Every group of given order is a permutable complement for symmetric groups
Statement
Symbolic statement
Let be a finite group of order
. Then the following are true:
- Via the regular representation, we can view
as a subgroup of the symmetric group on
elements (this is the idea behind Cayley's theorem).
- Under this representation,
is a permutable complement to the symmetric group on any
of the elements
In particular this means that every isomorphism class of a group of order occurs as a permutable complement of
in
. Since there can exist non-isomorphic groups of the same order, this shows that the isomorphism type of the permutable complement of
in
is not uniquely determined.
Analysis
In fact, define the following equivalence relation on groups:
Two groups and
are equivalent if there is a group
with embeddings of
and
in
, such that both
and
have a common permutable complement.
A priori, it is clear that two finite groups can be equivalent only if they have the same order. The above result shows that the converse is also true: any two finite groups that have the same order are equivalent.
Related facts
- Complement to normal subgroup is isomorphic to quotient: This rules out a variant of the result where the other subgroup is normal.
- Complements to abelian normal subgroup are automorphic
- Complements to normal subgroup need not be automorphic
- Retract not implies normal complements are isomorphic
- Direct product is cancellative for finite groups
- Semidirect product is not left-cancellative for finite groups
Facts used
- Cayley's theorem
- Group equals product of transitive subgroup and isotropy of a point (get more related information at proving product of subgroups)
Proof
Given: A group of order
.
To prove: is isomorphic to a permutable complement to the symmetric group on
(which is the isotropy of the point
, in the symmetric group on
.
Proof:
- By Cayley's theorem,
embeds inside
-- the group of permutations on the underlying set of
, via the action by left multiplication.
- Under this embedding, every non-identity element of
acts as a fixed point-free permutation. In particular, the intersection of
with the isotropy subgroup of any point is trivial.
- Also, the action of
on itself by left multiplication is transitive. Thus, by fact (2),
is the product of the subgroup
and the isotropy of any point.
- Combining steps (2) and (3), we see that
is a permutable complement to the isotropy of any point in
.
- A set-theoretic bijection between
and the set
thus shows that
is isomorphic to a permutable complement to the symmetric group on
in
.