Cosets

Let H be a subgroup of a group G. If a ∈ G, then the set 
                                       Ha = {ha : h ∈ H}
is called a right coset of H in G determined by a and the set
                                       aH = {ah : h ∈ H}
is called a left coset of H in G determined by a.
If the operation in addition, then the right coset becomes
                                       H + a = {h + a : h ∈ H}
and the left coset becomes
                                      a + H = {a + h : h ∈ H}.

If e is the identity element of group G, then He and eH are right and left cosets of H in G.
Also    He = {he : h ∈ H} = {h : h ∈ H} = H.
            eH = {eh : h ∈ H} = {h : h ∈ H} = H.
∴   If H is a subgroup of a group G, then H itself is a right coset as well as left coset of H in G determined by e.

Remark: When G is an abelian group then there is no distinction between a left coset and a right coset.
i.e., left coset = right coset.
i.e., aH = Ha.

General Properties: The identity is in precisely one left or right coset, namely H itself. Thus H is both a left and right coset of itself.
A coset representative is a representative in the equivalence class sense. A set of representatives of all the cosets is called a transversal. There are other types of equivalence relations in a group, such as conjugacy, that form different classes, which do not have the properties discussed here.

Example: Find the right cosets of the subgroup {1, -1} of the group {1, -1, i, -i} under multiplication.
Sol. G = {1, -1, i, -i} is a group under multiplication.
         H = {1, -1} is a subgroup of G.
The right cosets of H in G are H1, H(-1), Hi, H(-i)
         H · 1  = { 1(1), -1(1) } = { 1, -1 } = H.
         H(-1) = { 1(-1), -1(-1) } = { -1, 1 } = H.
         Hi      = { 1(i), -1(i) }
         H(-i)  = { 1(-i), -1(-i) } = {-i , i}.

Example: Let <S₃ , 0> be the symmetric group on 
{1, 2, 3} and H = {I, (12)} is subgroup of <S₃ , 0>. Find all the left cosets of H in S₃ and also find all the right cosets of H in S₃.
Sol. Let S₃ = {i, (12), (13), (23), (123), (132)} be a symmetric group on {1, 2, 3} and H = {I, (12)}.
        The Left cosets of H in S₃ are
           I · H = H
         (12)H = {(12)I, (12)(12)} = {(12), I} = H
         (13)H = {(13)I, (13)(12)} = {(13), (132)}
         (23)H = {(23)I, (23)(12)} = {(23), (123)}
        (123)H = {(123)I, (123)(12)} = {(123), (23)}
        (132)H = {(132)I, (132)(12)} = {(132), (13)}
∴  The distinct left cosets of H in S₃ are H, (13)H and (23)H.
The right cosets of H in S₃ are
          H · I = H
        H(12)  = {I(12), (12)(12)} = {(12), I} = H
        H(13)  = {I(13), (12)(13)} = {(13), (123)}
        H(23)  = {I(23), (12)(23)} = {(23), (132)}
        H(123) = {I(123), (12)(123)} = {(123), (13)}
        H(132) = {I(132), (12)(132)} = {(132), (23)}
The distinct right cosets of H in S₃ are H, H(13) and H(23).

Example: Let G be the group of integers under addition. Let H be the subgroup of G having even integers. Find the right cosets of H in G.
Sol. G = Group of all integers under addition.
                H = {o, ±2, ±4, ...} is subgroup of G having even integers.
         H + 0 = {h + 0 : h ∈ H} = {h : h ∈ H} = H
         H + 1 = {h + 1 : h ∈ H} = {±1, ±3, ...}
Similarly,
        H + 2 = {o, ±2, ±4, ...}
        H + 3 = {±1, ±3, ...}  and so on.
∴ The only distinct right cosets of H in G are H and H + 1.

Normalizer

Let G be a group, a ∈ G be any element. then the subset 
N(a) = {x ∈ G : xa = ax} is called the Normalizer or Centralizer of a in G.

Theorem: (i) The Normalizer of an element in a group G is a subgroup of G.
                     (ii) Prove that Z(G) ⊆ N(a).

Proof: (i) Let a ∈ G be any element of a group G. Then the set
 N(a) = {x ∈ G : xa = ax} is the normalizer of a in G.
To show that N(a) is a subgroup of G.
Clearly, N(a) ⊆ G.
Since ea = ae
⇒ e ∈ N(a).
Therefore N(a) is a non-empty subset of G.
Let x, y ∈ N(a) be any two elements
∴ xa = ax      and       ya = ay
⇒ ay^-1 = y^-1a
Now a(x y^-1) = (ax)y^-1 = x (ay^-1)
                       = x(y^-1a) = (xy^-1) a
i.e., a(x y^-1) = (xy^-1) a
So, x y^-1 ∈ N(a).
Hence N(a) is a subgroup of G.

(ii) ∀ x ∈ Z(G)
⇒ xg = gx  ∀ g ∈ G
Also a ∈ G
⇒ xa = ax
⇒ x ∈ N(a).
∴ ∀ x ∈ Z(G)
⇒ x ∈ N(a).
⇒Z(G) ⊆ N(a).

Example: Let G = S₃, the symmetric group on three numbers 1, 2 and 3.
Find  (i)  N(a), where a = (12).
          (ii) Z(G), the centre of G.
Sol. Here G = S₃ = {i, (12), (13), (23), (123), (132)}.
(i) N(a) = N(12) = {x ∈ S₃: x(12) = (12)x}.
Since i(12) = (12)i
∴ i ∈ N(12).
Also (12)(12) = i = (12)(12)
∴ (12) ∈ N(12).
But 
(13)(12) = (132) ≠ (123) = (12)(13)
∴ (13) ∉ N(12).
(23)(12) = (123) ≠ (132) = (12)(23)
∴ (23) ∉ N(12).
(123)(12) = (23) ≠ (13) = (12)(123)
∴ (123) ∉ N(12).
(132)(12) = (13) ≠ (23) = (12)(132)
∴ (132) ∉ N(12).
Thus N(12) = {i, (12)}.

(ii) Since Z(G) = Z(S₃) = {x ∈ S₃: xz = zx for all z ∈ S₃}
We know
                Z(G) ⊆ N(a), for each a ∈ G.
∴            Z(G) ⊆ {i, (12)}.
But (12)(13) (123) ≠ (132) = (13)(12)
∴ (12) ∉ Z(G)
Hence Z(G) = (i).       [ix = x = xi, ∀ x  S₃].

Example: Cent(G) = 4 if, and only if, G/Z ≅ Z₂ ⊕ Z₂;
i.e., G modulo its center id isomorphic to the Klein four group.
Proof: If G/Z ≅ Z₂ ⊕ Z₂, then there are non central elements p, q and s of G such that
                 G = Z ∪ Z_p ∪ Z_r ∪ Z_s. It follows that the three proper subgroups of G containing Z are
      P = Z ∪ Z_p, R = Z ∪ Z_r and S = Z ∪ Z_s.
Let x be one of p, r and s and
Let X be the corresponding subgroup. 
Notice that for zx ∈ ZX, G ⊃ C(zx) ⊇ X.
So,  [G : X] = [G : C(zx)][C(zx) : X] = 2 and [G : C(zx)] ≠ 1,
thus C(zx) = X.
Therefore the proper centralizers of G are precisely P, R and S; i.e., Cent(G) = 4.
For the converse, 
it is sufficient to show that [G : Z] = 4 because then
           either G/Z ≅ Z₂ ⊕ Z₂ or G/Z ≅ Z₄.
As G is non-abelian, G/Z can not be cyclic which means the latter case is impossible.
So, suppose Cent(G) = 4. and 
Let P = C(p), R = C(r) and S = C(s) be the three proper centralizers of G.
Since G cannot be written as the union of two proper subgroups and since an element must belong to its centralizer,
we may choose ip, r and s in G - (R ∪ S), G - (P ∪ S) and 
G - (P ∪ R) respectively.
Moreover, atleast one of the proper centralizers, say, P has index two in G. For otherwise
|G| ≤ |P| + |R| + |S| - 2|Z| ≤ |G|/3 + |G|/3 + |G|/3 - 2 < |G|.
Further,
              P ∩ R = P ∩ R ∩ S = Z.
because if x ∈ (P ∩ R) - Z, then
              (i) C(x) ≠ G because x ∉ Z,
             (ii) C(x) ≠ P and C(x) ≠ R because p, r ∈ C(x),
            (iii) C(x) ≠ S because x ∉ S,
which means that Cent(G) must be atleast 5.
Now we can compute |Z| using the fact that for subgroup x and Y of G,
                   |X ∩ Y| = (|X||Y|)/ |XY|
                                 ≥ (|X||Y|)/ |G|
Indeed,
                  |Z| = |P ∩ R|
                        ≥ (|P||R|)/ |G|
                        = |R|/2
since |P| = |G|/2.
But Z ≠ R, So |Z| = |R|/2.
Similarly, |Z| = |S|/2.
Thus |G| = |P| + |R| + |S| -2|Z| = |G|/2 + 2|Z| + 2|Z| -2|Z|
                 = |G|/2 + 2|Z|
implies |G|/2 = 2|Z|; i.e, [G : Z] = 4, as desired.
Groups for which G/Z ≅ Z₂ ⊕ Z₂ are as abelian as an non-abelian group can be in the probabilistic sense also.
To see this, recall that the order of the conjugacy class of an element is the index of the centralizer of that element.
Thus, each conjugacy class of G is of order one or two.
Therefore the number of conjugacy classes in G is
  k = |Z| + (|G| - |Z|)/2 = |G|/4 + 3|G|/8 = 5|G|/8
and prob(G) = 5/8.
This computation also suggests why prob(G) ≤ 5/8 for non-abelian groups: prob(G) is as large as possible when the centre is as large as possible and all of the non-central elements are in conjugacy classes of size two.
This occurs when [G : Z] = 4;
i.e., when G/Z ≅ Z₂ ⊕ Z₂.
The threshold of 'abelianness', as measured by Cent(G) or Prob(G), is occupied by the same class of groups.

Centre of a group

The Centre of a group G is denoted by Z(G) or C(G) or Z and is defined as
             Z = C(G) = Z(G) = {g ∈ G | gx = xg, ∀ x ∈ G}.
The centre of a subgroup of G, which by definition is abelian (i.e., commutative). As a subgroup, it is always normal, and indeed characteristic, but it need not be fully characteristic. The quotient group G/Z(G) is isomorphic to the group of inner automorphisms of G.
A group G is abelian if and only if Z(G) = G. At the other extreme, a group is said to be centerless if Z(G) is trivial, 
i.e., consists only of the identity element.
The elements of the centre are sometimes called central.

Examples:

• The centre of an abelian group G is all of G.
• The centre of a non-abelian simple group is trivial.
• The centre of the dihedral group D_n is trivial when n is odd. When n is even, the centre consists of the identity element together with the 180 degree rotation of the polygon.
• The centre of the quaternion group Q₈ = {1, -1, i, -i, j, -j, k, -k} is {1, -1}.
• The centre of the symmetric group S_n is trivial for n ≥ 3.
• The centre of the alternating group A_n is trivial for n ≥ 4.
• The centre of the orthogonal group O(n, F) is {I_n, -I_n}.
• The centre of the multiplicative group of non-zero quaternions is the multiplicative group of non-zero real numbers.
• The quotient group G/Z(G) is not isomorphic to the quaternion group Q₈.

Theorem: The centre Z(G) of a group G is a subgroup of G.

Proof: Let Z(G) = {g ∈ G | gx = xg, ∀ x ∈ G} be the centre of a group G.
Clearly Z(G) ⊆ G.
Since ex = xe, ∀ x ∈ G
∴ e ∈ Z(G).
∴ Z(G) is a non-empty subset of G.
Let g₁ , g₂ ∈ Z(G) be any two elements, then
           g₁x = xg₁       and   g₂x = xg₂ ,   ∀ x ∈ G
⇒       xg₂^-1 = g₂^-1x.
Now, x(g₁ g₂^-1) = (xg₁) g₂^-1= (g₁x)g₂^-1
                           = g₁ (xg₂^-1) = g₁ (g₂^-1x)
                           = (g₁ g₂^-1)x
i.e.,   x(g₁ g₂^-1)  =  (g₁ g₂^-1)x,      ∀ x ∈ G
So, g₁ g₂^-1 ∈ Z(G),      ∀ g₁ , g₂ ∈ Z(G).
Hence, Z(G) is a subgroup of G.

Theorem: G is abelian group iff Z(G) = G.

Proof: Firstly, let Z(G) = G
i.e., Z (G) = {g ∈ G | gx = xg, ∀ x ∈ G} = G
⇒   xy = yx,         ∀ x, y ∈ G
⇒ G is an abelian group.

Conversely: Let G be abelian.
⇒   xy = yx,         ∀ x, y ∈ G.
To show that Z(G) = G.
Since Z(G) is a subgroup of G
∴ Z(G) ⊆ G
Now, let x ∈ G be any element
∵ G is abelian
∴ xy = yx,         ∀ y ∈ G.
⇒ x ∈ Z(G)
∴ G ⊆ Z(G)
Hence Z(G) = G.

Theorem: Let G be a group, then Z(G) ≤ G. 
Moreover, G/Z(G) ≅ Inn(G) ≤ Aut(G) (where Inn(G) is the set of inner automorphism).
Proof: We know that Φ : G → Aut(G) : g → i_g is a homomorphism.
Note that 
                ker Φ = {g ∈ G : i_g = idG}
                           = {g ∈ G : g^-1hg = h for all h ∈ G}
                           = {g ∈ G : hg = gh for all h ∈ G}
                           = Z(G)
from where the conclusion follows from our earlier characterization of normality. 
the fact that  G/Z(G) ≅ Inn(G) now follows immediately from the First Isomorphism Theorem.