In these examples we'll explicitly construct some quotient groups. Recall that if N is a normal subgroup of the group G, then the elements of the quotient G/N are the cosets Ng.  The important thing to remember is that cosets can be viewed in two ways: as subsets of G given by Ng = { xg | x Î N } and as the elements of G/N. In the second case each coset is a single element of the set G/N.

Example 1. Let G = ( Z[x] , + ) be the set of polynomials with integer coefficients, with addition as the group operation. Each element of G can be written as a_nxⁿ + ... + a₁x + a₀ for a_iÎ Z and some n ÎN. Let H = {a_nxⁿ + ... + a₁x + a₀ | a_n + ... + a₁ + a₀ = 0}.  You should check that H is a subgroup of G  and is (since G is abelian) normal in G .

What are the cosets of H ? Let f = a_nxⁿ + ... + a₁x + a₀ and g = c_mx^m + ... + c₁x + c₀ be in G. These elements are in the same coset of H exactly when
f - g Î H, or equivalently when ( a_n + ... + a₁ + a₀ ) - ( c_m + ... + c₁ + c₀ )  = 0, or a_n + ... + a₁ + a₀  = c_m + ... + c₁ + c₀.  Thus two elements are in the same coset when the sums of their coefficients are equal. Now such a sum is an integer and each integer occurs as such a sum. Regarding kÎ Z as a constant polynomial, we get the coset H + k of all polynomials whose coefficients add to k. Clearly, H + k = H + m exactly when k = m, and if  f = a_nxⁿ + ... + a₁x + a₀ with   a_n + ... + a₁ + a₀ = k, then f Î H + k. Consequently, the collection of constant polynomials gives a complete set of representatives for the cosets, i.e. G/H = { H + k | kÎ Z}. Although each coset H + k contains many elements of G, as an element of G/H, the coset H + k is a single object. We emphasize this by setting H + k = [k]. Note that in G/H  we have [k] + [n] = (H + k) + (H + n) = H + (k + n) = [k + n].

What group is G/H? Clearly, there is a bijection t between Z and G/H given by (n)t = [n] = H + n. The map t is certainly surjective, and if [n] = [m], then n - m Î H. But this means n-m = 0 or n = m, so t is injective. Also (n + m)t = [n + m] = [n] + [m] = (n)t + (m)t, proving that t is an isomorphism of Z onto G/H.  ·   ×

Example 2. Let G = R²  = { ( a , b ) | a,b Î R } be the Euclidean plane with group operation given by ( a , b ) + ( c , d ) = ( a + c , b + d ). Note that G is an abelinan group.  If L Ì R²  is any straight line through the origin, then L = { ( x , y ) |  y = m×x } where m  is the slope of  the line (excluding the
possibility that L is the y-axis). Thus,  L =  { ( x , m×x ) | x Î R }.   You should now verify that  L  is a normal subgroup of  G.  What are the cosets of L in G?  Two elements  ( a , b ) and  ( c , d ) Î G  are in the same coset exactly when ( a , b ) - ( c , d ) ÎL, or more precisely, when ( a - c , b - d ) = ( x , m×x ) for some x Î R. To see how ( a , b ) and ( c , d ) are related, we equate coordinates to obtain a - c = x  and  b - d = m×x, or  b - d = m×(a- c).

If m = 0, then L is the x-axis, and  b = d is the necessary and sufficient condition; that is, two points in R² are in the same coset of L exactly
when their second coordinates are the same, and thereby lie on the same line y = d  parallel to the x-axis. When m ¹ 0, then b - d = m×(a -c) implies either a = c and so b = d, or ( b - d )/( a- c ) = m in which case the line through the points ( a , b ) and ( c , d ) has slope m.

In each case the cosets of L are made up to the points on the straight lines which are parallel to L, i.e. those having  slope m.  How can we describe G/L? First we need a way to represent the cosets. Since m Î R,  L is not the y-axis. Since each coset is a line with slope m, no coset is parallel to the y-axis, either.  Therefore, each coset of L has a unique point on the y-axis, that is, the coset L + ( a , b ) contains a unique point of the form ( x , 0 ). Since ( x , 0 ) Î L + ( a , b ), we have L + ( a , b )  =  L + ( x , 0 ).  Consequently, we can regard G/L = {L + ( a , 0 ) | a Î R}.

The addition in G/L is, as usual, ( L + ( a , 0 ) ) + ( L + ( b , 0 ) ) = L + ( a + b , 0 ). (As an exercise, verify that if ( s , t ) Î L + ( a , 0 ) and ( u , v ) Π L + ( b , 0 ) then ( s + u , t + v ) Î L + ( a + b , 0 ).)   It should be clear there is a bijection t between the elements of G/L and those of R (identified as the y-axis) via ( r )t = L + ( r , 0 ). Since ( r + s )t = L + ( r + s , 0 ) = L + ( ( r , 0 ) + ( s , 0 ) ) = ( L + ( r , 0 ) ) + ( L + ( s , 0 ) ) = ( r )t + ( s )t, we have that  R  is isomorphic to G/L. ·