Page 73 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Proof: We have already observed that the product of two cosets is a coset.
This multiplication is also associative, since

((Hx) (Hy)) (Hz) = (Hxy) (Hz)
= Hxyz, as the product in G is associative,
= Hx (yz)
= (Hx) ( Hyz)

= (Hx) ((Hy) (Hz)) for x, y, z  G.
Now, if e is the identity of G, then Hx, He = Hxe = Hx and He. Hx = Hex = Hx for every x  G.
Thus, He = H is the identity element of G/H.

Also, for any x  G, Hx Hx = Hx x = He = Hx x = Hx .Hx.
-1
-1
-1
-1
Thus, the inverse of Hx is Hx .
-1
So, we have proved that G/H, the set of all cosets of a normal subgroup H in G, forms a group
with respect to the multiplication defined by Hx.Hy = Hxy. This group is called the quotient
group (or factor group) of G by H.
Note that the order of the quotient group G/H is the index of H in G. Thus, by Lagrange’s
theorem you know that if G is a finite group, then
o(G)
o(G/H) 
o(H)
Also note that if (G, +) is an abelian group and H  G, then H  G. Further, the operation on
G/H is defined by (H + x) + (H + y) = H + (x + y).
Let us look at a few examples of quotient groups.

Example: Obtain the group G/H, where G = S and H = A = {I, (1 2 3), (1 3 2)}.
3 3
Solution: Firstly, note bat A  S , since |S : A | = 2.
3
3
3
3
You know that G/H is a group of order 2 whose elements are H and (1 2) H.
Example: Show that the group Z/nZ is of order n.
Solution: The elements of Z/nZ are of the form a + nZ = {a + kn | k  Z).
Thus, the elements of Z/nZ are precisely the congruence classes modulo n, that is, the elements
of Z .
n
Thus, Z/nZ = {0,1,2,....,n 1}.

 o(Z/nZ) = n.
Note that addition in Z/nZ is given a + b = a + b
Definition: Let G be a group and x, y  G. Then x y xy is called the commutator of x and y. It is
-1 -1
denoted by [x, y].
The subgroup of G generated by the set of all commutators is called the commutator subgroup
of G. It is denoted by [G, G].

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