In summary: H) = \rho(g_i)##So ##\ker(\phi) = \{g_i H \in G/H : \phi(g_i H) = e'\}## So ##\ker(\phi) = \{g_i H \in G/H : \rho(g_i)=e'\}## which is only true if ##g_i=e##So ##\phi## is an injection.These are all good points, but the reasoning could be a bit clearer.For 2, suppose ##\phi(g
O.K, let's give it another try: the statement I will use is similar to the one you had.
Here is a sketch . Fill in the details and ask if you need to:
Let h: G-->G' be a homomorphism with kernel K. Then G/K is isomorphic to h(G) ( in case h is not onto G').
Remember an isomorphism is a homomorphism that is both 1-1 and onto. We already have that h is a
homeomorphism--given-- and the map is (tautologically) onto h(G) . So we only need to show, as JBunii
pointed out :
1)h^: G/K -->h(G) is well-defined; since this is a map defined on equivalent classes, we need to show that
members of the same euivalence class have the same image.
(How do we define f^ ?)
2) h^: G/K -->h(G) is 1-1.
First, we need to define the map h^:
h^: G/K -->G' takes [ a] in G/K into h^(a).
Note that a~a' means h(a)=h(a')
Now, we need to see if it is true that a~a' implies h^(a)=h^(a'). So we have,
in the Abelian case (try the non-Abelian), : h(a)-h(a')= h(a-a')=0 .
Then show h^([a]) =h^([a'])
Then h([a]):= h(a) and h([a']):=h(a') , and h(a)-h(a')=...
We can get 2) out of the way quickly: use that G/K collapses exactly those elements with the same image in
G' . Let me do the Abelian case, and you do the non-Abelian one:
Assume g~g' in G/K . Then h(g)-h(g') is in K, which means h(g)-h(g')=...