Results 1 to 3 of 3

Thread: Automorphism - question

  1. #1
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1

    Automorphism - question

    G - Finite Group
    T - Automorphism of G
    For all x in G, T(x) = x IFF x=1
    Also, T(T(x))=x for all x in G.

    Prove G is abelian.

    Struggling a lot with this. I suspect $\displaystyle T(x) = x^{-1}$.
    And once I establish this, proving G is abelian is simple.

    But not going anywhere with the hunch of $\displaystyle T(x) = x^{-1}$.

    Any pointers please?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by aman_cc View Post
    G - Finite Group
    T - Automorphism of G
    For all x in G, T(x) = x IFF x=1
    Also, T(T(x))=x for all x in G.

    Prove G is abelian.

    Struggling a lot with this. I suspect $\displaystyle T(x) = x^{-1}$.
    And once I establish this, proving G is abelian is simple.

    But not going anywhere with the hunch of $\displaystyle T(x) = x^{-1}$.

    Any pointers please?
    define the function $\displaystyle f: G \longrightarrow G$ by $\displaystyle f(g)=g^{-1}T(g).$ now if $\displaystyle f(g_1)=f(g_2),$ then $\displaystyle T(g_1g_2^{-1})=g_1g_2^{-1}$ and thus $\displaystyle g_1=g_2,$ because we're given that $\displaystyle T(x)=x \Longrightarrow x = 1.$ so $\displaystyle f$ is injective and thus

    surjective, because $\displaystyle G$ is finite. now let $\displaystyle x \in G.$ then $\displaystyle x=f(g)=g^{-1}T(g),$ for some $\displaystyle g \in G.$ thus $\displaystyle T(x)=T(g^{-1}T(g))=T(g^{-1})T(T(g))=(T(g))^{-1}g=(g^{-1}T(g))^{-1}=x^{-1}.$ hence $\displaystyle G$ is abelian.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by NonCommAlg View Post
    define the function $\displaystyle f: G \longrightarrow G$ by $\displaystyle f(g)=g^{-1}T(g).$ now if $\displaystyle f(g_1)=f(g_2),$ then $\displaystyle T(g_1g_2^{-1})=g_1g_2^{-1}$ and thus $\displaystyle g_1=g_2,$ because we're given that $\displaystyle T(x)=x \Longrightarrow x = 1.$ so $\displaystyle f$ is injective and thus

    surjective, because $\displaystyle G$ is finite. now let $\displaystyle x \in G.$ then $\displaystyle x=f(g)=g^{-1}T(g),$ for some $\displaystyle g \in G.$ thus $\displaystyle T(x)=T(g^{-1}T(g))=T(g^{-1})T(T(g))=(T(g))^{-1}g=(g^{-1}T(g))^{-1}=x^{-1}.$ hence $\displaystyle G$ is abelian.
    Thanks !
    I could never have guessed it - to define $\displaystyle f(g)=g^{-1}T(g)$ and then do it !!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Automorphism
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: Aug 22nd 2010, 11:39 PM
  2. Automorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jan 21st 2009, 11:48 AM
  3. AutoMorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Jul 31st 2008, 05:28 AM
  4. Automorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 19th 2008, 07:03 AM
  5. automorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 1st 2008, 07:11 PM

Search Tags


/mathhelpforum @mathhelpforum