If N is a subgroup of G then N is a normal subgroup if:
g N = N g
for all g in G
This does not necessarily mean that g n = n g for all gG, nN but it means that the set {g N} is the same as the set {N g} or, in other words,
g n = n' g
That is, if we choose an n in N then n' will also be a member of N (although not necessarily the same member).
If we multiply the above equation by g^{1} we get:
n' = g n g^{1}
where:
 gG
 nN
 n'N
This second form is probably the best way to test if N is a normal subgroup.
Calculation of G/N
we calculate the set:
g N  gG
This will be the group G/N under the operation
(a N) (b N) = a b N
Example 1
Lets take the example of C_{3}×C_{2} :
generator  cayley graph  table  

<m,r  m²,r³,rm=mr> 

In this example we will test if C_{3} is a normal subgroup and then divide by it to check that the result is C_{2}.
First we need to know the inverse function, to do this we take a particular column, go down until we get to the identity element '1' and then read out the row:
x  x^{1} 

1  1 
r  r² 
r²  r 
m  m 
rm  r²m 
r²m  rm 
let R = the elements of C_{3}, that is, 1 , r and r²:
x  x 1 x^{1}  x r x^{1}  x r² x^{1} 

1  1 1 1 = 1  1 r 1 = r  1 r² 1 = r 
r  r² 1 r² = r  r² r r² = r²  r² r² r² = 1 
r²  r 1 r = r²  r r r = 1  r r² r = r 
m  m 1 m = 1  m r m = r  m r² m = r² 
rm  r²m 1 rm = 1  r²m r rm = r  r²m r² rm = r² 
r²m  rm 1 r²m = 1  rm r r²m = r  rm r² r²m = r² 
So the result is always a member of C_{3} so it is a normal subgroup. We can now try divideing by C_{3} as follows:
The result of this division has elements R and Rm:
R  Rm 
Rm  R 
Example 2  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 3 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} 