Why does ker(ψ)=G imply ψ(G)={eN} in this group homomorphism proof?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
31 replies · 5K views
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
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.