My Math Forum Small Subgroup Problem

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 October 25th, 2009, 04:32 PM #1 Member   Joined: Oct 2009 Posts: 85 Thanks: 0 Small Subgroup Problem G and H are groups. Let $\phi : G \mapsto H$ be a homomorphism. Show that $\phi (G)$ is a subgroup of $H$. 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 $\phi (G)$ equivalent to saying $\phi (g)= h$ ? I am assuming $\phi(G) \not= \emptyset$ or this problem would be pointless. All I know is that since $\phi$ is a homomorphism we have $\phi : (g_z g_2)= \phi(g_1) \phi(g_2)$ for all $g_1, g_2 \in G$. To show for all $a,b \in H$, $ab^{-1} \in H$ I have no clue, maybe do I need to create another function $\psi: H \mapsto G$ 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, 05:11 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 $\phi$ is a homomorphism we have $\phi : (g_z g_2)= \phi(g_1) \phi(g_2)$ for all $g_1, g_2 \in G$.
And there's the first part. Now inverses.

Quote:
 To show for all $a,b \in H$, $ab^{-1} \in H$ I have no clue, maybe do I need to create another function $\psi: H \mapsto G$ 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.

 October 25th, 2009, 05:29 PM #3 Member   Joined: Oct 2009 Posts: 85 Thanks: 0 Re: Small Subgroup Problem I think $h^{-1}$ would be the element of H that is the inverse of $\phi (g)$. So we know H is a group, and $g, h \in H$ and since there exists an $h^{-1} \in H$ we have that $gh^{-1} \in H$. Is this what I should be saying (in the previous sentence) to prove the last axiom?
October 25th, 2009, 11:22 PM   #4
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Re: Small Subgroup Problem

Quote:
 Originally Posted by HairOnABiscuit I think $h^{-1}$ would be the element of H that is the inverse of $\phi (g)$. So we know H is a group, and $g, h \in H$ and since there exists an $h^{-1} \in H$ we have that $gh^{-1} \in H$. 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) $\phi:G\rightarrow H$.
We want to show that the set $\{h\in H|\exist g\in G, \phi(g)=h\}$ is in fact a group under the operation of H.
If it makes things more comfortable, call $\phi(G)$ 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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