Lemma on inverse flower arrangement and product of subgroups
From Groupprops
History
This lemma appeared in Feit and Thompson's paper proving the odd-order theorem.
Definition
Suppose are three subgroups of a group such that:
where denotes the Product of subgroups (?) and , then , and in particular, the products are subgroups. Thus, permute, as do and .
Proof
Given: Subgroups of a group such that , , and .
To prove: .
Proof:
- , and : We first observe that , i.e., the set of inverses of elements of , is given as the set . Since the inverse map is bijective on a subgroup, we get .
- , and : Since , , yielding . Similar arguments show that and .
- , , : We have . Thus, . Similarly, . Thus, .
- : We have . Similarly, and . 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 page}^{More info}