# Group homomorphisms

We want to study different maps between groups, so called homomorphisms, in this section. Homomorphisms (and especially isomorphisms) are very useful and important tools for understanding structures and properties of groups, for understanding how we can take one group structure and see whether we can move it to the other group, or for classifying groups up to isomorphisms. The last few words basically mean that two groups can be considered the same, although they may have different elements and different operations, while the whole structures are identical.

## Sources:

• Drapal part 1.4
• Wiki post dealing with group homomorphisms

## Theory:

We already mentioned group homomorphisms in the previous post on cyclic groups. Recall that if you have two groups and , and a map , then is a homomorphism, if for any two . If is also a bijection, then we say that is an isomorphism (denoted by ).

## Properties:

• If is a group homomorphism, then and .
• is an isomorphism if and only if is a homomorphism and there exists a homomorphism such that and .
• Between any two groups and there always exists at least one homomorphism , namely the zero homomorphism sending each element to .
• Let . Then .
• For every , a map , is a group homomorphism.
• Every group homomorphism is equal to for some .
• is an isomorphism (i.e., an automorphism) .
• For every we have , where .

## Examples:

### Example 1:

Let be a cyclic group generated by an element , such that and . Show that is isomorphic to .

### Solution:

Realise that . Consider a map given by . We will show that is an isomorphism, i.e., a bijective homomorphism.

• Homomorphism: Take arbitrary two elements of , namely and consider their product . Then  By definition this proves that is a homomorphism.
• Bijection: is obviously surjective – given any , then . Since is a surjective map between two finite sets of the same cardinality, it is also injective, and hence, a bijection.

### Example 2:

Find all homomorphisms .

### Solution:

We know there is at least the zero homomorphism, but what about some other homomorphisms? Well, by the first property, and for some . Also, we have So, . However, the equation mod has only one solution, namely . So, we see that there is no homomorphism other than the zero homomorphism.

## Problems:

### Problem 1:

Prove the first of the properties of group homomorphism. (Hint: the identity element and inverses are unique in any group.)

### Problem 2:

Let be a set . Show that can be viewed as a multiplicative group and find a group , such that .