Naive Isomorphism

     As one of my imaginary friends says "the mathematics are the science of form and quantity; mathematical reasoning is merely logic applied to observation upon form and quantity. The great error lies in supposing that even the truths of what is called pure algebra, are abstract or general truths". In this post I will try to observe and deduce properties of algebraic objects, escaping the usual structure theorem1 => theorem2 => theorem3. So, we will play in the backyard of group theory focusing on the first isomorphism theorem. 

     The simplest example of a group is the set of integers ℤ with addition. Another relatively simple group is the additive group of integers modulo n. For the particular case when n = 3, we have the group G = (ℤ3 ,*). We can see that these groups are somehow related, each element from ℤ can be put under the modulus operator,that is we divide it by 3 and take the remainder, and will yield an element of ℤ3. Formally, we write this:
let f : ℤ ->; ℤ3 be a mapping such that f(x) = x mod 3
It can also be noticed that f(x+y) = ( (x+y) mod 3) = (x mod 3) + (y mod 3) = f(x)+f(y), that is, the operation is the same(preserved) on both sides of the mapping. This is structure preserving property is important because if we know something about the structure of a group we can say something about the structure of all the groups homomorphic with it.
A more careful look at the structure of those groups will uncover another useful facts:

• for every element of ℤ3 we have a corresponding element of ℤ which means the function is surjective
• every element of ℤ which is a divisible by 3 is mapped on the identity of  ℤ3

     If we now look at the additive group of all elements in ℤ divisible by 3 A={...-9,-6,-3,0,3,6,9...} and ℤ3 we can come up with a different homomorphism (which is both surjective and injective) from the set of sets A'={g+A/ for every g in ℤ} and ℤ3.
     I will let you think what this homomorphism might be and in the mean time lets see what other useful information can be extracted from this. Suppose we've found this mapping from A' to ℤ3 ,lets call it θ, then we have the following homomorphic mappings:
- f from ℤ to ℤ3
- g from  ℤ3 to ℤ3 the identity function ( g(y) = y )
- θ from A' to ℤ3 
- h from  ℤ to A' , h(x) = x+A

 So we can decompose the mapping f in f = g * θ * h.
  

     Now, that we've quite enjoyed ourselves with this concrete example, lets state these properties in an abstract form so they could be applied for all other groups with different mappings between them(i.e. complex numbers, symmetries, dihedral groups, etc).
 

              The first isomorphism theorem

Let G and H be groups and φ : G → H a group homomorphism. Then:

1. Im φ  is a subgroup of H
   
2. Ker φ  is a normal subgroup of H
3. Im φ is isomorphic with the quotient group G/Ker φ  

     The first statement makes use of the basic notions of groups and homomorphisms. Being groups, G and H are sets with binary operations(the binary operation on G can be different from the one on H). For the group to exist the binary operation  must respect the following laws:

1. closure law:  let x,y be elements of the group G, then x • y is still in G
   
2. associative law: (x • y) • z = x • (y • z), for all x,y,z in G
3. identity law: there exist e in G such that e • x = x • e = x, for all x in G
4. inverse law: for all x in G there exists y in G such that x • y = y • x = e

        A subgroup of a group G is a subset G' ⊆ G such that G' is a group relative to the binary operation in G.
The binary operation • is a function from G × G to G, that is, the elements with the binary operation can't spawn elements that are not in G. It can be easily seen that there are n^(n^2) possible binary operations, where n = |G|, because |G × G| = n^2 and for every element of G × G  there are n possible mappings in G. For example for two sets |A| = 2 and |B| = 3 we have for the first element of A 3 choices in B and for the second element in A another 3 choices  in B, but the second time those 3 choices can be combined with the previous 3 choices of the first element resulting in 9 choices, that is, 3^2.

The homomorphism φ is a mapping from G to H having the structure-preserving property: φ(x • y) = φ(x) • φ(y). If the mapping is also bijective then it is called isomorphism. It's origins are in the Greek language: homos morphe meaning "similar shape", and isos morphe meaning "equal shape".

The image and the kernel of a mapping are both subsets, the first of H and the second one of G:    
Im φ  = {y in H / y = φ(x) for x in G} ⊆ H;
Ker φ = {x in G / φ(x) = e, where e is the identity in H} ⊆ G. 

      It is called a normal subgroup of a group G ⊆ G', a subgroup G' if x • G' • x^-1 ⊆ G' for every x in G. If G is abelian then every subgroup of G is normal since x • g' • x^-1 = x • x^-1 • g' = g', for each g' in G' and x in G. The converse, however, is not true: a group can have it's subgroups normal and not necessarily be commutative. A trivial counterexample to sustain this is the quaternion group which is a non-abelian group with 8 elements under multiplication:    
Q={-1, i, j, k / (-1)^2 =1, i^2=j^2=k^2=ijk=-1}.
For any subgroup G' we can define an equivalence relation x ~ y if and only if x = y • g', for g' in G'. The equivalence classes for this relation are called cossets and defined as follows:     
xG' = {x • g' / g' is an element of G'} -- left cosset of G' in G , similarly :
G'x = {g' • x / g' is an element of G'} -- right cosset of G' in G. 

 If G' is a normal subgroup of G we have xG' = G'x, for every x in G because:
x • G' = x • G' • e = x • G' • x^-1 • x = (x • G' • x^-1 ) • x = G' • x ■ 

Which means that the left cossets of a normal subgroup are the same with the right cossets. And since we no longer distinguish between them we write G/G' to denote the set of all cossets of G' in G. This forms a group under multiplication called the quotient group of G in G'.

      I've put it all together in a small drawing for a better overview. I have denoted by θ the isomorphism between the quotient group G/Kerφ and Imφ.

      At this point the first two conclusion of the theorem should be clear and the reader should be able to prove them easily. For the third conclusion, if you haven't already figured out what the second isomorphism in the example was, you might want to check this mapping: 
                          θ(xG) = φ(x)