Subgroup lattice and quotient lattice of finite abelian group are isomorphic

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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. 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 \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.