5
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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).

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  • $\begingroup$ 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. $\endgroup$ – Marty Isaacs May 7 '15 at 23:37
  • $\begingroup$ @MartyIsaacs: you're right, your last sentence is what I had in mind. I've edited that. $\endgroup$ – Sebastien Palcoux May 8 '15 at 2:56
  • $\begingroup$ See the sequel: Generalization of the fundamental theorem of cyclic groups 2 $\endgroup$ – Sebastien Palcoux May 10 '15 at 4:55
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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.

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  • 1
    $\begingroup$ 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). $\endgroup$ – Sebastien Palcoux May 8 '15 at 4:57

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