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 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.

References

Journal references

• Solvability of groups of odd order by Walter Feit and John Griggs Thompson, Pacific Journal of Mathematics, Volume 13, Page 775 - 1029(Year 1963): This 255-page long paper gives a proof that odd-order implies solvable: any odd-order group (i.e., any finite group whose order is odd) is a solvable group.^{Project Euclid pageMore info}