MHB Correspondence Theorem for Groups .... Another Question ....

[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:
https://www.physicsforums.com/attachments/7995
View attachment 7996
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

 

[steenis]

Q1: if $f:G \longrightarrow H$ is a group-homomorphism. If $B$ is a subgroup of $H$ then $\pi^{-1}B$ is a subgroup of $G$, if $A$ is a subgroup of $G$, then $\pi A$ is a subgroup of $H$. You can prove that easily.
Now if $k \in K=\pi^{-1}1$, then $\pi k =1 \in U$, and so on ...

Q2: In set-theory we have this property: Let $X$ en $Y$ be sets, and $f:X \longrightarrow Y$ is a function, if $B \subset Y$, then $ff^{-1}B \subset B$. If, furhermore, $f$ is surjective, then $ff^{-1}B = B$. Conversely, you can easily prove that if $ff^{-1}B = B$ then $f$ is surjective.
You know that $\pi$ is surjective. You can apply this now to your problem.

 

[Math Amateur]

Q1: if $f:G \longrightarrow H$ is a group-homomorphism. If $B$ is a subgroup of $H$ then $\pi^{-1}B$ is a subgroup of $G$, if $A$ is a subgroup of $G$, then $\pi A$ is a subgroup of $H$. You can prove that easily.
Now if $k \in K=\pi^{-1}1$, then $\pi k =1 \in U$, and so on ...

Q2: In set-theory we have this property: Let $X$ en $Y$ be sets, and $f:X \longrightarrow Y$ is a function, if $B \subset Y$, then $ff^{-1}B \subset B$. If, furhermore, $f$ is surjective, then $ff^{-1}B = B$. Conversely, you can easily prove that if $ff^{-1}B = B$ then $f$ is surjective.
You know that $\pi$ is surjective. You can apply this now to your problem.

Hi Steenis ... great to hear from you again ...

Thanks for your help ...

... just now reflecting on what you have written ...

Peter