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....
Thanks in advance
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....
Thanks in advance
This is worded weird. Maybe you mean given $\displaystyle K\leqslant G$ we have that $\displaystyle \phi(K)\leqslant \phi(G)$?
If that is the case:
Clearly $\displaystyle e\in K\implies \phi(e)\in \phi(K)$
Also, if $\displaystyle a,b\in\phi(K)$ then $\displaystyle a=\phi(c),b=\phi(d)$ for some $\displaystyle c,d\in K$. But, since $\displaystyle K$ is a group we have that $\displaystyle cd\in K\implies \phi(cd)=\phi(c)\phi(d)=ab\in \phi(K)$.
Lastly, if $\displaystyle a\in \phi(K)$ then $\displaystyle a=\phi(c)$ for some $\displaystyle c\in K$ and so $\displaystyle c^{-1}\in K$ since it is in a group. Thus, $\displaystyle \phi(c^{-1})=\phi(c)^{-1}=a^{-1}\in\phi(K)$
"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?