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Math Help - automorphism proof

  1. #1
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    automorphism proof

    If G is a group for which Aut(G)={1}, prove that |G| \leq 2
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  2. #2
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    Caveat: I am new to abstract algebra.

    I can't think of a slick proof. Essentially I show this case by case.

    Case 1: |G|=1
    Because the order of G is 1, then the only element of G is the identity element, G={1}. Further more the only automorphic map \phi:G\rightarrow G is then the identity map.

    Case 2: |G|=2
    Because the order of G is 2, then the only two elements of G are the identity element and some other element, G={1,a}. Because a homomorphism must contain all the inverses for its elements, a must be its own inverse, a^2 = 1. Furthermore, because a homomorphism maps identities to identities and inverses to inverses the only possible homomorphism (and therefore automorphism) is again the identity map.

    Case 3: |G|>2
    We need to think up an automorphism that is not the identity map. We need to show that this map is an automorphism. That is as far as I got.

    Sorry,

    hope this helps some
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  3. #3
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    My thinking was for non-abelian groups, conjugation by an element not in the centre. For abelian groups for which not all elements are their inverses, \theta (x) = x^{-1} and for groups where all elements are self inverse, permuting elements should work.
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  4. #4
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    Quote Originally Posted by n0083 View Post
    Caveat: I am new to abstract algebra.

    I can't think of a slick proof. Essentially I show this case by case.

    Case 1: |G|=1
    Because the order of G is 1, then the only element of G is the identity element, G={1}. Further more the only automorphic map \phi:G\rightarrow G is then the identity map.

    Case 2: |G|=2
    Because the order of G is 2, then the only two elements of G are the identity element and some other element, G={1,a}. Because a homomorphism must contain all the inverses for its elements, a must be its own inverse, a^2 = 1. Furthermore, because a homomorphism maps identities to identities and inverses to inverses the only possible homomorphism (and therefore automorphism) is again the identity map.

    Case 3: |G|>2
    We need to think up an automorphism that is not the identity map. We need to show that this map is an automorphism. That is as far as I got.

    Sorry,

    hope this helps some
    Thanks. I'm new to it also. I had a proof by using cases but i'm stuck on the last case also.
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  5. #5
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    so, we may assume that G is an abelian group and every element \neq 1 of G has order 2. looking at G as an additive group, now we can consider G as a vector space over \mathbb{F}_2.

    if |G| > 2, then \dim_{\mathbb{F}_2} G > 1. let H=\{g_{\alpha}: \ \alpha \in I \} be a basis for G. since |H| \geq 2, we may choose g_{\alpha} \neq g_{\beta} \in H. now define \varphi:G \longrightarrow G by \varphi(g_{\alpha})=g_{\beta}, \ \varphi(g_{\beta})=g_{\alpha} and

    \varphi(g_{\gamma})=g_{\gamma}, for any \gamma \in I - \{\alpha, \beta \}. then \varphi is a non-trivial automorphism of G. Q.E.D.
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