# Lattice of Subgroups forming Totally Ordered Set is Indecomposable

## Theorem

Let $\struct {G, \circ}$ be a Group.

Let $\mathbb G$ be the set of subgroups of $G$.

Let $\struct {\mathbb G, \subseteq}$ be the complete lattice formed by the subset ordering on $\mathbb G$.

Let $\struct {\mathbb G, \subseteq}$ be totally ordered.

Then $\struct {G, \circ}$ is an indecomposable group.

## Proof

First we note that from Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ is indeed a complete lattice.

As postulated, let $\struct {\mathbb G, \subseteq}$ be totally ordered.

Aiming for a contradiction, suppose there exists a decomposition of $\struct {G, \circ}$.

Then:

$\exists H, K \in \mathbb G: H \cap K = \set e$

where $\struct {H, \circ}$ and $\struct {K, \circ}$ are non-trivial normal subgroups of $\struct {G, \circ}$.

But then neither $H \subseteq K$ nor $K \subseteq H$.

That is, $\struct {\mathbb G, \subseteq}$ is not totally ordered.

Hence the result by Proof by Contradiction.

$\blacksquare$