Cayley's Theorem
If
is a group of order
then
is isomorphic to a subgroup of the permutation group![]()
Example: Let
be the Klein group
of order 4,
with every element self inverse, so of order 2.
Example: We can define
then
is an isomorphism from
onto ![]()
Proof of Cayley's Theorem
Let the
elements of
be
For an element
we find the permutation of the elements of
formed by multiplying each element on the left by
Each
must be equal to
for some
if and only if![]()
We must show
-
is a uniquely defined element of
for each
-
is one to one. -
has the morphism property,
By construction
maps the set
to itself.
is one to one since if
and
then
and
and 2 is proved. Hence
is a one to one map of the set
to itself , so is a permutation of this set and
so 1 is satisfied.
To show 3, we show that
and
have the same effect on each of the elements in![]()
Assume
and
so that![]()
By the definition of
if and only if
and
if and only if
by associativity in
but
if and only if
for each
hence
and
have the same effect on every element of
so
and Cayley's Theorem is proved.