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Homework Help: Abstract Algebra Proof Help

  1. Feb 10, 2005 #1
    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.

    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.
    Last edited by a moderator: Feb 10, 2005
  2. jcsd
  3. Feb 10, 2005 #2


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    You seem to be assuming G is a group of functions, with group operation composition, and that Aut(G) is finite. Was that intentional?

    You might want this notation for a function: I think it's clearer, and more standard:

    \alpha_e : G \rightarrow G : x \rightarrow x

    Don't you need to prove that the identity element is both a left and a right identity?

    Why did you appeal to the associativity of G?
  4. Feb 10, 2005 #3
    Alright I'm looking at what you said, and I guess I screwed up the wording of the question a little so I'll edit it to be verbatim out of the book. I corrected the definition of [tex]\alpha_e[/tex]. It doesn't specify what G is, but how would the argument change if G is just a bunch of random numbers? Also, is it safe to just remove the constraint n from the set to make it infinite or is that not right? As for the third line, I am writing that in now because I accidentally assumed the group was Abelian.

    The fourth line has me a little stumped though. How should I approach proving associativity then? I did sort of just get lazy at the end and blurt out 3.

    Edit: I changed 3. a little because it didn't seem to make sense to bring up the fact that associativity holds between elements of G when I was talking about compositions of functions in Aut(G). Is it any less wrong?
    Last edited by a moderator: Feb 10, 2005
  5. Feb 11, 2005 #4


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    It changes because you must speak of the "group operation" instead of "function composition". That's it, though.

    Your way of saying something is in Aut(G) is somewhat awkward... why not say things like:

    If [itex]\pi, \phi, \theta \in \mathrm{Aut}(G)[/itex], then [itex]\pi (\phi \theta) = (\pi \phi) \theta[/itex]?
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