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October 25th, 2009, 05:32 PM  #1 
Member Joined: Oct 2009 Posts: 85 Thanks: 0  Small Subgroup Problem
G and H are groups. Let be a homomorphism. Show that is a subgroup of . Ok so I am kind of stumped since I am new to abstract algebra and my professor assumes way too much for me so bear with me Anyways, first of all is equivalent to saying ? I am assuming or this problem would be pointless. All I know is that since is a homomorphism we have for all . To show for all , I have no clue, maybe do I need to create another function and do something with the composition? Or does should I have the kernel involved in showing this axiom? For any help I am extremely grateful for, have a great night. 
October 25th, 2009, 06:11 PM  #2  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: Small Subgroup Problem
phi(G) is just the whole image of G under phi. So phi(G)={hexist g, phi(g)=h}. You know phi(G) is a subset of H, so just show that it is closed under the operation and taking inverses. Quote:
Quote:
Cheers.  
October 25th, 2009, 06:29 PM  #3 
Member Joined: Oct 2009 Posts: 85 Thanks: 0  Re: Small Subgroup Problem
I think would be the element of H that is the inverse of . So we know H is a group, and and since there exists an we have that . Is this what I should be saying (in the previous sentence) to prove the last axiom?

October 26th, 2009, 12:22 AM  #4  
Senior Member Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3  Re: Small Subgroup Problem Quote:
I think you are a little confused on what's going on. Maybe I can clarify. We have two groups, G and H. There is a homomorphism (structure preserving map) . We want to show that the set is in fact a group under the operation of H. If it makes things more comfortable, call F. So we have a subset F of H, and we want to show that this subset F: Contains the identity of H. Is closed under multiplication Is closed taking inverses. We have the identity: phi(1)=phi(1*1)=phi(1)*phi(1). The only element in a H satisfying x=x*x is the identity, so phi(1_G)=1_H We have closure under multiplication by definition of homomorphism. Does this make sense to you? If not, I can explain in more detail. So we need to show that for any g, phi(g)^1 is in F=phi(G). Obviously the inverse element is in H, H is a group, but we need to show that there is some element f such that phi(f)phi(g) = 1_H. Can you think of what f should be? If so, then you know phi(g)^1 = phi(f).  

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