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Math Help - Automorphism of the Dihedral Group D8

  1. #1
    Super Member Bernhard's Avatar
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    Automorphism of the Dihedral Group D8

    Dummit and Foote Section 4.4 Automorphisms, Exercise 3 reads as follows:

    ==================================================

    Prove that under any automomorpism of , r has at most two possible images and s has at most 4 possible images. Deduce that |Aut()| 8

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    In a post on Project Crazy Project nbloomf gives a proof:

    First the following Lemma is stated and proved:

    Lemma: Let be a group homomorphism. Then , with equality if is bijective

    Then the proof for the D&F exercise follows:


    Let be an automorphism of . Automorphisms must preserve order, so that has order 4 and has order 2. There are 2 elements of order 4 in , so that we have at most 2 choices for . Similarly, there are 5 elements of order 2 in . However, one of these is in . By the lemma, we have at most 4 choices for

    ================================================== ==

    I cannot follow the reassoning:

    Similarly, there are 5 elements of order 2 in . However, one of these is in . By the lemma, we have at most 4 choices for

    Can anyone please give a much more explicit proof of this assertion - why does one of the elements being in mean that we have at most 4 choices for ???

    Appreciate some help.

    Peter
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    Re: Automorphism of the Dihedral Group D8

    because Z(D8) = {1,r^2}. so the center has to map to itself. an automorphism is bijective, meaning s can't map to r^2, because r^2 already is mapped to r^2. so we have exactly 4 choices:

    s-->s
    s-->rs
    s-->r^2s
    s-->r^3s

    homomorphisms preserve commutativity, if ab = ba in G, then φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a) in H. so φ(Z(G)) is always contained in Z(φ(G)).

    if φ is an isomorphism, then φ(Z(G)) has to BE Z(φ(G)), if you have a bijective map f:A-->B then f(A) = B, right?
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  3. #3
    Super Member Bernhard's Avatar
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    Re: Automorphism of the Dihedral Group D8

    Thanks Deveno

    Still struggling a bit ... when you say "so the center has to map to itself" - why ... because of the Lemma?

    Peter
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  4. #4
    Super Member Bernhard's Avatar
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    Re: Automorphism of the Dihedral Group D8

    Concerning the centre mapping to itself, I really need someone to prove explictly that \phi( r^2) = r^2

    Can anyone help?

    I am assuming that \phi(1)= 1 is true because \phi is a hmomorphism

    Peter
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    Re: Automorphism of the Dihedral Group D8

    Quote Originally Posted by Bernhard View Post
    Concerning the centre mapping to itself, I really need someone to prove explictly that \phi( r^2) = r^2

    Can anyone help?

    I am assuming that \phi(1)= 1 is true because \phi is a hmomorphism

    Peter
    Let G be a group, \phi an automorphism on G. Suppose a\in Z(G) and g\in G. Then \phi (ab) = \phi (ba). Since \phi is an automorphism it follows that \phi(a)\phi(g)=\phi(g)\phi(a). You can see that the image of an element of the center commutes with every element of G (since \phi(g) \in G).

    You may ask why couldn't \phi(g) be in the center instead. The reason is that if g,g' are both elements of G but are not in the center, you know for sure that \phi (gg')\not = \phi(g'g), so \phi(g)\phi(g')\not = \phi(g')\phi(g).
    Last edited by Gusbob; December 8th 2011 at 03:05 PM.
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  6. #6
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    Re: Automorphism of the Dihedral Group D8

    we know that φ is a homomorphism (every automorphism is).

    this means, in particular, for ANY g in D8, φ(r^2)φ(g) = φ(r^2g) = φ(gr^2) = φ(g)φ(r^2).

    so whatever φ(r^2) is, it commutes with every φ(g), for every g in D8. but φ is an automorphism which means it is ONTO.

    so every element of D8 IS φ(g) for some g. so φ(r^2) is in the center of D8. well, we only have two choices:

    φ(r^2) = r^2, or φ(r^2) = 1, because those are the only two elements the center has.

    but φ(1) = 1, so 1 is already taken. so φ(r^2) has to be r^2.
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