Solving Simple Group Homomorphism Problem: Proving phi(G) is Subgroup of N

Click For Summary

Homework Help Overview

The discussion revolves around a problem in group theory, specifically regarding homomorphisms from a simple group \( G \) to another group \( G' \) and the implications of such mappings when \( G' \) contains a normal subgroup \( N \) of index 2. The original poster seeks clarification on the reasoning behind certain steps in a proof that asserts \( \phi(G) \subset N \).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of the kernel of the homomorphism \( \psi \) being equal to \( G \), questioning how this leads to the conclusion that \( \psi(G) = \{eN\} \). They also discuss the relationship between the mappings and the containment of \( \phi(G) \) within \( N \).

Discussion Status

Several participants have provided insights and analyses regarding the steps in the proof, particularly focusing on the transition from \( \psi(G) = \{eN\} \) to the conclusion that \( \phi(G) \subset N \). There is ongoing exploration of the definitions and properties of the mappings involved, with no explicit consensus reached yet.

Contextual Notes

Participants express uncertainty about the definitions and implications of the mappings, particularly regarding the nature of the identity elements in the context of the homomorphisms and the normal subgroup \( N \). There is also mention of the simplicity of group \( G \) and its implications for the kernel of the homomorphism.

  • #31
Dick said:
If ##A## isn't a group then ##A^{-1}## might be a different set. If you want one more question to test your understanding, then if ##A## is a group, what is ##AA##?
I think ##AA = A^2 = A##, because all elements are closed by operation. Correct?
 
Physics news on Phys.org
  • #32
A.Magnus said:
I think ##AA = A^2 = A##, because all elements are closed by operation. Correct?

Correct. Another way to see it is that ##e## is an element of ##A##. So ##A=eA \subset AA##, and, of course ##AA## can't contain any elements that aren't in ##A## since group multiplication is a closed operation. As you say. I think you've got it.
 

Similar threads

Normal group of order 60 isomorphic to A_5
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Question about "A Group Epimorphism is Surjective"
  • · Replies 13 ·
Replies
13
Views
2K
Solving Simple Group Problem: Subset of Normal Subgroup of Index 2
  • · Replies 2 ·
Replies
2
Views
1K
No group of order 10,000 is simple
  • · Replies 1 ·
Replies
1
Views
2K
Order of image of homomorphism divides order of G
  • · Replies 2 ·
Replies
2
Views
4K
Show injectivity, surjectivity and kernel of groups
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Dfns group action & group, written in terms of the other
  • · Replies 26 ·
Replies
26
Views
1K
Group Homomorphism & Group Order
  • · Replies 2 ·
Replies
2
Views
2K
Prove ##1 + a=s(a)=a+1## for ##a \in \mathbb{N}##
  • · Replies 1 ·
Replies
1
Views
1K
Prove ##a \cdot 1 = a = 1 \cdot a## for ##a \in \mathbb{N}##
  • · Replies 1 ·
Replies
1
Views
2K