Final answer:
There exists a one-to-one correspondence between subgroups H of G containing Ker(f) and subgroups H' of G', where H' is the image of H under the homomorphism f. This is due to homomorphisms preserving group structure and the kernel being included in H, ensuring the subgroup property is respected in G'. The correspondence is given by H mapping to f[H] and H' mapping to the pre-image of H' under f.
Step-by-step explanation:
To show that there is a one to one correspondence between the subgroups H of G such that Ker(f) ⊆ H and the subgroups H' of G', with H' given by f[H] = H', we start by noting that a group homomorphism f from a group G to another group G' maps subgroups of G to subgroups of G'. The kernel of a homomorphism f, Ker(f), is the set of elements in G that map to the identity element in G'.
A subgroup H of G that contains Ker(f) will be mapped by f onto a subgroup H' of G'. This is because homomorphisms preserve the group structure, and since Ker(f) is in H, no additional relations are introduced that could break the subgroup structure in G'. For each such H, its image under f, denoted f[H], is indeed a subgroup of G'. Conversely, for every subgroup H' of G', the pre-image under f, which is f-1[H'], will be a subgroup of G that contains Ker(f), since the identity element's pre-image under any homomorphism includes the kernel.
Thus, we established a correspondence: for every subgroup H of G with Ker(f) ⊆ H, there's a unique H' in G' that is f[H], and vice versa. This one to one correspondence is also known as an isomorphism of the lattice of subgroups.