MHB The Correspondence Theorem for Groups .... Rotman, Proposition 1.82 ....

[Math Amateur]

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:
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In the above proof by Rotman we read the following:

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

$$\pi (a) = \pi (s) $$

$$\Longrightarrow aK = sK$$

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

$$\Longrightarrow s^{-1} a = k$$

$$\Longrightarrow as^{-1} = k$$

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

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

 

[GJA]

Hi, Peter.

The reasoning in your EDIT section copied below is correct, including the step where you ask if $a=sk$ is legitimate (it is because $K$ is a (sub)group and so must contain the identity, which implies $a\in aK$).

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

My thoughts are as follows:

$$\pi (a) = \pi (s) $$

$$\Longrightarrow aK = sK$$

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

$$\Longrightarrow s^{-1} a = k$$

$$\Longrightarrow as^{-1} = k$$

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

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

Indeed, your reasoning is expressed a little different from Rotman's (with yours on the level of elements from $G$ throughout and with Rotman's given on the level of cosets and group homomorphisms), but it is equally valid and yields the same conclusion, well done!

Can someone please explain exactly how/why $$\pi (a) = \pi (s)$$ implies that $$as^{-1} \in \text{ ker } \pi = K$$ ... ...

For any group homomorphism, say $h:G_{1}\rightarrow G_{2},$ $h(g^{-1})=\left[h(g)\right]^{-1}$ (a short but good exercise to try for yourself if you're not already familiar with this fact). Hence, $\pi(a)=\pi(s)\Longrightarrow\left[\pi(a)\right]^{-1}=\left[\pi(s)\right]^{-1}.$

Now we compute: $\pi(as^{-1})=\pi(a)\pi(s^{-1})=\pi(a)\left[\pi(s)\right]^{-1}=\pi(a)\left[\pi(a)\right]^{-1}=\text{Identity in}~G/K=K.$

 

[Math Amateur]

Hi, Peter.

The reasoning in your EDIT section copied below is correct, including the step where you ask if $a=sk$ is legitimate (it is because $K$ is a (sub)group and so must contain the identity, which implies $a\in aK$).
Indeed, your reasoning is expressed a little different from Rotman's (with yours on the level of elements from $G$ throughout and with Rotman's given on the level of cosets and group homomorphisms), but it is equally valid and yields the same conclusion, well done!
For any group homomorphism, say $h:G_{1}\rightarrow G_{2},$ $h(g^{-1})=\left[h(g)\right]^{-1}$ (a short but good exercise to try for yourself if you're not already familiar with this fact). Hence, $\pi(a)=\pi(s)\Longrightarrow\left[\pi(a)\right]^{-1}=\left[\pi(s)\right]^{-1}.$

Now we compute: $\pi(as^{-1})=\pi(a)\pi(s^{-1})=\pi(a)\left[\pi(s)\right]^{-1}=\pi(a)\left[\pi(a)\right]^{-1}=\text{Identity in}~G/K=K.$

Thanks GJA ...

... most helpful !

Peter