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Abstract algebra: elements of fiber writable as

  1. Nov 26, 2013 #1

    For a homomorphism [itex]\varphi[/itex], I'm trying to show that elements of a fiber, say the fiber above [itex]a[/itex], [itex]X_a[/itex], are writable as a given element of [itex]X_a[/itex] times an element of the kernel [itex]K[/itex]. So, if [itex]a\in X_a[/itex] and [itex]b\in X_a[/itex], then [itex]\exists k\in K[/itex] such that [itex]b=ak[/itex].

    I want to do this without using the theorem that [itex]\{[/itex]left cosets of [itex]K[/itex] in [itex]G\} =G/K[/itex] - in fact, one of my motivations for looking for this is that I want a different proof of this theorem then the ones that I have seen.

    Does anyone know of a way to do this?

    Thanks for any help that you can give.

    -HJ Farnsworth
  2. jcsd
  3. Nov 27, 2013 #2


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    By fiber I assume you mean preimage, i.e. "the fiber above ##a##" means ##\varphi^{-1}(a)##.

    Choose ##x,y \in \varphi^{-1}(a)##. We can always write ##x = y(y^{-1}x)##, so it suffices to show that ##y^{-1}x \in \ker \varphi##.

    But this is easy: ##\varphi(y^{-1}x) = \varphi(y^{-1})\varphi(x) = \varphi(y)^{-1} \varphi(x) = a^{-1}a = 1##.
  4. Nov 27, 2013 #3
    Hi jbunniii,

    That is indeed what I mean by fiber. Thanks for the help, that was exactly the kind of thing I was looking for!

    -HJ Farnsworth
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