Definition of an isomorphism: $f: G \rightarrow H$ for two groups $G, H$ is an isomorphism if 1) $f$ is bijective, 2) $f(1_G) \rightarrow 1_H$ and 3) For any $a, b \in G$, we have $f(ab) = f(a)f(b)$. Where part 2 and 3 define a homomorphism (right?) and so I actually haven't shown its bijective? –  Kaish Dec 30 '12 at 15:12
    
As far as I can tell, you have not shown anything other than the third criterion. (It is noteworthy that the second criterion for isomorphisms is superflous: $f(1_G)=f(1_G1_G)=f(1_G)f(1_G)$, so applying $f(1_G)^{-1}$ to both sides gives $f(1_G)=1_H$.) –  peoplepower Dec 30 '12 at 15:26
    
@peoplepower Great. So I have to show its bijective? Lol I still can't do that :( I'll try it an put an edit with what I have done. Once I've shown its bijective, then in my OP, I can keep part 3 as it is as now I'm referring to an isomorphism and not a homomorphism right? –  Kaish Dec 30 '12 at 15:30
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To answer your question in the EDIT section, yes, it follows easily from the definition of bijective map that the composition of two bijective maps is bijective, but at this level I think you are meant to do the simple calculation and prove it carefully. To demonstrate that the composition is one-to-one, take $x_1 \neq x_2$ and explain why $\phi(\psi(x_1))\neq\phi(\psi(x_2))$. To show that the composition is onto, take an arbitrary $y$ and explain why there exists $x$ such that $\phi(\psi(x)) = y$. –  William DeMeo Dec 30 '12 at 16:30
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