# 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.

## 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.