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I am reading the book: "Advanced Modern Algebra" (Second Edition) by Joseph J. Rotman ...
I am currently focused on Chapter 1: Groups I ...
I need help with an aspect of the proof of Proposition 1.82 (Correspondence Theorem) ...
Proposition 1.82 reads as follows:
In the above proof by Rotman we read the following:
" ... ... To see that ##\Phi## is surjective, let ##U## be a subgroup of ##G/K##. Now ##\pi^{-1} (U)## is a subgroup of ##G## containing ##K = \pi^{-1} ( \{ 1 \} )##, and ##\pi ( \pi^{-1} (U) ) = U## ... ... "My questions on the above are as follows:
Question 1
How/why is ##\pi^{-1} (U)## is a subgroup of ##G## containing ##K##? And further, how does ##\pi^{-1} (U) = \pi^{-1} ( \{ 1 \} )## ... ... ?
Question 2
How/why exactly do we get ##\pi ( \pi^{-1} (U) ) = U##? Further, how does this demonstrate that ##\Phi## is surjective?
Help will be much appreciated ...
Peter
I am currently focused on Chapter 1: Groups I ...
I need help with an aspect of the proof of Proposition 1.82 (Correspondence Theorem) ...
Proposition 1.82 reads as follows:
In the above proof by Rotman we read the following:
" ... ... To see that ##\Phi## is surjective, let ##U## be a subgroup of ##G/K##. Now ##\pi^{-1} (U)## is a subgroup of ##G## containing ##K = \pi^{-1} ( \{ 1 \} )##, and ##\pi ( \pi^{-1} (U) ) = U## ... ... "My questions on the above are as follows:
Question 1
How/why is ##\pi^{-1} (U)## is a subgroup of ##G## containing ##K##? And further, how does ##\pi^{-1} (U) = \pi^{-1} ( \{ 1 \} )## ... ... ?
Question 2
How/why exactly do we get ##\pi ( \pi^{-1} (U) ) = U##? Further, how does this demonstrate that ##\Phi## is surjective?
Help will be much appreciated ...
Peter