Results 1 to 2 of 2

Math Help - Isomorphism of a surjective group homomorphism

  1. #1
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2

    Isomorphism of a surjective group homomorphism

    Let p be a prime and let  \mathbb {Z} (p^ \infty ) = \{ \hat { \frac {a}{b} } \in \frac { \mathbb {Q} }{ \mathbb {Z} } : a,b \in \mathbb {Z} \ , \ b=p^i \ , \ i \geq 0 \} .

    Prove that if  f: \mathbb {Z} (p^ \infty ) \rightarrow G is a surjective group homomorphism and G is not the trivial group  \{ 1_G \} , then G is isomorphic to  \mathbb {Z} (p^ \infty ) .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by tttcomrader View Post
    Let p be a prime and let  \mathbb {Z} (p^ \infty ) = \{ \hat { \frac {a}{b} } \in \frac { \mathbb {Q} }{ \mathbb {Z} } : a,b \in \mathbb {Z} \ , \ b=p^i \ , \ i \geq 0 \} .

    Prove that if  f: \mathbb {Z} (p^ \infty ) \rightarrow G is a surjective group homomorphism and G is not the trivial group  \{ 1_G \} , then G is isomorphic to  \mathbb {Z} (p^ \infty ) .
    this is one of interesting properties of \mathbb{Z}_{p^{\infty}}. suppose \ker f = K. then since f is surjective, we have \frac{\mathbb{Z}_{p^{\infty}}}{K} \simeq G. since G is nontrivial, we have K \neq \mathbb{Z}_{p^{\infty}}. so what we need to prove is that for any

    subgroup I of \mathbb{Z}_{p^{\infty}} with I \neq \mathbb{Z}_{p^{\infty}} we have \frac{\mathbb{Z}_{p^{\infty}}}{I} \simeq \mathbb{Z}_{p^{\infty}}. we have proved before that we must have I=<\hat{\frac{1}{p^k}}>, for some k \geq 0. now define \varphi: \mathbb{Z}_{p^{\infty}} \longrightarrow \mathbb{Z}_{p^{\infty}} by \varphi(\hat{\frac{a}{p^n}})=\hat{\frac{a}{p^{n-k}}}. see that \varphi is a

    well-defined homomorphism. it's also surjective. finally \ker \varphi = I, which is again obvious because: \hat{\frac{a}{p^n}} \in \ker \varphi \Longleftrightarrow \frac{a}{p^{n-k}} \in \mathbb{Z} \Longleftrightarrow \frac{a}{p^n} \in \frac{1}{p^k} \mathbb{Z} \Longleftrightarrow \hat{\frac{a}{p^n}} \in <\hat{\frac{1}{p^k}}>=I. so we've proved that \varphi is a

    surjective homomorphism and its kernel is I. thus by the first isomorphism theorem for groups we must have \frac{\mathbb{Z}_{p^{\infty}}}{I} \simeq \mathbb{Z}_{p^{\infty}}. \ \Box
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Well-defined, surjective ring homomorphism.
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 16th 2011, 10:02 PM
  2. Homomorphism & Isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 28th 2010, 08:41 AM
  3. Surjective Ring Homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: March 3rd 2010, 12:44 AM
  4. proving surjective homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: November 29th 2009, 08:04 PM
  5. surjective homomorphism
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 6th 2009, 01:27 PM

Search Tags


/mathhelpforum @mathhelpforum