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 pageMore info