Quotient Group:
Let G be a 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]
xmH = H
⇒ xm ∈ H.
Thus ∀ x ∈ G, we have xm ∈ H.
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]
xmH = H
⇒ xm ∈ H.
Thus ∀ x ∈ G, we have xm ∈ H.
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