Abstract Algebra/Group Theory/Homomorphism
We are finally making our way into the meat of the theory. In this section we will study structure-preserving maps between groups. This study will open new doors and provide us with a multitude of new theorems.
Up until now we have studied groups at the "element level". Since we are now about to take a step back and study groups at the "homomorphism level", readers should expect a sudden increase in abstraction starting from this section. We will try to ease the reader into this increase by keeping one foot at the "element level" throughout this section.
From here on out the notation will denote the identity element in the group unless otherwise specified.
Definition 1: Let and be groups. A homomorphism from to is a function such that for all ,
Thus, a homomorphism preserves the group structure. We have included the multiplication symbols here to make explicit that multiplication on the left side occurs in , and multiplication on the right side occurs in .
Already we see that this section is different from the previous ones. Up until now we have, excluding subgroups, only dealt with one group at a time. No more! Let us start by deriving some elementary and immediate consequences of the definition.
Theorem 2: Let be groups and a homomorphism. Then . In other words, the identity is mapped to the identity.
Proof: Let . Then, , implying that is the identity in , proving the theorem. ∎
Theorem 3: Let be groups and a homomorphism. Then for any , . In other words, inverses are mapped to inverses.
Proof: Let . Then implying that , as was to be shown. ∎
Theorem 4: Let be groups, a homomorphism and let be a subgroup of . Then is a subgroup of .
Proof: Let . Then and . Since , , and so is a subgroup of . ∎
Theorem 5: Let be groups, a homomorphism and let be a subgroup of . Then is a subgroup of .
Proof: Let . Then , and since is a subgroup, . But then, , and so is a subgroup of . ∎
From Theorem 4 and Theorem 5 we see that homomorphisms preserve subgroups. Thus we can expect to learn a lot about the subgroup structure of a group by finding suitable homomorphisms into .
In particular, every homomorphism has associated with it two important subgroups.
Definition 6: A homomorphism is called an isomorphism if it is bijective and its inverse is a homomorphism. Two groups are called isomorphic if there exists an isomorphism between them, and we write to denote " is isomorphic to ".
Theorem 7: A bijective homomorphism is an isomorphism.
Proof: Let be groups and let be a bijective homomorphism. We must show that the inverse is a homomorphism. Let . then there exist unique such that and . Then we have since is a homomorphism. Now apply to all equations. We obtain , and , so is a homomorphism and thus is an isomorphism. ∎
Definition 8: Let be groups. A homomorphism that maps every element in to is called a trivial homomorphism (or zero homomorphism), and is denoted by
Definition 9: Let be a subgroup of a group . Then the homomorphism given by is called the inclusion of into . Let be a group isomorphic to a subgroup of a group . Then the isomorphism induces an injective homomorphism given by , called an imbedding of into . Obviously, .
Definition 10: Let be groups and a homomorphism. Then we define the following subgroups:
- i) , called the kernel of , and
- ii) , called the image of .
Theorem 11: The composition of homomorphisms is a homomorphism.
Proof: Let be groups and and homomorphisms. Then is a function. We must show it is a homomorphism. Let . Then , so is indeed a homomorphisms. ∎
Theorem 12: Composition of homomorphisms is associative.
Proof: This is evident since homomorphisms are functions, and composition of functions is associative. ∎
Corollary 13: The composition of isomorphisms is an isomorphism.
Proof: This is evident from Theorem 11 and since the composition of bijections is a bijection. ∎
Theorem 14: Let be groups and a homomorphism. Then is injective if and only if .
Proof: Assume and . Then , implying that . But by assumption then , so is injective. Assume now that and . Then there exists another element such that . But then . Since both and map to , is not injective, proving the theorem. ∎
Corollary 15: Inclusions are injective.
Proof: The result is immediate. Since for all , we have . ∎
The kernel can be seen to satisfy a universal property. The following theorem explains this, but it is unusually abstract for an elementary treatment of groups, and the reader should not worry if he/she cannot understand it immediately.
Theorem 16: Let be groups and a group homomorphism. Also let be a group and a homomorphism such that . Also let is the inclusion of into . Then there exists a unique homomorphism such that .
Proof: Since , by definition we must have , so exists. The commutativity then forces , so is unique. ∎
Definition 17: A commutative diagram is a pictorial presentation of a network of functions. Commutativity means that when several routes of function composition from one object lead to the same destination, the two compositions are equal as functions. As an example, the commutative diagram on the right describes the situation in Theorem 16. In the commutative diagrams (or diagrams for short, we will not show diagrams which no not commute) shown in this chapter on groups, all functions are implicitly assumed to be group homomorphisms. Monomorphisms in diagrams are often emphasized by hooked arrows. In addition, epimorphims are often emphasized by double headed arrows. That an inclusion is a monomorphism will be proven shortly.
Remark 18: From the commutative diagram on the right, the kernel can be defined completely without reference to elements. Indeed, Theorem 16 would become the definition, and our Definition 10 i) would become a theorem. We will not entertain this line of thought in this book, but the advanced reader is welcome to work it out for him/herself.
In this subsection we will take a look at the homomorphisms from a group to itself.
Definition 19: A homomorphism from a group to itself is called an endomorphism of . An endomorphism which is also an isomorphism is called an automorphism. The set of all endomorphisms of is denoted , while the set of all automorphisms of is denoted .
Theorem 20: is a monoid under composition of homomorphisms. Also, is a submonoid which is also a group.
Proof: We only have to confirm that is closed and has an identity, which we know is true. For , the identity homomorphism is an isomorphism and the composition of isomorphisms is an isomorphism. Thus is a submonoid. To show it is a group, note that the inverse of an automorphism is an automorphism, so is indeed a group. ∎
Groups with OperatorsEdit
An endomorphism of a group can be thought of as a unary operator on that group. This motivates the following definition:
Definition 21: Let be a group and . Then the pair is called a group with operators. is called the operator domain and its elements are called the homotheties of . For any , we introduce the shorthand for all . Thus the fact that the homotheties of are endomorphisms can be expressed thus: for all and , .
Example 22: For any group , the pair is trivially a group with operators.
Lemma 23: Let be a group with operators. Then can be extended to a submonoid of such that the structure of is identical to .
Proof: Let include the identity endomorphism and let be a generating set. Then is closed under compositions and is a monoid. Since any element of is expressible as a (possibly empty) composition of elements in , the structures are identical. ∎
In the following, we assume that the operator domain is always a monoid. If it is not, we can extend it to one by Lemma 23.
Definition 24: Let and be groups with operators with the same operator domain. Then a homomorphism is a group homomorphism such that for all and , we have .
Definition 25: Let be a group with operators and a subgroup of . Then is called a stable subgroup (or a -invariant subgroup) if for all and , . We say that respects the homotheties of . In this case is a sub-group with operators.
Example 26: Let be a vector space over the field . If we by denote the underlying abelian group under addition, then is a group with operators, where for any and , we define . Then the stable subgroups are precisely the linear subspaces of (show this).
Problem 1: Show that there is no nontrivial homomorphism from to .