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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:

How/why is ##\pi^{-1} (U)## is a subgroup of ##G## containing ##K##? And further, how does ##\pi^{-1} (U) = \pi^{-1} ( \{ 1 \} )## ... ... ?

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