QUOTIENT GROUP

Quotient Group:
Let G be a group.

Let N be a normal subgroup.

Then the left coset space G/N is a group, where the group product is defined as:

         (aN)(bN) = (ab)N

G/N is called the quotient group of G by N.

It is proven to be a group in Quotient group is Group.

A quotient group is also known as a factor group.
In other words, We can say 
If H is a normal subgroup of a group G, then the group G/H of all the right cosets of H in G under the composition
         (Ha)(Hb) = Hab is called a quotient group or a factor group.

Note: If the composition in G/H is addition, then the composition in G/H is defined by
           (H + a) + (H + b) = H + (a + b).

Remark: If H is a normal subgroup of a finite group G, then G/H form a group of order O(G)/O(H).

Theorem: If H is a subgroup of an abelian group G, then the group G/H of all right cosets of H in G forms an abelian group under the composition defined by Ha.Hb = Hab.
Proof: If H is a subgroup of an abelian group G, then H is normal subgroup of G.
∴        G/H forms a quotient group.
Let Ha, Hb ∈ G/H      so that a,b ∈ G.
(Ha)(Hb) = Hab = Hba,  since G is abelian.   ∴   ab = ba
                  = (Hb)(Ha).
Hence G/H is an abelian group.

Converse: The converse of the above result is not true that is, the quotient group may be abelian even if G may not be abelian.

Theorem: Every quotient group of a cyclic group is cyclic.
Proof: Let G = <a> be a cyclic group generated by a.
∴ G is an abelian group.
∴ Each subgroup of G is normal subgroup.
Let h be any subgroup of G.
∴ H is a normal subgroup of G.
So G/H form a quotient group.
We prove that G/H is a cyclic group generated by Ha.
Let Hx ∈ G/H be arbitrary element, where x ∈ G.
∴ But G = <a>
∴ x = aⁿ for some integer n.
∴ Hx = Haⁿ = H a . a ... a (n times)
                       = Ha. Ha ... Ha (n times) 
                       = (Ha)ⁿ 
∴ Hx = (Ha)ⁿ , ∀ Hx ∈ G/H.
∴ G/H is a cyclic group generated by Ha.
So, each quotient group of a cyclic group is cyclic.

Converse: The converse of the above theorem may not be true.
i.e, quotient group may be cyclic even if the group may not be cyclic.

Example: If H be a normal subgroup of a group G and [G : H] = m, then show that for any x ∈ G, xⁿ ∈ H.
Sol. Since H is a normal subgroup of group G such that 
[G : H] = m.
∴ O(G/H) = m.
∴ ∀ xH ∈ G/H,    where x ∈ G, we have
      (xH)^m = H    [∵ If O(G) = n then aⁿ = e, ∀ a ∈ G]
      x^mH = H
⇒  x^m ∈ H.
Thus ∀ x ∈ G,   we have  x^m ∈ H.

THEOREMS ON NORMAL SUBGROUP

Theorem: A subgroup H of a group G is a normal subgroup of G iff ghg^-1 ∈ H for every h ∈ H, g ∈ G.
Proof: Firstly, let H be a normal subgroup of G.
∴               gH = Hg,     ∀ g ∈ G
Let h ∈ H and g ∈ G be any element. Then
                  gh ∈ gH = Hg
⇒              gh ∈ Hg
⇒              gh = h₁g  for some h₁ ∈ H
⇒              ghg^-1 = h₁ ∈ H
⇒              ghg^-1 ∈ H

Conversely: Let H be a subgroup of G, such that
                  ghg^-1 ∈ H,      ∀ h ∈ H, g ∈ G.
We show that H is a normal subgroup
i.e.,            aH = Ha,       ∀ a ∈ G.
Let a ∈ G be any element. Then by given hypothesis
                  aha^-1 ∈ H,      ∀ h ∈ H
Let ah ∈ aH be any element. Then
                  ah = (aha^-1)a ∈ Ha
⇒              ah ∈ Ha
∴              aH ⊆ Ha.                                                                 ... (1)
Again, Let b = a^-1 be any element of G.
Then by given hypothesis bhb^-1 ∈ H.
But            bhb^-1 = a^-1h(a^-1)^-1 = a^-1ha ∈ H.
Let ha ∈ Ha be any element. Then
                  ha = (aa^-1)ha = a (a^-1ha) ∈ aH
⇒              ha ∈ aH
∴              Ha ⊆ aH.                                                                ... (2)
From (1) & (2) we get
                  aH = Ha,      ∀ a ∈ G.
Hence, H is a normal subgroup of G.

Theorem: Let H be a subgroup of G. Then the following statements are equivalent
      (i) ghg^-1 ∈ H,           ∀ h ∈ H, g ∈ G.
     (ii) ghg^-1 = H,           ∀ g ∈ G.
    (iii) gH = Hg,             ∀ g ∈ G.

Proof: (i) ⇒(ii) Since ghg^-1 ∈ H,      ∀ h ∈ H, g ∈ G.
Let ghg^-1 = h₁   for some h₁ ∈ H
⇒       ghg^-1 = H,           ∀ g ∈ G.

(ii) ⇒ (iii) Let ghg^-1 = H,           ∀ g ∈ G.
⇒       (ghg^-1)g = Hg
⇒       gH(gg^-1) = Hg
⇒               gHe = Hg
⇒                 gH = Hg.                                                   (∵ He = H)

(iii) ⇒ (i) Let  gH = Hg,       ∀ g ∈ G.
⇒         gh = h₁g   for some h,h₁ ∈ H
⇒         ghg^-1 = h₁ ∈ H
⇒         ghg^-1 ∈ H,    ∀ h ∈ H, g ∈ G.
Hence (i) ⇒ (ii) ⇒ (iii) ⇒ (i).
Hence the given statements are equivalent.

Theorem: The centre Z(G) of a group G is a normal subgroup of G.
Proof: We know Z(G) = {g ∈ G ; xg = gx   ∀ x ∈ G}.
Clearly          Z(G) ⊆ G.
Since     ex = xe,    ∀ x ∈ G⇒ e ∈ Z(G).
∴  Z(G) is a non-empty subset of G.
Let a, b ∈ Z(G) be any two elements, then
              ax = xa, ∀ x ∈ G            and 
              bx = xb, ∀ x ∈ G 
⇒          xb^-1 = b^-1x
Now      x(ab^-1) = (xa)b^-1 = (ax)b^-1
                                 = a(xb^-1) = a(b^-1x)
                           = (ab^-1)x
⇒          x(ab^-1) = (ab^-1)x      ∀ x ∈ G.
∴          ab^-1 ∈ Z(G)     ∀ a, b ∈ Z(G).
So, Z(G) is a subgroup of G.
Now, we show that Z(G) is a normal subgroup of G.
Let     h ∈ Z(G)        and     g ∈ G , then
              ghg^-1 = (gh)g^-1 = (hg)g^-1 = h(gg^-1)
                        = he = h ∈ Z(G)
∴           ghg^-1 ∈ Z(G),          ∀ g ∈ G,  h ∈ Z(G)
Hence, Z(G) is a normal subgroup of G.

NORMAL SUBGROUPS (OR INVARIANT SUBGROUPS OR SELF CONJUGATE SUBGROUPS)

A subgroup H of a group G is called a normal subgroup of G if every left coset of H in G is equal to the corresponding right coset of H in G.
i.e.,           aH = Ha,       ∀ a ∈ G.

If the composition defined on G be addition, then H will be a normal subgroup of G iff.
                a + H = H + a,        ∀ a ∈ G.

In general, if H is a subgroup of a group G, then the left coset aH of H in G may not be equal to the corresponding right coset Ha. In this section, our aim is to study a particular class of subgroups H for which each left coset of H in G is equal to the corresponding right coset of H in G. We call such subgroups as normal subgroups.

Remark: (i) When G is an abelian group. Then every subgroup H of G is a normal subgroup, for 
                           aH = Ha,        ∀ a ∈ G.
(ii) The subgroups {e} and G of any group G are always normal subgroups of G. These are called trivial normal subgroups.
(iii) If H is a normal subgroup of G, then we write it as 
H Δ G.

Properties:

• Normality is preserved upon surjective homomorphism, and is also preserved upon taking inverse images.
• Normality is preserved on taking direct products.
• Every subgroup of index 2 is normal.
• A normal subgroup of a central factor is normal. In particular, a normal subgroup of direct factor is normal.

                                  Examples

Example: Let G = S₃, the symmetric group on three numbers 1, 2, 3.Show that the subgroup 
                           H = {i, (123), (132)}
is a normal subgroup of G but the subgroup
                           K = {i, (12)}
is not a normal subgroup of G.
Sol. We know that Ha = H = aH   if a ∈ H.
Since i, (123), (132) ∈ H
∴ iH = Hi, (123)H = H(123), (132)H = H(132).
Now       (12)H = {(12)i, (12)(123),(12)(132)}
                          = {(12)(23)(13)} .
and        H(12) = {i(12), (123)(12), (132)(12)}
                          = {(12)(13)(23)}.
∴           (12)H = H(12).
Again    (23)H = {(23)i, (23)(123),(23)(132)}
                          = {(23)(13)(12)} .
and        H(23) = {i(23), (123)(23), (132)(23)}
                          = {(23)(12)(13)}.
∴           (23)H = H(23).
Also       (13)H = {(13)i, (13)(123),(13)(132)}
                          = {(13)(23)(12)} .
and        H(13) = {i(13), (123)(13), (132)(13)}
                          = {(13)(12)(23)}.
∴           (13)H = H(13).
Thus      xH = Hx,       ∀ x ∈ S₃.
∴  H is a normal subgroup of S₃.
But        (13)K = {(13)i, (13)(12)}
                         = {(13)(132)}
and        K(13) = {i(13), (12)(13)}
                         = {(13)(123)}
Clearly   (13)K ≠ K(13).
Hence K is not a normal subgroup of G.

Example: If H is a subgroup of G of index 2 in G. Then H is normal subgroup of G.
Sol. Let H be a subgroup of G such that [G : H] = 2.
∴  The number of distinct left (or right) cosets of H in G is 2.
To show that H is a normal subgroup of G.
It is sufficient to prove that xH = Hx,    ∀ x ∈ G.
Let x ∈ G be arbitrary element of G.

Case I. When  x ∈ H.
Since x ∈ H           So,    xH = H = Hx
Hence  xH = Hx.

Case II. When x ∉ H.
∴          xH ≠ H         and   Hx ≠ H.
Also     [G : H] = 2.
∴         H ∪ xH = G = H ∪ Hx
⇒         xH = Hx.
Combining the two cases, we find that
            xH = Hx      ∀ x ∈ G.
∴   H is a normal subgroup of G.

SOME IMPORTANT THEOREMS ON CYCLIC GROUP

Theorem: Let G be a finite group of order n. If G contains an element of order n, then G must be cyclic.
Proof: Let a ∈ G such that O(a) = n.
Let H = {a^r : r ∈ 1 } be a subgroup of G.
But O(a) = n
⇒ H = {e, a, a², ... , a^n-1} = <a>
i.e., H is a cyclic subgroup of G generated by a.
Also O(H) = O(G)
⇒ G = H = <a>.
i.e., G is a cyclic group.

Theorem: Every cyclic group is abelian.
Proof: Consider a cyclic group G  generated by a.
i.e.,  G = <a>.
Let x, y ∈ G be arbitrary element.
∴ x = aⁿ and y = a^m for some integers n and m.
Then xy = aⁿa^m = a^n+m =a^m+n = a^maⁿ = yx.
∴ G is an abelian group.

Theorem: Prove that a subgroup of a cyclic group is cyclic.
Proof: Let G = <a> be a cyclic group generated by a.
Let H be a subgroup of G.
If H = G, then H = <a> is a cyclic group generated by a.
If H = {e}, then H = <e> is a cyclic group generated by e.
So, let H ≠ G, {e} i.e.,  h is a proper subgroup of G.
∴ ∃ an element x ∈ H such that x ≠ e.
Now x ∈ H
⇒ x ∈ G.
⇒ x = a^r for some non-zero integer r
∴ x ∈ H
⇒ x^-1 ∈ H, since H is a subgroup of G.
⇒ a^-r ∈ H
∴ a^r , a^-r ∈ H.
Since r ≠ 0, therefore atleast one of r, -r is a positive integer.
So, positive integral powers of a belong  to H.
Let q be the least positive integer such that a^q ∈ H.
We shall prove that H is a cyclic group generated by a^q.
Let x ∈ H be arbitrary element
⇒ x ∈ G                                                                           [∵ H ⊆ G]
⇒ x = aⁿ  for some integer n.                                                 ... (1)
By division algorithm, ∃ integers s and r such that
           n = sq + r     where 0 ≤ r ≤ q
⇒      aⁿ = a^{sq + r}

⇒      aⁿ = a^sq a^r
⇒     a^{n- sq} = a^r .                                                                       ... (2)
Since a^q ∈ H and s is an integer, so a^sq∈ H.
Also aⁿ ∈ H
∴ aⁿ (a^sq)^-1 ∈ H
⇒ aⁿ a^{- sq} ∈ H
⇒ a^n-sq ∈ H
⇒ a^r ∈ H,                                                                           From (2)
∴ a^r ∈ H    0 ≤ r ≤ q-1.                                                           ... (3)
But q is the least positive integer such that a^q ∈ H.
∴ r = 0,                                                              [From (2) and (3)]
∴ From (1), (2) and (3) we get
     x = aⁿ = a^{sq + r} = a^{sq + 0} = (a^q)^s
This is true for all x ∈ H.
∴ Each element of H is an integral power of a^q.
So, H is cyclic group generated by a^q.
Hence subgroup of a cyclic group is cyclic.

Theorem: Every group of prime order is cyclic.
Proof: Let G be a group of order p, a positive prime.
Since p ≥ 2
∴ G has atleast two elements.
Consider a ≠ e ∈ G and let H = <a> be the cyclic subgroup of G.
Therefore O(H) = O(a) > 1.
By Lagrange's theorem O(H) | O(G)
i.e.,  O(H) | p.
But p is prime.
Therefore O(H) = p = O(G).
Hence G must be cyclic group.