# Generalization of the fundamental theorem of cyclic groups

Let $$G$$ be a finite group then the fundamental theorem of cyclic groups can be formulated as follows:
Theorem: $$G$$ is cyclic iff it admits no two different subgroups with the same order.
proof: see here p44 together with Lagrange theorem. $$\square$$

But there is an other characterization of the finite cyclic groups, using the lattice theory:
Theorem: $$G$$ is cyclic iff its subgroups lattice $$\mathcal{L}(G)$$ is distributive.
proof: see here theorem 4 p267. $$\square$$

So we get immediately the following statement for a finite group $$G$$:
If $$G$$ admits no two different subgroups with the same order, then its subgroups lattice is distributive.

We ask about a generalization of this statement for an inclusion $$(H \subset G)$$ of finite groups:

Question: Is it true that if $$(H \subset G)$$ admits no two different intermediate subgroups $$H \subset K \subset G$$ with the same order, then its intermediate subgroups lattice $$\mathcal{L}(H \subset G)$$ is distributive?

Remark: It is checked (by GAP) for $$[G:H] \le 31$$.
The converse is false because the inclusion $$(S_2 \times S_2 \subset S_3 \times S_3)$$ is a counter-example (see here).

• What does "at most one subgroup of fixed order" mean? It is not true that if a group G has only one subgroup of some given order, then G must be cyclic. Perhaps what is wanted is that there are no two different subgroups with the same order. May 7, 2015 at 23:37
• @MartyIsaacs: you're right, your last sentence is what I had in mind. I've edited that. May 8, 2015 at 2:56
• See the sequel: Generalization of the fundamental theorem of cyclic groups 2 May 10, 2015 at 4:55

Say $G=S_n$ and $H$ is a Young subgroup with three orbits, no two of which have the same size and no two of which have sizes summing to $n/2$. Then the only subgroups between $H$ and $G$ should be three Young subgroups with two orbits, no one of which contains any other and no two of which have the same order.
• I've checked the smallest example $(S_1 \times S_2 \times S_4 \subset S_7)$ of index $105$, and it runs as you state. To prove your statement in general you use the result that $(S_n \times S_m \subset S_{n+m})$ is a maximal inclusion (I believe it's proved somewhere on MO). May 8, 2015 at 4:57