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.


 * Proof by Counterexample

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:


 * Hasse-Diagram-Symmetry-Groups-of-Square.png

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.

Also see

 * Lattice of Subgroups forming Totally Ordered Set is Indecomposable: the converse