Indecomposable Lattice of Subgroups does not necessarily form Totally Ordered Set
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 $\struct {G, \circ}$ be an indecomposable group.
Then it is not necessarily the case that $\struct {\mathbb G, \subseteq}$ is totally ordered.
Proof
First we note that from Set of Subgroups forms Complete Lattice, $\struct {\mathbb G, \subseteq}$ is indeed a complete lattice.
Let $D_4$ denote the dihedral group of order $8$, also known as the symmetry group of the square.
From Internal Group Direct Product Examples: $D_4$, it is seen that $D_4$ is indecomposable group.
But from the Hasse diagram, we see:
it is seen that $\struct {\mathbb G, \subseteq}$, where $\mathbb G$ denotes the set of subsets of $D_4$, is not totally ordered.
Hence the result.
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.16 \ \text {(b)}$