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Thread: Abelian and isomorphic groups, homomorphisms

  1. #1
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    Abelian and isomorphic groups, homomorphisms

    Suppose $\displaystyle A$ and $\displaystyle B$ are Abelian groups and $\displaystyle \varphi :A \rightarrow B$ is a group homomorphism. Suppose that there exists another group homomorphism $\displaystyle \psi : B \rightarrow A$ such that $\displaystyle \psi \circ \varphi = id_A$. Prove that $\displaystyle B$ is isomorphic to $\displaystyle A \oplus M$ for some other group $\displaystyle M$.
    Hint: Set $\displaystyle M=B/A$ (note that $\displaystyle A \rightarrow B$ is injective, so viewing $\displaystyle A$ as a subgroup of $\displaystyle B$ is essentially harmless).


    This was an extra credit question on my last exam. I just want to know what the proof is.
    Last edited by Ackbeet; Mar 1st 2011 at 02:30 AM. Reason: Fixed grammar in title.
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  2. #2
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    Quote Originally Posted by joestevens View Post
    Suppose $\displaystyle A$ and $\displaystyle B$ are Abelian groups and $\displaystyle \varphi :A \rightarrow B$ is a group homomorphism. Suppose that there exists another group homomorphism $\displaystyle \psi : B \rightarrow A$ such that $\displaystyle \psi \circ \varphi = id_A$. Prove that $\displaystyle B$ is isomorphic to $\displaystyle A \oplus M$ for some other group $\displaystyle M$.
    Hint: Set $\displaystyle M=B/A$ (note that $\displaystyle A \rightarrow B$ is injective, so viewing $\displaystyle A$ as a subgroup of $\displaystyle B$ is essentially harmless).


    This was an extra credit question on my last exam. I just want to know what the proof is.

    If you know a little about short exact sequences then the hint is huge:

    as $\displaystyle \psi\circ\phi=1_a$ , then $\displaystyle \psi$ is onto and $\displaystyle \phi$ is 1-1 , and mentioned there, so that

    $\displaystyle 0\rightarrow A\,\,\xrightarrow{\phi}\,\,B\,\,\xrightarrow {\pi}\,\,B/A\,\,\rightarrow 0\,,\,\,\pi=$ the canonical projection,

    is an exact sequence of abelian groups, and we know this sequence splits iff

    there's a homom. $\displaystyle f:B\rightarrow A$ s.t. $\displaystyle f\circ\phi=1_A$ ...well, put now $\displaystyle \phi=f$ and we're done!

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    If you know a little about short exact sequences then the hint is huge:
    I don't know anything about short exact aequences.
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