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Math Help - Automorphisms

  1. #1
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    Automorphisms

    Let a,b be automorphisms of G, and let ab be the product of a and b as mappings on G. Prove that ab is an automorphism of G, and that the inverse of a is an automorphism of G, so that the set of all automorpisms of G is itself a group.
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  2. #2
    Super Member Gamma's Avatar
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    Aut(G)

    Are you sure you want your group operation in Aut(G) to be pointwise multiplication? I think you would want composition to be the group operation.
    For instance, let \sigma^1, \sigma^2 \in Aut(G). And consider the following.

    \sigma^1(xy)\sigma^2(xy)=\sigma^1(x)\sigma^1(y)\si  gma^2(x)\sigma^2(y)

    And what you want to say is that this should be
    \sigma^1(x)\sigma^2(x)\sigma^1(y)\sigma^2(y)

    But this requires G to be commutative since you need to switch the two inner terms if it is to be a homomorphism.

    If instead you use composition
    \sigma^1(\sigma^2(xy))=\sigma^1(\sigma^2(x)\sigma^  2(y))=\sigma^1(\sigma^2(x))\sigma^1(\sigma^2(y)) which is a homomorphism. Furthermore the composition of bijections is a bijection, so it is an isomorphism and clearly it still goes from G to G so it is an automorphism.

    Inverses are particularly clear for function composition as \sigma \in Aut(G) necessarily means it is an isomorphism from G to G and therefore has an inverse that is an isomorphism \exists \sigma^{-1}\in Aut(G), this clearly goes from G to G in the oposite direction and gives you the identity map on G. That is \sigma(\sigma^{-1}(x)=Id(x)=\sigma^{-1}(\sigma(x))

    Associativity and Identity are clear. (composition of functions is associative and the identity is clearly an automorphism from G to G.)

    I hope this helps.
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  3. #3
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    Quote Originally Posted by mpryal View Post
    Let a,b be automorphisms of G, and let ab be the product of a and b as mappings on G. Prove that ab is an automorphism of G, and that the inverse of a is an automorphism of G, so that the set of all automorpisms of G is itself a group.
    Let a be an automorphism of G such that a:G \rightarrow G defined by a(g) = \sigma_a \cdot g where \sigma_a,g \in G; let b be an automorphism of G such that b:G \rightarrow G defined by b(g) = \sigma_b \cdot g where \sigma_b,g \in G.

    Let ab:G \rightarrow G defined by ab(g) = \sigma_{ab} \cdot g where \sigma_{ab},g \in G and \sigma_{ab}=\sigma_{a}\cdot \sigma_{b}. If g_1 = g_2, then ab(g_1)=\sigma_{ab}(g_1) = \sigma_{a}(\sigma_{b}(g_1))=\sigma_{a}(\sigma_{b}(  g_2))=ab(g_2) for g_1, g_2 \in G. Thus, ab is well-defined and it is an automorphism since (ab)(ab)^{-1} = 1_{G} and (ab)^{-1}(ab) = 1_{G}.

    Let a^{-1}:G \rightarrow G defined by a^{-1}(g) = \sigma_a^{-1} \cdot g where \sigma_a^{-1}, g \in G. Since a is an automorphism, a^{-1} is clearly automorphism.

    Now it remains to show that the set of all automorphisms of G itself a group. The identity element of this group is 1_{G}, which maps an element of G to itself. As shown above, the inverse of an automorphism of a is a^{-1}. It is left to you to show the associative law holds for this group.
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