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Question about automorphisms

  1. Nov 12, 2012 #1


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    So I've been reading a bit about automorphisms today and I was wondering about something. I'm particularly talking about groups in this case, say a group G.

    So an automorphism is a bijective homomorphism ( endomorphism if you prefer ) of a group G.

    We use Aut(G) to denote the set of all automorphisms of G.

    There is also an inner automorphism induced by a in G. The map f := G → G such that f(x) = axa-1 for all x in G is called the inner automorphism of G induced by a.

    We use Inn(G) to denote the set of all inner automorphisms of G.

    My question now :

    If |G| is infinite, are the orders of Aut(G) and Inn(G) also infinite? It makes sense to me that they would be.
    Last edited: Nov 12, 2012
  2. jcsd
  3. Nov 12, 2012 #2


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    No. It is easy to check that [itex]\mathrm{Aut}(\mathbb{Z}) \cong \mathbb{Z}/(2)[/itex] and that [itex]\mathrm{Inn}(\mathbb{Z}) = 0[/itex].
  4. Nov 12, 2012 #3


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    No, not always. The integers, Z, is a group under addition. |Z| is infinite, but there are only two automorphisms on Z, given by f(x)=x (identity map) and g(x)=-x, respectively.
  5. Nov 12, 2012 #4


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    Ahh yes, that makes sense now because every other map doesn't generate all of Z.

    So f(x) = 2x wouldn't be an automorphism for example. While it is a homomorphism, it is not injective, only surjective I believe.

    Short proof to make sure : So f : Z → Z is already well defined for us. So we assume f(x) = f(y) → 2x = 2y → 2(x-y) = 0. Since 2|0, clearly x = y and f is injective. Now notice that the n elements of Z get mapped to fewer elements, so f is not surjective. Easy to show it's a homomorphism.

    So f(x) = x and g(x) = -x are the only things that could be in Aut(Z) since they are the only maps which turn out to be automorphisms.

    As for Inn(Z) = 0. That confused me a bit.

    EDIT : Wouldn't it be Inn(Z) = { 0 } ?
  6. Nov 12, 2012 #5


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    They mean the same thing. Both of them are shorthand for the statement that [itex]\mathrm{Inn}(\mathbb{Z})[/itex] is the trivial group.
  7. Nov 12, 2012 #6
    You switched the two around. It is injective but not surjective, which is what you proved next.
  8. Nov 12, 2012 #7


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    Oh whoops, mistype. Thanks for noticing.

    Thanks for clarifying jgens I think I understand this now.
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