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October 25th, 2009, 05:32 PM   #1
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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.
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October 25th, 2009, 06:11 PM   #2
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Re: Small Subgroup Problem

phi(G) is just the whole image of G under phi. So phi(G)={h|exist 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:
Originally Posted by HairOnABiscuit
All I know is that since is a homomorphism we have for all .
And there's the first part. Now inverses.

Quote:
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.
You already know H is a group. You need to show phi(G) is a group. You know the inverse is unique, so what element of H is the inverse of phi(g)?

Cheers.
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October 25th, 2009, 06:29 PM   #3
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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?
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October 26th, 2009, 12:22 AM   #4
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Re: Small Subgroup Problem

Quote:
Originally Posted by HairOnABiscuit
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?
Everything you said is true, but it doesn't really get us anywhere...

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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