Let Z be the group of integers under addition and let f map Z onto Z mod 7. Note that we can take x+y mod 7, and that's the same as taking x mod 7 and adding it to y mod 7. In other words, f(xy) = f(x)f(y). That's it; that's all we need for a group homomorphism.
Every normal subgroup defines a group homomorphism.
If K is a normal subgroup, and H the resulting quotient group,
let the function f take an element x in G to the coset of K represented by x.
As described in the previous section,
f is a function from G onto H, where G and H are both groups;
but is f a homomorphism?
Does f commute with *?
Well the coset of x times the coset of y is,
by its very definition, the coset of xy.
Thus f is a group homomorphism.
Conversely, let f be a homomorphism from the group G onto the set H, which possesses a preexisting operator *. Since f is a homomorphism, f(x)f(y) = f(xy). In other words, f commutes with *. It is not hard to show that H is a group, inheriting all the group properties through f. If G were a ring, H would be a ring, and so on.
Let 1 (the identity) in G map to an element we will call 1 in H, and verify that 1 acts as an identity element in H. That is, f(1)f(x) = f(1x) = f(x), hence 1 leaves H where it is.
Let K be the set of elements in G that map to 1 in H. For any x and y in K, f(x) = f(y) = 1 in H. Their product is 1 in H, hence f(xy) = 1, hence xy is still in K. Inverses must also map to 1 in H, and are present in K, hence K is a subgroup of G.
It is easy to show that xK/x remains in K, as f(xK/x) = 1, hence K is a normal subgroup of G.
Finally, the elements of H correspond to the cosets of K, and are multiplied acccordingly. The image group H is indistinguishable from the quotient group G/K. A group homomorphism defines, and is defined by, a normal subgroup and quotient group.
If the kernel of a homomorphism is the identity element, f is a monomorphism. This comes from the Greek word mono, meaning one; the map is 1-1.
If f carries G onto the entire group H, rather than a piece of H, f is an epimorphism. This is usually what we mean when we say f maps G to H. Sometimes we say f maps G "onto" H - same thing. If f maps G "into" H, it is not always an epimorphism. Take the integers mod 7, then multiply by 2. This maps Z onto the even numbers mod 14. It is a group homomorphism into Z14, but it is not an epimorphism, because nothing maps to the odd numbers.
If f is an epimorphism and a monomorphism it is an isomorphism. Elements map 1-1, and all of G maps to all of H. The map can be reversed to produce an inverse homomorphism. The two groups are indistinguishable. We have merely assigned different labels to the elements of G. Some people would say these are in fact the same group; where the structure defines the group, without regard to the underlying set. This abstract definition is valid, because group isomorphism is an equivalence relation. We can clump all the isomorphic groups together and declare them a single group.
For example, if p is prime there is one group of order p, namely Zp. Let x be an element of G, other than the identity element, and consider the powers of x. If xn becomes 1 before n reaches p, x generates a proper subgroup, which is forbidden by Lagrange's theorem. Hence G is merely the powers of x, and the resulting exponents form the integers mod p. After some judicious relabeling, the group becomes Zp.
By the Numbers: Accounting Certification Resources
Which Certification Do I Need | Accounting Certifications Online | Analysis of Accounting Principles