Lemma on inverse flower arrangement and product of subgroups

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


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.


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.


  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.


Journal references