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Math Help - Question about isomorphism 4?

  1. #1
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    Question about isomorphism 4?

    Given a group G, let A(G) be the set of all isomorphism from G to itself. For every g∈G, define the map φg : G-->G by φg(x) = g^ -1xg
    (a) Show that A(G) is a group under composition.
    (b) Prove thatφg ∈A(G) for every g ∈G.
    (c) Let φ: G-->A(G) be the map g-->φg. Verify that φ is a homomorphism.

    Note φg, g is subscript to the φ (i do not know how to type)
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  2. #2
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    Quote Originally Posted by koukou8617 View Post
    Given a group G, let A(G) be the set of all isomorphism from G to itself. For every g∈G, define the map φg : G-->G by φg(x) = g^ -1xg
    (a) Show that A(G) is a group under composition.
    (b) Prove thatφg ∈A(G) for every g ∈G.
    (c) Let φ: G-->A(G) be the map g-->φg. Verify that φ is a homomorphism.

    Note φg, g is subscript to the φ (i do not know how to type)
    For (a), you need to verify the group axioms for A(G).
    The identity element of A(G) is the trivial automorphism sending each element to itself and the inverse of an automorphism \phi is its inverse \phi^{-1} as a set map.

    For (b) and (c), to show that \phi_g is an automorphism, you need to show first that \phi_g is well-defined. Then, you need to show that \phi_g is homomorphism such that \phi_g(xy)=\phi_g(x) \phi_g(y). To show injectivity of \phi_g, you need to show that if \phi_g(x)=\phi_g(y), then x=y, where x,y \in G. To show surjectivity, it suffice to show that x=\phi_g(gxg^{-1}), where x, g \in G.

    =======================
    This is some additional stuff.
    Those maps in (b) is called an inner automorphism of G. If \sigma \in Aut(G) and \phi_g \in Inn(G), then \sigma \phi_g \sigma^{-1} = \phi_{\sigma(g)}, which shows Inn(G) \triangleleft Aut(G) . Additionally, outer automorphism can be defined as Out(G)=Aut(G)/Inn(G) .
    Last edited by aliceinwonderland; October 15th 2009 at 06:17 PM. Reason: Some additions and rectifications.
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