Subgroup lattice and quotient lattice of finite abelian group are isomorphic

Statement
Let $$G$$ be a finite abelian group and let $$L(G)$$ be the lattice of subgroups (i.e., the set of subgroups with the partial order of containment) of $$G$$.

Statement in terms of anti-automorphism of the subgroup lattice
There is a set map $$\alpha:L(G) \to L(G)$$ such that:


 * 1) $$\! H \le K \iff \alpha(K) \le \alpha(H)$$.
 * 2) $$\alpha(H)$$ is isomorphic to $$G/H$$.
 * 3) $$\alpha^2$$ is the identity map.

The map is not canonical, but it is canonical up to pre-composition by automorphisms.

Note that (2) in particular shows that $$H$$ is the kernel of an endomorphism of $$G$$ with image the subgroup $$\alpha(H)$$ (specifically, take the quotient map to $$G/H$$ and compose with the isomorphism to $$\alpha(H)$$. Thus, it shows that every subgroup of a finite abelian group is an endomorphism kernel. Similarly, it shows that every subgroup of a finite abelian group is an  endomorphism image.

Facts used

 * 1) uses::Finite abelian group is isomorphic to its Pontryagin dual
 * 2) uses::Pontryagin duality theorem

Proof
Denote by $$\hat{G}$$ the Pontryagin dual of $$G$$, i.e., the group of homomorphisms from $$G$$ to the circle group under pointwise multiplication.

By fact (1), $$G$$ is isomorphic to $$\hat{G}$$. Let $$\sigma: \hat{G} \to G$$ be one such isomorphism.

Then, we define:

$$\alpha(H) = \sigma(\beta(H))$$

where:

$$\beta(H) = \left \{ f \in \hat{G} \mid H \subseteq \operatorname{ker}(f) \right \}$$.

We now check the three conditions:


 * 1) $$\! H \le K \iff \alpha(K) \le \alpha(H)$$: If $$H \le K$$, then any element of $$\hat{G}$$ that has $$K$$ in its kernel also has $$H$$ in its kernel. Thus, $$\beta(H) \le \beta(K)$$. Applying $$\sigma$$ preserves the containment.
 * 2) $$\alpha(H)$$ is isomorphic to $$G/H$$: $$\beta(H)$$ is the group of homomorphisms from $$G$$ to the circle group with $$H$$ in the kernel. These are naturally identified with the group of homomorphisms from $$G/H$$ to the circle group, which is $$\hat{G/H}$$. By fact (1), this is isomorphic to $$G/H$$. Thus, $$\alpha(H) = \sigma(\beta(H))$$ is isomorphic to $$G/H$$.
 * 3) $$\alpha^2$$ is an automorphism: First, consider the map $$\beta:L(G) \to L(\hat{G})$$. Define the map $$\hat{\beta}$$ as the analogue of $$\beta$$ from $$\hat{G}$$ to $$\hat{\hat{G}}$$. Denote by $$l_\sigma$$ the map induced by $$\sigma$$ on $$L(\hat{G}) \to L(G)$$. Then, $$\alpha = l_\sigma \circ \beta = \hat{\beta} \circ l_{\sigma}^{-1}$$. Then, $$\alpha^2 = \hat{\beta} \circ \beta$$, which is the identity map.