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Abstract Algebra: Automorphisms

  1. Sep 23, 2015 #1
    I have a question about Automorphisms. Please check the following statement for validity....

    An automorphism of a group should map generators to generators. Suppose it didn't, well then the group structure wouldn't be preserved and since automorphisms are homomorphisms this would be a contradiction.

    If this is valid is there an example of a homomorphism (not an automorphism) of groups, say ##\phi:G\to H## that doesn't map a generator of ##G## to a generator of ##H##?
     
  2. jcsd
  3. Sep 23, 2015 #2

    WWGD

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    Isnt the constant map to the identity a homomorphism? Or do you mean for non-constant maps?

    EDIT: Actually, the identity does generate the subgroup Im(G).
     
    Last edited: Sep 23, 2015
  4. Sep 23, 2015 #3
    An automorphism of a group with a generating set will map the generating set to a generating set, but not necessarily the same one. For a concrete example, we could map the free abelian group G on two symbols [itex]\{x,y\}[/itex] to itself by mapping [itex]x\mapsto x-y[/itex] and [itex]y\mapsto 2y-x[/itex].

    More generally, homomorphisms [itex]G\rightarrow H[/itex] don't have any choice other than to map generators of G to the generators of the image in H. [itex]\mathbb{Z}[/itex] makes a good example since it only has one generator. If I'm looking at any morphism [itex]\mathbb{Z}\rightarrow G[/itex], I need only define [itex]\varphi(1)[/itex]. As then, [tex]\varphi(n) = \varphi(\underbrace{1+1+\ldots +1}_\text{n}) = \underbrace{\varphi(1)+\varphi(1)+\ldots \varphi(1)}_\text{n} \in G[/tex] So [itex]\varphi(1)[/itex] generates [itex]\textrm{Im}(\varphi)\leq G[/itex]. This works similarly for any other group with a generating set, since every element will break down into combinations of those in the generating set. The only caveat is that its possible to map generators to the identity of the codomain group.. in which case it's not much of a generator anymore.
     
  5. Sep 24, 2015 #4
    OK, thank you both for your reply. It's a little more clear now.
     
  6. Sep 24, 2015 #5

    WWGD

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    Like Fire Garden said, this is the defining ( Universal) property of free "objects".
     
  7. Sep 28, 2015 #6

    lavinia

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    Map any proper subgroup of a group to itself by the inclusion homomorphism. Then any set of generators of the subgroup can not be a set of generators of the group.

    More generally the homomorphism must be surjective.

    However, if the kernel is non-trivial then a generator may be mapped to the identity. In this case one might ask whether a set that contains the identity is really a set of generators. Certainly if you take set of generators to mean a minimal set i.e. a set that can not be made smaller and still generate the group, then a set containing the identity is not a set of generators.
     
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