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Group Homomorphism

Let ϕ:A→B\phi: A \rightarrow B be a homomorphism of groups. Show that ϕ\phi induces an order-preserving one-to-one correspondence between the set of all subgroups of A that contain Ker(ϕ)Ker(\phi) and the set of all subgroups of B that are contained in Im(ϕ)Im(\phi) .

So I have a query, what does the question mean when it says "order-preserving"? Does it mean the order of the set that contains all the subgroups, or is it referring to the actual order of the subgroups themselves?

Secondly, is my proof sound?

Let Ker(ϕ)≤N≤AKer(\phi) \leq N \leq A, and also let M≤Im(ϕ)≤BM \leq Im(\phi) \leq B.

Also since ϕ(A)=Im(ϕ)\phi(A) = Im(\phi) and since N≤AN \leq A, ϕ(N)≤Im(ϕ)\phi(N) \leq Im(\phi), thus assume that ϕ(N)=M\phi(N) = M, as homomorphisms of groups preserve subgroups.

Also let X:={N≤A:Ker(ϕ)≤N}X := \{N \leq A : Ker(\phi) \leq N\}, and let Y:={M≤B:M≤Im(ϕ)}Y := \{M \leq B : M \leq Im(\phi)\}.

We will first prove that ϕ:X→Y\phi: X \rightarrow Y is injective

Let ϕ(N1)=ϕ(N2)\phi(N_1) = \phi(N_2), and let ϕ(N):={x∈Im(ϕ):ϕ−1(x)∈N}\phi(N) := \{x \in Im(\phi) : \phi^{-1}(x) \in N\}. Now if x∈ϕ(N1)x \in \phi(N_1), it is clear that ϕ−1(x)∈N1\phi^{-1}(x) \in N_1, via the definition and since every x∈Im(ϕ)x \in Im(\phi) has an inverse ϕ−1(x)\phi^{-1}(x). Also since x∈ϕ(N1)x \in \phi(N_1), it is also in ϕ(N2)\phi(N_2), thus ϕ−1(x)∈N2\phi^{-1}(x) \in N_2, and N1⊆N2N_1 \subseteq N_2.

We can prove the converse like so, and thus N1=N2N_1 = N_2, and ϕ:X→Y\phi: X \rightarrow Y is injective.

Surjectivity for ϕ:X→Y\phi: X \rightarrow Y is easy, as all subsets M≤Im(ϕ)M \leq Im(\phi), has an inverse for every element as it is a subgroup of the image of ϕ\phi and every subgroup of the form ϕ(N)=M≤Im(ϕ)\phi(N) = M \leq Im(\phi), has 1∈M1 \in M, as MM is a subgroup. Thus Ker(ϕ)≤ϕ−1(M)=NKer(\phi) \leq \phi^{-1}(M) = N, and thus the homomorphism ϕ:X→Y\phi: X \rightarrow Y is surjective.

Hence X≅YX \cong Y.


Now if this is asking whether XX and YY, are of the same size, then we have shown it by showing that X≅YX \cong Y, but if it's asking for the order of the actual subgroups in XX and YY, then there is more to do.
(edited 9 years ago)
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