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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:

View attachment 7993

View attachment 7994

In the above proof by Rotman we read the following:

" ... ... For the reverse inclusion let \(\displaystyle a \in \pi^{-1} \pi (S)\), so that \(\displaystyle \pi (a) = \pi (s)\) for some \(\displaystyle s \in S\). It follows that \(\displaystyle as^{-1} \in \text{ ker } \pi = K\) ... ... "Can someone please explain exactly how/why \(\displaystyle \pi (a) = \pi (s)\) implies that \(\displaystyle as^{-1} \in \text{ ker } \pi = K\) ... ... Peter

===========================================================================================***EDIT***

Just had some thoughts ... BUT ... unfortunately my logic does not seem to line up with Rotman's logic ...

My thoughts are as follows:

\(\displaystyle \pi (a) = \pi (s) \)

\(\displaystyle \Longrightarrow aK = sK\)

\(\displaystyle \Longrightarrow a = sk\) for some \(\displaystyle k \in K\) since \(\displaystyle a\) must belong to \(\displaystyle sK\) ... ... (is this a legitimate step ...)

\(\displaystyle \Longrightarrow s^{-1} a = k\)

\(\displaystyle \Longrightarrow as^{-1} = k\)

\(\displaystyle \Longrightarrow as^{-1} \in \text{ker } \pi = K\) Is that correct?

BUT note ... logic is different from Rotman's set of steps ... ... what is Rotman's logic ...?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:

View attachment 7993

View attachment 7994

In the above proof by Rotman we read the following:

" ... ... For the reverse inclusion let \(\displaystyle a \in \pi^{-1} \pi (S)\), so that \(\displaystyle \pi (a) = \pi (s)\) for some \(\displaystyle s \in S\). It follows that \(\displaystyle as^{-1} \in \text{ ker } \pi = K\) ... ... "Can someone please explain exactly how/why \(\displaystyle \pi (a) = \pi (s)\) implies that \(\displaystyle as^{-1} \in \text{ ker } \pi = K\) ... ... Peter

===========================================================================================***EDIT***

Just had some thoughts ... BUT ... unfortunately my logic does not seem to line up with Rotman's logic ...

My thoughts are as follows:

\(\displaystyle \pi (a) = \pi (s) \)

\(\displaystyle \Longrightarrow aK = sK\)

\(\displaystyle \Longrightarrow a = sk\) for some \(\displaystyle k \in K\) since \(\displaystyle a\) must belong to \(\displaystyle sK\) ... ... (is this a legitimate step ...)

\(\displaystyle \Longrightarrow s^{-1} a = k\)

\(\displaystyle \Longrightarrow as^{-1} = k\)

\(\displaystyle \Longrightarrow as^{-1} \in \text{ker } \pi = K\) Is that correct?

BUT note ... logic is different from Rotman's set of steps ... ... what is Rotman's logic ...?Peter

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