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

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SUMMARY

The discussion revolves around proving that if ##\phi## is a homomorphism from a simple group ##G## to another group ##G'## with a normal subgroup ##N## of index 2, then ##\phi(G) \subset N##. The proof involves analyzing the kernel of the composition of homomorphisms ##\psi = \sigma \circ \phi##, where ##\sigma## is the canonical homomorphism from ##G'## to the quotient group ##G'/N##. It is established that since ##|G'/N|=2## and ##|G| \neq 2##, the kernel of ##\psi## must be the entire group ##G##, leading to the conclusion that ##\psi(G) = \{eN\} = N##, thus confirming that ##\phi(G) \subset N##.

PREREQUISITES
  • Understanding of group theory, specifically simple groups and homomorphisms.
  • Familiarity with normal subgroups and quotient groups.
  • Knowledge of kernel of a homomorphism and its implications.
  • Basic concepts of cosets and their properties in group theory.
NEXT STEPS
  • Study the properties of simple groups and their implications in group theory.
  • Learn about homomorphisms and their kernels in detail.
  • Explore the concept of normal subgroups and their role in quotient groups.
  • Investigate the relationship between cosets and subgroup containment.
USEFUL FOR

Mathematicians, particularly those specializing in abstract algebra, group theorists, and students studying advanced group theory concepts.

  • #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?
 
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  • #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.
 

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