Chapter 2 introduced groups, subgroups, and homomorphisms. In this chapter we learn to divide a group by a subgroup. The cosets of a subgroup cut into pieces of equal size, which gives Lagrange’s theorem, and when is normal the cosets themselves form a group . Quotient groups lead to the isomorphism theorems, which describe how quotients, subgroups, and homomorphisms fit together. Along the way we study groups acting on sets, the symmetric group in some detail, and exact sequences; the chapter ends with free groups and presentations, which describe a group by generators and relations.
Consider . Congruence mod is an equivalence relation on , and we may regard as its quotient set, whose elements are the classes . Addition is compatible with congruence: if and , then . So the rule does not depend on the representatives chosen (this is the criterion of Proposition 1.6.4 for a map to descend to a quotient set). It makes the set of classes a group, and the quotient map ,, is a homomorphism with kernel .
Why stop at the integers? Given a group and a subgroup , we can define congruence mod and form the set of equivalence classes. When does the operation of descend to these classes, so that the quotient map is a homomorphism? The answer, normality, is one of the foundational ideas of group theory.
3.1.1 Congruence Mod and Cosets
Let . For define if and only if . This is an equivalence relation. We have since . If , then , so and . Finally, if and , then and are both in , so since is closed under the group operation; thus . We call congruence mod .
Definition 3.1.1 (Cosets and index). Let and . The left coset of containing is , and the right coset is . The set of left cosets is denoted . The index of in , written , is the number of left cosets of in (possibly infinite).
In additive notation the cosets are written . For instance, the cosets of in are the congruence classes , and .
Proposition 3.1.2 (Cosets partition ). Let . For we have if and only if . Hence the left cosets of are exactly the equivalence classes of congruence mod : they form a partition of , every lies in exactly one left coset, namely , and if and only if .
Proof. We have if and only if for some , that is, if and only if , which is . So is the equivalence class of . The equivalence classes of an equivalence relation partition the set (Proposition 1.2.8), and two classes are equal exactly when their representatives are equivalent (Lemma 1.2.5). ∎
Here is the key observation that makes quotient theory work: all cosets have the same size.
Proposition 3.1.3 (Cosets are equinumerous). For any , the map given by is a bijection. In particular, for all .
Proof. Surjectivity is immediate from the definition of . For injectivity, suppose . Left-multiplying by gives . (This is left translation, which is always a bijection in a group; we will meet it again in Section 3.2.) ∎
Theorem 3.1.4 (Lagrange’s Theorem). Let be a finite group and a subgroup. Then In particular, divides , and .
Proof. By Proposition 3.1.2, the left cosets of partition , and each has elements. Hence .∎
Corollary 3.1.5 (Consequences of Lagrange’s theorem). Let be a finite group.
The order of every element divides .
for every .
If is prime, then is cyclic, generated by any element , and .
Proof. (a) The order of equals (Proposition 2.6.1), and is a subgroup of , so its order divides by Lagrange’s theorem.
(b) By (a), for some integer , so .
(c) Let . Then divides and is not , so . Thus has elements and equals . A cyclic group of order is isomorphic to (Theorem 2.6.2). ∎
Note. The converse of Lagrange’s theorem is false: a divisor of need not be the order of any subgroup. The alternating group , of order , has no subgroup of order ; we prove this in Example 3.4.17 once has been defined. Partial converses do hold. For a finite abelian group every divisor of is the order of a subgroup, as the classification of finite abelian groups shows (Theorem 10.2.1). For an arbitrary finite group, the Sylow theorems of Chapter 11 (Theorem 11.5.3) give, for each prime , a subgroup whose order is the largest power of dividing .
Example 3.1.6 (Subgroups of ). The group has order 6 and is non-abelian; in particular . Write and . Composing right to left, and , so By Lagrange’s theorem a subgroup of has order 1, 2, 3, or 6. A subgroup of order 2 or 3 has prime order, so by Corollary 3.1.5 it is generated by an element of order 2 or 3, respectively. The elements of order 2 are the three transpositions, and the elements of order 3 are and , which generate the same subgroup. So the subgroups of are:
the trivial subgroup and itself;
one subgroup of order 3, , the rotations of the triangle;
three subgroups of order 2, ,, and , each generated by a reflection of the triangle.
In the diagram below (the subgroup lattice), a line joins two subgroups when the lower one is contained in the upper one with no subgroup strictly between them.
Now we ask when is a group. The natural candidate operation is . For this to make sense it must be well defined: if and , we need . This does not always hold.
Example 3.1.7 (Coset multiplication can fail). Let and , composing right to left (so ). The three left cosets are ,, and . Fix and take two representatives and of the coset . We compute: So far the answer does not depend on the representative. Now vary instead: take and , which lie in the same coset, with : Since , coset multiplication is not well defined for this .
What went wrong? The subgroup is not “symmetric” enough: its left cosets and right cosets differ. For instance, but .
3.1.2 Normal Subgroups and Quotient Groups
For and , write .
Definition 3.1.8 (Normal subgroup). A subgroup is normal, written , if for all ; that is, is closed under conjugation by elements of .
Proposition 3.1.9 (Equivalent conditions for normality). For a subgroup , the following are equivalent:
for all ;
for all (left cosets equal right cosets);
for all (one inclusion suffices).
Proof. (a) (b): Each element of equals , so .
(b) (c): If , then for we have , so for some , giving .
(c) (a): We have by assumption. For the reverse inclusion, take . Applying (c) to gives , so .∎
Every subgroup of an abelian group is normal, since there . In any group, and are normal.
Theorem 3.1.10 (Quotient groups exist exactly for normal subgroups). Let be a subgroup of a group . The following are equivalent:
for some homomorphism to some group ;
is normal in ;
the rule is a well-defined operation on ; that is, and imply .
When these hold, is a group under this operation, called the quotient group of by , and the map ,, is a surjective homomorphism with , called the canonical projection (or quotient map).
Proof. (a) (b): If , then for and , so for all , and is normal by Proposition 3.1.9.
(b) (c): Suppose and , that is, and . Then The first factor lies in because is normal, and the second lies in . So , which says .
Now suppose (c) holds. The group axioms for follow from those of :; the coset is an identity, since ; and is inverse to . The map is surjective, and it is a homomorphism because . Its kernel is by Proposition 3.1.2.
(c) (a): We have just seen that .∎
In particular, is a group under if and only if is normal, and the normal subgroups of are exactly the kernels of homomorphisms out of . The direction (c) (b) can also be seen directly: if the operation is well defined, then for and we have , so , that is, . In additive notation the operation on is ; for this recovers .
Proposition 3.1.11 (Subgroups of index two are normal). If and , then .
Proof. If , then . Let . The two left cosets are and , and they partition , so . The right coset is disjoint from : if with , then . Hence . The same argument applied to gives , and taking inverses of all elements turns this into . So .∎
Example 3.1.12 (Normal and non-normal subgroups of ). In , the subgroup has index 2, so it is normal and is a group of order 2. The subgroup is not normal: , so . Similarly, conjugation by carries to and to , so none of the three subgroups of order 2 is normal. Hence the normal subgroups of are ,, and .
Definition 3.1.13 (Center). The center of a group is .
Proposition 3.1.14 (The center). is an abelian normal subgroup of , and is abelian if and only if .
Proof. Clearly . If , then for all , so ; multiplying by on both sides gives , so . Elements of commute with everything, in particular with each other, so is abelian. For and ,, so and is normal. Finally, says exactly that all elements commute. ∎
Question 3.1.15. Let be a group.
Prove that if is cyclic, then is abelian.
Find and describe .
Show solution
(a) Write and suppose . Every coset is then for some , so every element of has the form with and . Take and . Since and commute with everything, Thus , and is abelian. (So in fact is cyclic only when it is trivial.)
(b) Recall (Proposition 2.2.15) that with and . From we get for all by induction. An element is central if and only if it commutes with the generators and , because the elements commuting with form a subgroup.
For rotations: commutes with if and only if , that is, , so . Thus and are the central rotations.
For reflections: commutes with if and only if . Now , while . Equality would give , so , which is false since has order 4. No reflection is central.
Therefore . The quotient has order , with cosets ,,,. Every non-identity element of the quotient has order 2 (since ,, and lie in ), so , the Klein four-group, and not .
Normal subgroups let us view a group as built from two pieces, a normal subgroup and the quotient . The groups that cannot be broken up this way are the basic building blocks.
Definition 3.1.16 (Simple groups). A group is simple if its only normal subgroups are and .
By Lagrange’s theorem, a group of prime order has no subgroups other than and itself, so is simple for every prime . The alternating groups with (Section 3.4) are simple as well; this is proved in Chapter 11 (Theorem 11.4.9).
Question 3.1.17. Let be a group of order , where is an odd prime.
Prove that has a unique subgroup of order , which is therefore normal.
Conclude that or , the dihedral group of order .
Show solution
(a)Existence. By Corollary 3.1.5, the order of every element divides , so it is ,,, or . If some has order , then has order . Suppose no element has order or . Then for every , which forces to be abelian: . Choose distinct non-identity elements (possible since ). Then is closed under multiplication and inverses, so it is a subgroup of order 4. But does not divide , contradicting Lagrange’s theorem. Hence some has order , and is a subgroup of order . It has index 2, so it is normal by Proposition 3.1.11.
Uniqueness. Suppose is a subgroup of order with . Then is a subgroup of (Proposition 2.4.4) whose order divides and is not , so . The map ,, is injective: if , then , so and . Hence , a contradiction.
(b) Let be the unique subgroup of order , and pick . The order of is not , and it is not , for then would be a second subgroup of order . If has order , then is cyclic and . It remains to treat the case .
Since is normal, , say . Conjugation by is a homomorphism, since ; using this and , so divides , and by Euclid’s lemma . As , the two cosets of are and , so every element of is uniquely or with .
If , then , so is abelian. Then equals if and only if , that is, if and only if both and divide . So has order , and .
If , then , so for all , and The dihedral group consists of the elements (,) with and (Proposition 2.2.15), so its elements multiply by the same rule. Hence is a bijective homomorphism, and .