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Order of Automorphism and Abelian Groups

  1. Nov 22, 2011 #1
    If |Aut H| = 1 then how can I show H is Abelian? I've shown a mapping is an element of Aut H previously but didn't think that would help.

    I have been looking through properties and theorems linked to Abelian groups but so far have had no luck finding anything that would help.

    The closest I have is that it may have something to do with "any subgroup of an Abelian group is normal"

    My line of argument was if |Aut G|=1 then Inn G = Aut G and so since Inn G is normal then G is normal?

    However something tells me this is incorrect.

    Thanks in advance for any help
     
  2. jcsd
  3. Nov 22, 2011 #2
    Here is my revised argument.
    To show |Aut H| = 1 implies H is Abelian, we can show H is not Abelian implies |Aut H| does not equal 1.
    If H is not Abelian then, we know firstly the identity map is always an automorphism of H. Furthermore we know that the mapping A: H ---> H given by A(x) = (h^-1)xh (where h,(h^-1),x are in H and h and (h^-1) are fixed). This tells us that we have |Aut H| does not equal to one. If H had been Abelian then A(x) = (h^-1)xh = (h^-1)x = x and this is just the identity map again.
     
  4. Nov 23, 2011 #3
    Your proof works, but try showing the result directly, i.e. without using contradiction.

    Here's an even shorter proof. We know [itex]G/Z(G) \cong \text{Inn}(G) \subseteq \text{Aut}(G) [/itex]. So if |Aut(G)|=1, then |G/Z(G)| = 1, implying Z(G)=G.
     
  5. Nov 24, 2011 #4

    Deveno

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    along the same lines:

    if |Aut(G)| = 1, then every inner automorphism is the identity.

    this means that for every x and g in G:

    gxg^-1 = x, so

    gx = xg.
     
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