Hall retract implies order-conjugate
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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., Hall retract) must also satisfy the second subgroup property (i.e., order-conjugate subgroup)
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This article states and (possibly) proves a fact that involves two finite groups of relatively prime order, requiring the additional datum that at least one of them is solvable. Due to the Feit-Thompson theorem, we know that for two finite groups of relatively prime orders, one of them is solvable. Hence, the additional datum of solvability can be dropped. However, the proof of the Feit-Thompson theorem is considered heavy machinery.
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Contents
Name
This result is often stated as the conjugacy part of the Schur-Zassenhaus theorem.
Statement
Suppose is a finite group and
is a Hall retract of
. In other words,
is a Hall subgroup of
that is also a retract: there exists a normal complement
to
in
. Note that
is thus a normal Hall subgroup. Assume, further, that either
or
is a solvable group.
Then, if there is any subgroup of
of the same order as
,
and
are conjugate subgroups.
Note that the assumption that either or
is a solvable group is superfluous because, as a corollary of the odd-order theorem, given two groups of coprime order, one of them is solvable.
Proof
We reduce the problem to basic cases by induction on the order.
Given: A finite group . A normal
-Hall subgroup
of
, and two
Hall subgroups
of
.
To prove: is conjugate to
.
Proof:
- Reduction to the case that
is minimal normal in
: Suppose
is a nontrivial normal subgroup of
properly contained in
. Let
be the quotient map.
is normal
-Hall in
, and
are
-Hall subgroups. By the inductive hypothesis,
and
are conjugate in
. Suppose
is such that
conjugates
to
. Then,
conjugates
to some
-Hall subgroup inside
. Thus, it suffices to show that any
-Hall subgroup of
is conjugate to
. This again follows by the inductive hypothesis on the order, since
is normal
-Hall in
.
- Reduction to the case that
is trivial; in other words,
has no nontrivial normal
-subgroups: Suppose
is a nontrivial normal
-subgroup of
. Then, by the product formula,
and
are
-subgroups, and since
are Hall, this forces
and
. Consider the quotient map
. Under this quotient map,
is a normal
-Hall subgroup, and
are
-Hall. The inductive hypothesis yields that
and
are conjugate in
. Since
and
both contain the kernel of
, this yields that
and
are conjugate in
.
- Resolution of the case where
is solvable: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
- Resolution of the case where
is solvable: PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
References
- Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 221, Theorem 2.1(ii), Chapter 6 (Solvable and pi-solvable groups), Section 6.2 (The Schur-Zassenhaus theorem), More info