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Problem of the Week #24 - November 12th, 2012

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Chris L T521

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Jan 26, 2012
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Here's this week's problem.

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Problem: Let $G$ be a group and let $f:G\rightarrow H$ be a group homomorphism. If $U\leq G$, show that $f^{-1}(f(U))=U\ker(f)$.

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Note: In this case $f^{-1}$ refers to the pre-image of $f$, not it's inverse!

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's question was correctly answered by Deveno. His answer can be found below.

we will show the two sets are equal by showing they contain each other.

suppose that \(\displaystyle g \in Uker( f)\).

then \(\displaystyle g = uk\) for some \(\displaystyle u \in U, k \in ker( f)\).

hence \(\displaystyle f(g) = f(uk) = f(u)f(k) = f(u)e_H = f(u)\) (because f is a homomorphism), so \(\displaystyle f(g) \in f(U)\),

so \(\displaystyle g \in f^{-1}(f(U))\).

on the other hand, suppose \(\displaystyle g \in f^{-1}(f(U))\).

this means that \(\displaystyle f(g) \in f(U)\), so \(\displaystyle f(g) = f(u)\) for some \(\displaystyle u \in U\).

hence \(\displaystyle (f(u))^{-1}f(g) = e_H\), and because f is a homomorphism:

\(\displaystyle (f(u))^{-1}f(g) = f(u^{-1})f(g) = f(u^{-1}g)\),

so \(\displaystyle f(u^{-1}g) = e \implies u^{-1}g \in ker(f)\),

so \(\displaystyle u^{-1}g = k\), for some \(\displaystyle k \in ker(f)\),

thus \(\displaystyle g = uk \in Uker(f)\), QED.
 
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