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Let
be a group and
a subgroup of
. For
belonging to
, the subset
of
is called a left coset
of
in
and the subset
of
is called a
right coset of
in
.
Example 5.12.1
If

and

as in
Example
5.8.2, then

(

) and

(

). Does

?
Notation: The set of all left cosets is written
and the set of all right cosets of
in
is
denoted
.
These two sets don't in general inherit a group
structure from G but they are useful none-the-less.
(
is a group with the ``obvious" multiplication
if and only if
is a
``normal" subgroup of
- we will define ``normal"
below.)
Let
be a finite set and let
be a finite subgroup of
the symmetric group
. The elements of
permute
. With this permutation in mind, we say that
acts on
. For
and
, let
denote
. We call the set of images
the orbit of
under
.
We call
the stabilizer of
in
.
As an example of their usefulness, we have
the following relationship between the orbits and the cosets of
the stabilizers.
Proposition 5.12.2
Let

be a finite group acting on a
set

. Then
for all

belonging to

.
proof:
The map
defines a function
.
The interested reader can easily check that
this function is a bijection
since it is both an injection and a surjection.
Corollary 5.12.3
Let

be a finite group acting on itself by
conjugation. Let

denote a complete
set of representatives of the conjugacy classes

in

and let

, i.e., the subset of

of those elements which are not central. Then
for all

belonging to

.
In particular,
proof:
In the first displayed equation:
The first equation is Theorem 5.11.5.
The second equality follows from the above proposition.
By taking cardinalities, the second displayed equation is a consequence
of the first.
proof:
The argument is by induction on
.
The result is trivial if
(since then no prime divides
).
Suppose
and let
be a prime dividing
.
By the induction hypothesis, we assume that the result is true for all
subgroups
of
with
.
Suppose
is not central.
Then its centralizer
is a proper subgroup of
.
If
then the result follows from the induction hypothesis.
If
does not divide
(for all non-central
) then since
, by Proposition 5.12.2,
it follows that
. By Corollary
5.12.3 above, we must have
.
If
is a proper subgroup of
then we are done by the
induction hypothesis.
Thus we may assume
is abelian.
If
is cyclic then we leave it to the reader to show that
contains an element of order
. We shall assume that
is not cyclic. Let
be a proper subgroup of
of
maximal order (this exists since
is not cyclic)
and let
. Then
(else
) and
(else
would not be
maximal). Thus
either divides
or
.
In either case, the result follows from the induction
hypothesis.
Theorem 5.12.5 (Lagrange)
: If

is
a finite group and

a subgroup then
Corollary 5.12.6
If

are
as above then the order of

divides the
order of

.
Now we prove the Theorem.
proof:
Let
be the set of left cosets of
in
and let
act on
by left multiplication.
Apply the previous lemma with
.
Definition 5.12.7
: Let

be a subgroup of

and let

be a left coset of

in

. We call an element

of

a
coset representative
of

if

. A
complete set of coset representatives
is a subset of

,

, such that
without repetition (i.e., all the

are disjoint).
Exercise 5.12.8
Let

be the
group of symmetries of the square. Using the
notation above, compute

and

.
Exercise 5.12.9
For

, define

if and only if

and

belong
to the same left coset of

in

.
(a) Show that

is an equivalence relation.
(b) Show that the left cosets of

in

partition

.
Exercise 5.12.10
If

is finite, show

.
Exercise 5.12.11
Let

and

(i.e.

is the set of multiples of 3).
(a) Prove that

.
(b) What is

?
Exercise 5.12.12
Suppose

and

is nonabelian and

is an
element of order

.
(a) What is

?
(b) What is

and what is

?
Exercise 5.12.13
Let

be the symmetry group of a cube. By this,
we mean the following: If

is the set
of vertices of a cube, e.g.

, consider
the subgroup of the full permutation group of

that
arises from physically possible rotations of space.
(This is the analog of the symmetry group of
the square,

, discussed earlier.)
(a)

can be considered to be a subgroup
of

. Find

,

and

.
(b) What is

?
Exercise 5.12.14
If

is a left coset of

in

and

is an element of

,
show that

is also a left coset of

in

.
Exercise 5.12.15
Let

,
the symmetric group on 3 letters, and let

, in the notation of §
5.2
above.
(a) Compute

using Lagrange's Theorem.
(b) Explicitly write down all the cosets of

in

.
Next: Functions between two groups
Up: An introduction to groups
Previous: Conjugation
  Contents
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David Joyner
2002-08-23