For a mapping F to be a homomorphism, it must be true that F(x o y) = F(x) * F(y) for the given operations o and *

For a mapping F to be a homomorphism, it must be true that F(x o y) = F(x) * F(y) for the given operations o and *. In other words, the element that x o y maps to must be the same element that is produced when the element that x maps to is operated with the element that y maps to under the operation *. Note that it is possible for o and * to be 2 different types of operations. For simplification purposes, I made the application problems on this page so that o and * represent the same operation (usually just multiplication).

For a mapping F to be an epimorphism, F must be a homomorphism that is surjective (onto).

If F is an epimorphism that is also injective (1-1), F can further be classified as an isomorphism. In other words, any homomorphism that is a bijection (surjective and injective) is also an isomorphism.

Application Problem (Created by Ben Mueller)

Let F be a mapping that maps integers (under multiplication) to integers (under multiplication) such that F(x) = x^2. Using the terms isomorphism, epimorphism, and homomorphism classify F in as many ways as possible and of course, verify your classifications.

Solution (Provided by Ben Mueller):

To check whether or not F is a homomorphism, we want to check whether or not

F(x)F(y) = F(xy)

Now,

F(x)F(y) = x^2*y^2 = (xy)^2 = F(xy)

This verifies that F is a homomorphism.

To check whether or not F is an epimorphism, we want to check whether or not the F represents a mapping that is surjective (onto). Since F maps integers to integers, the question of interest here is can every integer be expressed in the form x^2 where x is another integer or is the square root of every integer, another integer?. Of course, the answer to this question is “no”. For example, take the integer 3. For 3 to be expressed in the form x^2, x must equal the square root of 3 and the square root of 3 is not an integer. Thus the mapping F is not an epimorphism as it is not surjective (onto).

Furthermore the mapping F does not represent an isomorphism because an isomorphism must be bijective (onto and 1-1) and since F is not surjective (onto), F cannot be bijective.

In summary, the mapping F is a homomorphism, but not an epimorphism or an isomorphism.

Let F be a homomorphism from the group G to the group G’. If e denotes the identity element in G and e’ denotes the identity element in G’, then prove that F(e)=e’.

Proof (Written by Ben Mueller): Since e is the identity element, ee = e and F(e) = F(ee). Now, F is a homomorphism which means that F(ee) = F(e)F(e). From this, F(e) = F(e)F(e) and this implies that F(e) is the identity element in G’ (F(e) is an element in G’). Finally, since the identity element in a group must be unique, e’ = F(e)./(Proof complete)

In the same situation above, prove that F(x^-1) = F(x)^-1 for all x in G.

Proof (Written by Ben Mueller): F(x) is an element in G’ and since G’ is a group, the inverse of F(x) (denoted F(x)^-1) is also in G’. This implies that

F(x)F(x)^-1 = e’=F(x)^-1F(x). To prove that F(x^-1)=F(x)^-1 it is sufficient to show that F(x)F(x^-1) = e’ and that F(x^-1)F(x) = e’.

Since F is a homomorphism, F(x)F(x^-1) = F(xx^-1) = F(e) and we have already proven that F(e)=e’.

Also, F(x^-1)F(x) = F(x^-1x) = F(e) = e’.

From the work above, F(x^-1) = F(x)^-1/(Proof complete)