This should probably be posted in the math forums but I guess it would fall under here because it's a HW question due late tomorrow.(adsbygoogle = window.adsbygoogle || []).push({});

If G is a group, prove that Aut(G) and Inn(G) are groups, where Aut(G) is the set of automorphisms of G and Inn(G) is the set of inner automorphisms of G

For now I'm not concerned with proving Inn(G) is a group so I'm only going to post my attempt at a proof that Aut(G) is a group. Please critique it to death because my professor is very meticulous, ie if I label a variable i that represents the last character in a finite set as opposed to n I will get marked off because i classically represents an arbitrary element in that set:

Let Aut(G) = {[tex]\phi_1 , \phi_2 ... \phi_n[/tex]} where [tex]\phi_i[/tex] is an automorphism for 1<=i<=n

Aut(G) is a group iff:

1. [tex]{\phi_i}^{-1} \in Aut(G)[/tex]

2. [tex]\exists[/tex] an identity [tex]\phi_e[/tex]

3. [tex]\phi_i (\phi_j \phi_k) = (\phi_i \phi_j) \phi_k[/tex] for [tex]\phi_i , \phi_j , \phi_k \in Aut(G)[/tex]

(Aside: I have access to a theorem that says if [tex]\phi_i[/tex] is an isomorphism, so is [tex]{\phi_i}^{-1}[/tex])

1. By property 3 of Thm. 6.3, if [tex]\phi_i[/tex] is an isomorphism [tex]G\rightarrow G'[/tex], [tex]{\phi_i}^{-1}[/tex] is an isomorphism [tex]G' \rightarrow G[/tex]. Since [tex]G' = G[/tex] for an automorphism, [tex]{\phi_i}^{-1} \in Aut(G)[/tex]

2. There exists an identity [tex]\phi_e \in Aut(G)[/tex]:

Let [tex]\alpha_e : G \rightarrow G : x \rightarrow x[/tex]. [tex]\alpha_e[/tex] is obviously 1-1. Since for [tex]a, b \in G[/tex] [tex]\alpha_e(a)\alpha_e(b) = (a)(b) = (ab) = \alpha_e(ab)[/tex], [tex]\alpha_e[/tex] preserves function composition and therefore is isomorphic. Because [tex]\alpha_e[/tex] transforms [tex]G \rightarrow G[/tex], it is an automorphism so [tex]\alpha_e \in Aut(G)[/tex].

- Since [tex]\forall x \in G[/tex], [tex]\phi_i\alpha_e(x) = \phi_i(\alpha_e(x)) = \phi_i(x)[/tex] and [tex]\alpha_e\phi_i(x) = \alpha_e(\phi_i(x)) = \phi_i(x)[/tex] for 1<=i<=n, [tex]\alpha_e = \phi_e[/tex]

3. The function composition operation is itself associative, so associativity holds in Aut(G)

Thus Aut(G) is a group satisfying the 3 properties above

-Is it fixable? What do I need to change? I'm really trying to learn this.

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# Homework Help: Abstract Algebra Proof Help

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