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Math Help - Show 1-1 and onto between ker(phi) and phi(inverse){(a')}

  1. #1
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    Show 1-1 and onto between ker(phi) and phi(inverse){(a')}

    OK this question was on a hw assignment i had and noone in the class could answer it and the teacher wouldnt give us the answer so its been bugging me. I think it is kinda like Lagrange's Theorem based on the conclusion but have no idea how to prove it i think the notation and inverses are screwing me up.

    Let phi: G---->G' be a homomorphism of groups. Suppose a' = phi(a) element of Range(phi) Let f: ker(phi)--->G be defined as f(k) = ak. Show that f is a 1-1 onto correspondence between ker(phi) and phi(inverse)({a')}. Thus every element of phi(G) has the same sinze inverse image implying that |phi(G)||ker(phi)|=|G|

    The only hint he gave was to show a 1-1 for f and for onto to show f(ker(phi)) = phi(inverse){a'}

    I really dont have any work to show because i am so confused like i said all the inverses and stuff are throwing me off. I know everything in the kernel is mapped to e' and all a' go back to phi(inverse) of a' but that is all i got. Any help is greatly appreciated
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  2. #2
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    Quote Originally Posted by ChrisBickle View Post
    OK this question was on a hw assignment i had and noone in the class could answer it and the teacher wouldnt give us the answer so its been bugging me. I think it is kinda like Lagrange's Theorem based on the conclusion but have no idea how to prove it i think the notation and inverses are screwing me up.

    Let phi: G---->G' be a homomorphism of groups. Suppose a' = phi(a) element of Range(phi) Let f: ker(phi)--->G be defined as f(k) = ak. Show that f is a 1-1 onto correspondence between ker(phi) and phi(inverse)({a')}. Thus every element of phi(G) has the same sinze inverse image implying that |phi(G)||ker(phi)|=|G|

    The only hint he gave was to show a 1-1 for f and for onto to show f(ker(phi)) = phi(inverse){a'}

    I really dont have any work to show because i am so confused like i said all the inverses and stuff are throwing me off. I know everything in the kernel is mapped to e' and all a' go back to phi(inverse) of a' but that is all i got. Any help is greatly appreciated
    let k \in \ker \varphi. then \varphi(f(k))=\varphi(ak)=\varphi(a)\varphi(k)=\va  rphi(a)=a'. so f(k) \in \varphi^{-1}(a'). also if k' \in \ker \varphi, then f(k)=f(k') iff ak=ak' iff k=k'. thus f is one-to-one.

    to prove that f is onto, let b \in \varphi^{-1}(a'). then \varphi(b)=a'=\varphi(a) and hence \varphi (a^{-1}b)=1, which means a^{-1}b \in \ker \varphi. let a^{-1}b=k. then b=ak=f(k) and you're done.
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  3. #3
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    It all seems so easy once you see a solution...but thats math isnt it?
    And where can i learn to use the actual math symbols its quite annoying trying to write everything out in a form that makes sense
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by ChrisBickle View Post
    It all seems so easy once you see a solution...but thats math isnt it?
    And where can i learn to use the actual math symbols its quite annoying trying to write everything out in a form that makes sense
    Here to start.

    Yeah, I'll be honest with you. I took one look at that and was like "I am not spending the time to decipher the math from the plain text". LaTeX sounds like a good idea
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