Results 1 to 6 of 6

Math Help - Prove a group H is isomorphic to Z.

  1. #1
    Junior Member
    Joined
    Nov 2008
    Posts
    61

    Prove a group H is isomorphic to Z.

    Let H be a group containing an element h with the following property: given a group G and an element g in G, there is a unique homomorphism from H to G mapping h to g. Prove that H is isomorphic to Z (the set of integers).

    I know (or I think) that \psi :H \rightarrow Z such that \psi (h)=1 defines an isomorphism from H to Z. It's a homomorphism and it is surjective since 1 is a generator of Z. But I'm having a tough time proving that it is injective. Any hints or advice? Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by JD-Styles View Post
    Let H be a group containing an element h with the following property: given a group G and an element g in G, there is a unique homomorphism from H to G mapping h to g. Prove that H is isomorphic to Z (the set of integers).

    I know (or I think) that \psi :H \rightarrow Z such that \psi (h)=1 defines an isomorphism from H to Z. It's a homomorphism and it is surjective since 1 is a generator of Z. But I'm having a tough time proving that it is injective. Any hints or advice? Thanks.
    Instead of trying to find an isomorphism you can show that \mathbb{Z} satisfies this property, and also that because of the conditions imposed on H mean it must in fact be \mathbb{Z} (these conditions can be found by remembering that H must map onto \mathbb{Z}, the only infinite, one-generated group (up to isomorphism)).

    Are you familiar with Free Groups?
    Last edited by Swlabr; September 23rd 2009 at 01:20 AM. Reason: Making what I was saying a bit clearer
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2008
    Posts
    61
    I'm not familiar with free groups. But I'll give your suggestion a shot. Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    as Swlabr mentioned, that's just the universal property of a free group on one generator. but it doesn't mean that the proof is hard! :

    consider the maps H \overset{\psi} \longrightarrow \mathbb{Z} \overset{\phi} \longrightarrow H, where \psi is the homomorphism sending h to 1 (your map) and \phi is defined by \phi(n)=h^n, for all n \in \mathbb{Z}. note

    that \phi is one-to-one. (why?) now consider the maps \phi \psi : H \longrightarrow H and \text{id}: H \longrightarrow H, the identity map. we have \phi \psi(h)=\phi(1)=h=\text{id}(h).

    thus by the "uniqueness" property: \phi \psi = \text{id}. therefore for any x \in H we have \phi(\psi(x))=x, i.e. \phi is onto and the proof is complete.
    Last edited by NonCommAlg; September 23rd 2009 at 05:04 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2008
    Posts
    61
    Wow. Thanks. It's so simple yet I just kept stumbling on it. Thanks a lot.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by NonCommAlg View Post
    as Swlabr mentioned, that's just the universal property of a free group on one generator. but it doesn't mean that the proof is hard! :

    consider the maps H \overset{\psi} \longrightarrow \mathbb{Z} \overset{\phi} \longrightarrow H, where \psi is the homomorphism sending h to 1 (your map) and \phi is defined by \phi(n)=h^n, for all n \in \mathbb{Z}. note

    that \phi is one-to-one. (why?) now consider the maps \phi \psi : H \longrightarrow H and \text{id}: H \longrightarrow H, the identity map. we have \phi \psi(h)=\phi(1)=h=\text{id}(h).

    thus by the "uniqueness" property: \phi \psi = \text{id}. therefore for any x \in H we have \phi(\psi(x))=x, i.e. \phi is onto and the proof is complete.
    That's a very neat proof. I like it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove that the finite cyclic group is isomorphic
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 19th 2011, 02:20 AM
  2. isomorphic group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 5th 2011, 10:51 PM
  3. Any group of 3 elements is isomorphic to Z3
    Posted in the Advanced Algebra Forum
    Replies: 11
    Last Post: November 2nd 2010, 06:30 AM
  4. Replies: 4
    Last Post: April 13th 2010, 07:09 PM
  5. Prove a group is isomorphic to another
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: February 8th 2010, 04:19 PM

Search Tags


/mathhelpforum @mathhelpforum