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. $\!H\leq K\iff \alpha (K)\leq \alpha (H)$.
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. Finite abelian group is isomorphic to its Pontryagin dual
2. 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. $\!H\leq K\iff \alpha (K)\leq \alpha (H)$: If $H\leq K$, then any element of ${\hat {G}}$ that has $K$ in its kernel also has $H$ in its kernel. Thus, $\beta (H)\leq \beta (K)$. 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.