For any set we denote by
the group of bijective maps from
to
Problem. Let be any map which satisfies the following two conditions:
and
for all
Let
be a set with
Prove that if
is an abelian subgroup of
then
Solution. The proof is by induction on If
then
Let
be a set with
and suppose that the claim is true for any set of size
Let
be an abelian subgroup of
Clearly
defines an action of
on
Let
be the orbits corresponding to this action and consider two cases.
Case 1. : Fix an element
Then
Suppose that
for some
and let
Then
for some
Thus, since
is abelian, we have
Hence for all
and thus
So the stabilizer of
is trivial and therefore, by the orbit-stabilizer theorem,
Case 2. : Let
Clearly
and, since
we have
for all
For every
and
let
the restriction of
to
and put
Then and
is an abelian subgroup of
Thus, by the induction hypothesis
for all Now, define
by
for all
It is obvious that
is one-to-one and so
Remark. The map defined by
for all
satisfies both conditions in the above Problem. So if
and if
is an abelian subgroup of
then