Cosets
Let H be a subset of the group G. (Usually, H is
chosen to be a subgroup of G.) If a ∈ G, then we
denote by aH the subset {ah | h ∈ H }, the left coset
of H containing a. Similarly, Ha = {ha | h ∈ H} is
the right coset of H containing a. In both
€
situations, we call a the coset representative.
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Examples:
• The (left or right) cosets of the subgroup H = 11
in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are
1H = {1, 11} = 11H,
7H = {7, 17} = 17H,
13H = {13, 23} = 23H,
19H = {19, 29} = 29H.
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• The (left or right) cosets of the subgroup 4Z in
the additive group Z are
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€
€
4Z = 0 + 4Z = ±4 + 4Z = ±8 + 4Z = ±12 + 4Z = L ,
L = –3 + 4Z = 1 + 4Z = 5 + 4Z = 9 + 4Z = L ,
L = –2 + 4Z = 2 + 4Z = 6 + 4Z = 10 + 4Z = L ,
L = –1 + 4Z = 3 + 4Z = 7 + 4Z = 11 + 4Z = L .
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• The left cosets of the subgroup {R0 ,H } in D4 are
R0 { R0 ,H } = {R0 ,H } = H{ R0 ,H },
R90 { R0 ,H } = {R90 , D′} = D′{R0 ,H},
€ ,H},
R180 { R0 ,H } = {R€180 ,V} = V {R
0
R270 { R0 ,H } = {R270 , D} = D{R0 ,H}.
There are a number of patterns evidenced in these
examples that are true more generally.
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Theorem Let H be a subgroup of G and suppose
a,b ∈ G . Then the left cosets of H in G satisfy the
following properties (with correspondingly similar
results for the right cosets of H):
1. a ∈ aH ,
2. aH = H ⇔ a ∈ H ,
3. either aH = bH or aH ∩bH = ∅,
4. aH = bH ⇔ a −1b ∈ H ,
5. |aH| = |bH|,
6. aH = Ha €
⇔ H = aHa−1 ,
7. aH
€ is a subgroup of G ⇔ a ∈ H .
Proof
€
1. Trivial.
€
2. (⇒) a = ae ∈ aH = H . (⇐) a ∈ H ⇒ a −1 ∈ H and
for every h ∈ H , ah ∈ H by closure in H, whence
aH ⊆ H ; and h = a(a −1 h) ∈ aH , whence aH ⊇ H .
Thus, aH = H.
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€
3. Suppose aH ∩bH ≠ ∅. Then there is some x that
lies in both cosets; that is, there are elements
y,z ∈ H for which x = ay = bz. Thus,
−1
−1
aH
=
(bzy
)H
=
b(zy
H ) = bH (by #2 above, since
€ −1
zy ∈ H ).
−1
−1
4.
aH
=
bH
⇔
H
=
a
bH
⇔
a
b ∈ H by #2 above.
€
5. The function
from the set aH to the set bH given
€
€
by ah a bh is clearly onto. It is one-to-one since
€
bh1 = bh2 ⇒ h1 = h2 ⇒ ah1 = ah2 . So |aH| =
€
|bH|.
−1
−1
−1
6.
aH
=
Ha
⇔
aHa
=
(aH
)a
=
(Ha)a
= He = H .
€
7. aH ≤ G ⇒ e ∈ aH ⇒ aH ∩ H ≠ ∅ ⇒ aH = H ⇒
€
a ∈ H (using #3 and #2 above). Conversely, by
#2, a ∈ H ⇒ aH = H ≤ G. //
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Properties #1, #3, and #5 above combine to assert
€
that G is partitioned by the left cosets of H into
€
subsets of the same size. In fact, we can define a
relation on the elements of G by a ~ b when a and b
lie in the same coset, i.e., aH = bH. This relation is
reflexive (#1), symmetric, and transitive (by #3), so
it is an equivalence relation on G. (See pp. 17-20.)
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Examples:
Let Π be a plane through the origin in R3. Then Π
is an additive subgroup of R3. The cosets of Π
represent all the planes in R3 parallel to Π.
The cosets of SL(2, R) partition GL(2, R) into
€
subsets consisting of all matrices with a given
€
determinant.
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Lagrange’s Theorem
One of the most versatile results in all of group
theory is
Lagrange’s Theorem Let G be a finite group and
suppose that H < G. Then |H| divides |G|.
Moreover, the number of distinct left (or right)
cosets of H in G, called the index of H in G and
denoted |G:H|, equals |G|/|H|.
Proof As we mentioned in the comment following
the previous theorem, properties #1, #3 and #5
imply that G is the disjoint union of the distinct
cosets of H, all of which have the same number of
elements. Since H is one of these cosets, it follows
that G is the disjoint union of |G:H| cosets, all of
which have size |H|. That is, |G |=|G:H | ⋅ | H |, from
which the final claim of the theorem follows. //
Corollary The order of any element of a finite
€
group divides the order of the group.
Proof For any element a of a group, a = a .
Corollary Groups of prime order must be cyclic.
Proof Let a be an element of €
a group G of prime
order p which is not the identity element. Then
|a| divides p but |a|≠ 1. So |a| = p and G = a . //
€
Corollary If G is a finite group and a ∈ G, then
a |G| = e.
Proof |a| must divide |G|.
€ //
€
Corollary [Fermat’s Little Theorem (F lT)] If p
is a prime number and a is any integer, then
a p mod p = amod p.
€
Proof If a is a multiple of p, then both amod p and
a p mod p equal 0. If a is not a multiple of p, then
both these numbers are elements of the
multiplicative group U(p), which
€ has order p – 1.
p−1
Therefore, a mod p = 1. Multiplication by a
yields a p mod p = amod p. //
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€
€
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Lagrange’s Theorem says that a group of order n
can only have subgroups whose orders are divisors
of n. Note however that the converse of this
theorem need not be true:
Proposition |A4| = 12, but A4 has no subgroup of
order 6.
€
€
Proof Suppose H were a subgroup of order 6.
Then since | A4 :H |= 2, we can conclude that H has
exactly two distinct cosets in A4. Let α represent
any one of the eight elements of order 3 in A4
(α 5€
,α6 ,K,α 12 in the Cayley table of A4 on p. 105).
Then at least one pair of the three
cosets
€
2
H,αH,α H must coincide and regardless which
pair it is, we must then have that α ∈ H . But then
H must contain all eight elements of order 3, which
is impossible. //
€
Lagrange’s Theorem gives us a powerful tool for
investigating the nature of finite groups. For
example:
Theorem A group whose order is twice an odd
prime p must be isomorphic to either Z2p or Dp.
€
Proof If p is an odd prime and G has order 2p, we
have two cases to consider:
(1) G has an element of order 2p. Then clearly
G ≈Z2p.
(2) G has no element of order 2p. By
Lagrange’s Theorem, any element different from e
must have order 2 or order p. If every element had
order 2, then every element would be its own
inverse and we could write xy = (xy)−1 = y−1 x −1 = yx ,
showing that G is Abelian; moreover, this would
allow any two such elements to generate a
subgroup {e,x, y,xy} of order 4. But 4 does not
divide 2p, so this is€not possible. Therefore, not
every element has order 2 and there must be some
element a of order p.
€ Now let b be an element not in a . Then the
coset b a is distinct from a , and since G: a = 2,
these are the only two distinct cosets. So b 2 a
must conicide with one of them.
€
Since b 2 a = b a
€ implies b a = a ,€which is false,€we must have
b 2 a = a ⇒ b 2 ∈ a . So | b 2| divides
a = p. It
€
€
€
€
€
€
follows that | b 2|= 1, for if | b 2|= p, the fact that
|b|≠ 2p together with b 2 ≤ b < G means that
|b|= p, and since b 2 ∈ a ,
€
€
p +1
p +1
b=€
b p+1 = (b 2 ) 2 = (a k ) 2 ∈ a ,
€
which we know to be false. Thus | b 2|= 1, and b
must have order 2.
€
Next, ab ∉ a (since ab ∈ a ⇒ b ∈ a ), so the
same argument we used in€the last paragraph will
show that |ab|= 2. Also, ab = (ab) −1 = b−1 a −1 = ba −1 .
Therefore,
€
€
G = {e,a, a 2 ,…,a p−1 ,b,ba,ba 2 ,…,ba p−1 }
€
and the group operation is determined by the
relations between a and b. In particular,
€
a k b = a k−1 (ab) = a k−1 (ba −1 ) = a k−1 ba −1
= a k− 2 (ab)a −1 = a k− 2 (ba −1 )a −1 = a k− 2ba −2
=L
= ba −k
€
so the product of any two elements in G is
completely determined, as follows:
a k a l = a k+l ;
(ba k )a l = ba k+l ;
a k (ba l ) = ba l−k ;
(ba k )(ba l ) = a l−k .
In other words, every group of order 2p without an
element of order 2p must have this structure: all
€ are isomorphic. As D is a group of
such groups
p
this form, all such groups are isomorphic to Dp. //
Corollary S3 ≈ D3. //
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Orbits and Stabilizers
We saw from Cayley’s Theorem how every group is
a group of permutations on some set. Suppose that
G is a group of permutations of the objects in the
set S. Then for each s ∈ S, the set
orbG(s) = {ϕ (s) ∈ S | ϕ ∈ G},
€
called the orbit of s under G, is the subset of
elements in S to which s is moved under the action
€
of G. We also define
stabG(s) = {ϕ ∈ G | ϕ (s) = s},
called the stabilizer of s in G, to be the subset of
G consisting of those elements that fix s. As this
€
(nonempty) subset of G is clearly closed under
composition and under the taking of inverses, it is a
subgroup of G.
Theorem Let G be a finite group of permutations
of the set S. Then for any s ∈ S,
G = orb G (s) ⋅ stab G (s) .
€
Proof By Lagrange’s Theorem, |G|/|stabG(s)| is
the number of left cosets of stabG(s) in G. So we are
€
done if we can establish a bijection between the left
cosets of stabG(s) and the orbit of s under G. To this
end, define the function f as
ϕ stabG(s) a ϕ (s).
€
€
Before we continue, we must first show that this is
well-defined; that is, the coset ϕ stabG(s) may be
€
€
equal to the coset ψ stabG(s), so we must ensure
that ϕ (s) = ψ (s). But this is always true, for
ϕ stabG(s) = ψ stabG(s) €
⇒ ϕ −1ψ ∈stabG(s), which
−1
means that
€ ϕ ψ (s) = s, or ϕ (s) = ψ (s). The function
f is one-to-one because we can reverse this last
argument,
and
€
€ it is onto because if t is in the orbit
of s, then t = ϕ (s) for some ϕ ∈ G, whence f maps the
€ ϕ stab (s) to €
coset
t = ϕ (s). //
G
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