Let G be a group and K a normal subgroup in G. Let f : G → G/K be the canonical epimorphism given by x → xK and L a subgroup of G/K. Then
(1) There exists a subgroup H of G. H contains K, and L=H/K. Further, H is normal in G, if and only if L is normal in G/K. Furthermore, if G is finite, prove [G : H] = [G/K : H/K]=[G/K :L] and |H|=|K|·|L|.
(2) Suppose that H1, H2 are two subgroups of G and H1, H2 contain K. If H1/K=H2/K, then H1=H2.
I find this one quite different from the original form of Correspondence Principle, which is a lot easier to understand and to proof. I know I have to use the isomorphism theorems to prove this one, but can anyone give a sketch or a hint? Thanks!
Ideas for demonstration:
We have $\;K\lhd G\;$ and the quotient $\;G/K\;$. Say $\;\overline L\le G/K\;$ , and define
$$L:=\left\{g\in G\;:\;\;gK\in\overline L\right\}$$
(1) Prove that $\;L\;$ is a subgroup of G containing $\;K\;$
(2) Prove that $\;\overline L\lhd G/K\iff L\lhd G\;$
(3) Prove that $\;[G/K : \overline L]=[G:L]\;$
Now, show that if $\; T\le G\;$ contains $\;K\;$ , then
$$\overline T:=\left\{ gK\in G/K\;:\;\;g\in T\right\}$$
is a subgroup of $\;G/K\;$ .
Finally, show that both correspondences above: $\;\overline L\to L\;,\;\;T\to\overline T\;$ are inverse to each other, and this fact is precisely what allows us to write any subgroup of $\;G/K\;$ in the form $\;L/K\;$ , with $\;K\le L\le G\;$