# Thread: Subsets of an Image

1. ## Subsets of an Image

Hey,

I'm having a bit of trouble with this problem:

Let phi: (G,*)->(H, . ) be a homomorphism.

If S is a subset of im(Phi), prove that the group S generates is a subgroup of im(Phi)

I don't know where to go with this....

2. Originally Posted by bakerconspiracy
Hey,

I'm having a bit of trouble with this problem:

Let phi: (G,*)->(H, . ) be a homomorphism.

If S is a subset of im(Phi), prove that the group S generates is a subgroup of im(Phi)

I don't know where to go with this....

This is worded weird. Maybe you mean given $K\leqslant G$ we have that $\phi(K)\leqslant \phi(G)$?

If that is the case:

Clearly $e\in K\implies \phi(e)\in \phi(K)$

Also, if $a,b\in\phi(K)$ then $a=\phi(c),b=\phi(d)$ for some $c,d\in K$. But, since $K$ is a group we have that $cd\in K\implies \phi(cd)=\phi(c)\phi(d)=ab\in \phi(K)$.

Lastly, if $a\in \phi(K)$ then $a=\phi(c)$ for some $c\in K$ and so $c^{-1}\in K$ since it is in a group. Thus, $\phi(c^{-1})=\phi(c)^{-1}=a^{-1}\in\phi(K)$

3. I really appreciate the help, but that wasn't quite what I was asking.

What I was asking is, if S is a subset (not subgroup) of G, then prove that <S> is a subgroup of G.

<S> meaning the group generated by S.

4. Originally Posted by bakerconspiracy
I really appreciate the help, but that wasn't quite what I was asking.

What I was asking is, if S is a subset (not subgroup) of G, then prove that <S> is a subgroup of G.

<S> meaning the group generated by S.

But now I'm confused because you're talking about $S\subseteq G$ when earlier $S\subseteq\phi(G)$?

Also, do you consider $\left\langle S\right\rangle$ to the intersection of all groups containing $S$?

5. "Also, do you consider to the intersection of all groups containing ?"

yes, the intersection of all subgroups containing S.

"But now I'm confused because you're talking about when earlier ? "

Is there a difference though? I mean what if I defined L = the image of Phi(G). Then S is a subset of L, prove that the group generated by the set S is a subgroup of L. Same question just reduced right?

6. Well thank you anyways