Lemma on inverse flower arrangement and product of subgroups

History
This lemma appeared in Feit and Thompson's paper proving the odd-order theorem.

Definition
Suppose $$A, B, C$$ are three subgroups of a group $$G$$ such that:

$$A \subseteq BC, \qquad B \subseteq CA, \qquad C \subseteq AB$$

where $$BC$$ denotes the fact about::product of subgroups $$B$$ and $$C$$, then $$AB = BA = BC = CB = CA = AC$$, and in particular, the products $$AB, BC, CA$$ are subgroups. Thus, $$A,B$$ permute, as do $$B,C$$ and $$C,A$$.

Proof
Given: Subgroups $$A,B,C$$ of a group $$G$$ such that $$A \subseteq BC$$, $$B \subseteq CA$$, and $$C \subseteq AB$$.

To prove: $$AB = BA = BC = CB = CA = AC$$.

Proof:


 * 1) $$(BC)^{-1} = CB$$, $$(CA)^{-1} = AC$$ and $$(AB)^{-1} = BA$$: We first observe that $$(BC)^{-1}$$, i.e., the set of inverses of elements of $$BC$$, is given as the set $$\{ c^{-1}b^{-1} \mid c \in C, b \in B \}$$. Since the inverse map is bijective on a subgroup, we get $$BC^{-1} = CB$$.
 * 2) $$A \subseteq CB$$, $$B \subseteq AC$$ and $$C \subseteq BA$$: Since $$A \subseteq BC$$, $$A^{-1} \subseteq (BC)^{-1}$$, yielding $$A \subseteq CB$$. Similar arguments show that $$B \subseteq AC$$ and $$C \subseteq BA$$.
 * 3) $$AB = BA$$, $$BC = CB$$, $$CA = AC$$: We have $$AB \subseteq (BC)(CA) = BCA \subseteq B(BA)A = BA$$. Thus, $$AB \subseteq BA$$. Similarly, $$BA \subseteq (AC)(CB) = ACB \subseteq A(AB)B = AB$$. Thus, $$AB = BA$$.
 * 4) $$AB = BA = BC = CB = CA = AC$$: We have $$AB  = BA \subseteq (CA)A = CA$$. Similarly, $$CA \subseteq BC$$ and $$BC \subseteq AB$$. The chain of inclusions shows that all the members are equal.