Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook