On the previous pages we saw how to combine two or more groups into a bigger group, here we look at the reverse process, that is, breaking a group down into two or more simpler groups.
Lets assume that we have a group G and we have a subgroup of G which we call N. We want to find the other subgroup G/N. For G/N to be a group then N must be a normal subgroup.
Test that N is a normal subgroup
To test that N is a normal subgroup then:
x N x^{1} must be a member of N for every x in G
Calculation of G/N
we calculate the set:
aN  aG
This will be the group G/N under the operation
(a N) (b N) = a b N
Example 1  divide C by Z_{2}
G = Complex numbers = C
N = Z_{2}
Test for Normal Subgroup
x  x N x^{1} 

1  1 {1,1} 1 = {1,1} 
1  1 {1,1} 1 = {1,1} 
i  i {1,1} i = {1,1} 
i  i {1,1} i = {1,1} 
So x N x^{1} is a member of N for every x which means that G/N will be a group and we can go on to calculate it.
Calculation of G/N
Elements of G/N are a•N:
1•{1,1} = {1,1}
1•{1,1} = {1,1}
i•{1,1} = {i,i}
i•{1,1} = {i,i}
so the elements are:
{1,1} and {i,i}
Example 2 divide H by C
G is the group of quaternions H:
Cayley Table 
Cayley Graph 


Note: in this example I have not shown the negative elements so where i is shown we also have i and so on for the other elements.
We want to divide it by C to get H/C
Test for Normal Subgroup
x  x K x^{1} 

1  1 {1,i} 1 = {1,i} 
i  i {1,i} i = {1,i} 
j  j {1,i} j = {1,i} 
k  k {1,i} k = {1,i} 
So x K x^{1} is a member of K for every x which means that G/K will be a group and we can go on to calculate it.
Calculation of G/N
Elements of G/N are a•N:
1•{1,i} = {1,i}
i•{1,i} = {1,i}
j•{1,i} = {j,k}
k•{1,i} = {k,j}
So the elements are:
{±1,±i} and {±j,±k}
which gives the multipication table:
a*b  {±1,±i}  {±j,±k} 
{±1,±i}  {±1,±i}  {±j,±k} 
{±j,±k}  {±j,±k}  {±1,±i} 