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

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In the discussion about proving that the image of a homomorphism from a simple group G to a group G' is contained within a normal subgroup N of index 2, participants analyze the implications of the kernel of the composition of homomorphisms. They establish that since the kernel of the composition is the entire group G, the image under the composition must be the trivial coset, leading to the conclusion that the image of the homomorphism must be contained in N. Clarifications are sought regarding the logical steps that connect the kernel's properties to the conclusion about the subgroup. The conversation emphasizes the importance of understanding how mappings and cosets interact in group theory. Ultimately, the consensus is that if the image of the homomorphism maps to the trivial coset, it implies that the image itself is a subset of the normal subgroup N.
  • #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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