with a+ b = 1 = c + d. [There are also stochastic groups 6(2, ) and 6(2, ).]
Prove that the product of two stochastic matrices is again stochastic, and that the inverse of a stochastic matrix is stochastic.
2.43 Show that the symmetry group 6(C) of a circle C is infinite.
*2.44 Prove that every element in a dihedral group D2nhas a unique factorization of the form aibj, where 0≤ i < n and j = 0 or 1.
2.45 Let e1 = (1, 0) and e2 = (0, 1), If ϕ is an isometry of the plane fixing O, let ϕ(e1)= (a, b), ϕ(e2)= (c, d), and let A =a c
b d
. Prove that det( A)= ±1.
2.4 S
UBGROUPS ANDL
AGRANGE’
ST
HEOREMA subgroup of a group G is a subset which is a group under the same opera-tion as in G. The following definiopera-tion will help to make this last phrase precise.
Definition. Let∗ be an operation on a set G, and let S ⊆ G be a subset. We say that S is closed under∗ if x ∗ y ∈ S for all x, y ∈ S.
The operation on a group G is a function∗ : G × G → G. If S ⊆ G, then S× S ⊆ G × G, and to say that S is closed under the operation ∗ means that
∗(S × S) ⊆ S. For example, the subset of the additive group
of rational numbers is closed under +. However, if × is the multiplicative group of nonzero rational numbers, then × is closed under multiplication, but it is not closed under + (for example, 2 and−2 lie in ×, but their sum−2 + 2 = 0 /∈ ×).
Definition. A subset H of a group G is a subgroup if (i) 1∈ H;
(ii) if x , y∈ H, then x y ∈ H; that is, H is closed under ∗;
(iii) if x ∈ H, then x−1∈ H.
12The term stochastic comes from the Greek word meaning “to guess.” Its mathematical usage occurs in statistics, and stochastic matrices first arose in the study of certain statistical problems.
We write H ≤ G to denote H being a subgroup of a group G. Observe that{1} and G are always subgroups of a group G, where {1} denotes the subset consisting of the single element 1. We call a subgroup H of G proper if H 6= G, and we write H < G. We call a subgroup H of G nontrivial if H 6= {1}. More interesting examples of subgroups will be given below.
Proposition 2.64. Every subgroup H ≤ G of a group G is itself a group.
Proof. Axiom (ii) (in the definition of subgroup) shows that H is closed under the operation of G; that is, H has an operation (namely, the restriction of the operation∗ : G × G → G to H × H ⊆ G × G). This operation is associative:
since the equation (x y)z= x(yz) holds for all x, y, z ∈ G, it holds, in particular, for all x , y, z ∈ H. Finally, axiom (i) gives the identity, and axiom (iii) gives inverses. •
It is quicker to check that a subset H of a group G is a subgroup (and hence that it is a group in its own right) than to verify the group axioms for H , for as-sociativity is inherited from the operation on G and hence it need not be verified again.
Example 2.65.
(i) Recall that Isom( 2)is the group of all isometries of the plane. The sub-set O2( ), consisting of all isometries fixing the origin, is a subgroup of Isom(
2). If ⊆ 2, then the symmetry group 6() is also a subgroup of Isom( 2). If the center of gravity of exists and is at the origin, then 6()≤ O2( ).
(ii) The four permutations V=
(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)
form a group, because V is a subgroup of S4: (1) ∈ V; α2= (1) for each α∈ V, and so α−1= α ∈ V; the product of any two distinct permutations in V− {(1)} is the third one. One calls V the four-group (or the Klein group) (V abbreviates the original German term Vierergruppe).
Consider what verifying associativity a(bc) = (ab)c would involve:
there are 4 choices for each of a, b, and c, and so there are 43 = 64 equations to be checked. Of course, we may assume that none is (1), leaving us with only 33 = 27 equations but, plainly, proving V is a group by showing it is a subgroup of S4is obviously the best way to proceed.
(iii) If 2 is the plane considered as an (additive) abelian group, then any line L through the origin is a subgroup. The easiest way to see this is to choose
a point (a, b) on L and then note that L consists of all the scalar multiples (r a, r b). The reader may now verify that the axioms in the definition of subgroup do hold for L.
One can shorten the list of items needed to verify that a subset is, in fact, a subgroup.
Proposition 2.66. A subset H of a group G is a subgroup if and only if H is nonempty and, whenever x , y∈ H, then x y−1∈ H.
Proof. If H is a subgroup, then it is nonempty, for 1 ∈ H. If x, y ∈ H, then y−1∈ H, by part (iii) of the definition, and so x y−1 ∈ H, by part (ii).
Conversely, assume that H is a subset satisfying the new condition. Since H is nonempty, it contains some element, say, h. Taking x = h = y, we see that 1= hh−1 ∈ H, and so part (i) holds. If y ∈ H, then set x = 1 (which we can now do because 1 ∈ H), giving y−1 = 1y−1 ∈ H, and so part (iii) holds.
Finally, we know that (y−1)−1 = y, by Lemma 2.46. Hence, if x, y ∈ H, then y−1∈ H, and so x y = x(y−1)−1∈ H. Therefore, H is a subgroup of G. •
Since every subgroup contains 1, one may replace the hypothesis “H is nonempty” in Proposition 2.66 by “1∈ H.”
Note that if the operation in G is addition, then the condition in the proposi-tion is that H is a nonempty subset such that x , y∈ H implies x − y ∈ H.
For Galois, a group was just a subset H of Snthat is closed under composi-tion; that is, if α, β∈ H, then αβ ∈ H. A. Cayley, in 1854, was the first to define an abstract group, mentioning associativity, inverses, and identity explicitly.
Proposition 2.67. A nonempty subset H of a finite group G is a subgroup if and only if H is closed under the operation of G; that is, if a, b∈ H, then ab ∈ H.
In particular, a nonempty subset of Sn is a subgroup if and only if it is closed under composition.
Proof. Every subgroup is nonempty, by axiom (i) in the definition of subgroup, and it is closed, by axiom (ii).
Conversely, assume that H is a nonempty subset of G closed under the oper-ation on G; thus, axiom (ii) holds. It follows that H contains all the powers of its elements. In particular, there is some element a ∈ H, because H is nonempty, and an ∈ H for all n ≥ 1. As we saw in Example 2.52, every element in G has finite order: there is an integer m with am= 1; hence 1 ∈ H and axiom (i) holds.
Finally, if h ∈ H and hm = 1, then h−1= hm−1(for hhm−1 = 1 = hm−1h), so that h−1 ∈ H and axiom (iii) holds. Therefore, H is a subgroup of G. •
This last proposition can be false when G is an infinite group. For example, the subset of the additive group
is closed under +, but it is not a subgroup of .
Example 2.68.
The subset An of Sn, consisting of all the even permutations, is a subgroup be-cause it is closed under multiplication: even ◦ even = even. This subgroup of Snis called the alternating13group on n letters, and it is denoted by An.
Definition. If G is a group and a∈ G, write
hai = {an: n ∈ } = {all powers of a};
hai is called the cyclic subgroup of G generated by a.
A group G is called cyclic if there is some a∈ G with G = hai; in this case a is called a generator of G.
It is easy to see that hai is, in fact, a subgroup: 1 = a0 ∈ hai; anam = an+m ∈ hai; a−1∈ hai. Example 2.47(vi) shows, for every n ≥ 1, that the mul-tiplicative group 0n of all nth roots of unity is a cyclic group with the primitive nth root of unity ζ = e2π i /nas a generator.
A cyclic group can have several different generators. For example, hai = ha−1i.
Proposition 2.69. If G= hai is a cyclic group of order n, then akis a generator of G if and only if gcd(k, n)= 1.
Proof. If ak is a generator, then a ∈ haki, so there is s with a = aks. Hence, aks−1= 1, so that Lemma 2.53 shows that n | (ks −1); that is, there is an integer t with ks − 1 = tn, or sk − tn = 1. Hence, (k, n) = 1, by Exercise 1.51 on page 52.
Conversely, if 1= sk + tn, then a = ask+tn = ask(because at n = 1), and so a∈ haki. Hence G = hai ≤ haki, and so G = haki. •
13The alternating group first arose in studying polynomials. If f (x)= (x − u1)(x− u2)· · · (x − un), then the number D = Q
i< j(ui − uj)changes sign if one permutes the roots: if α is a permutation of{u1,u2, . . . ,un}, then it is easy to see thatQ
i< j[α(ui)− α(uj)] = ±D.
Thus, the sign of the product alternates as various permutations α are applied to its factors.
The sign does not change for those α in the alternating group.
Corollary 2.70. The number of generators of a cyclic group of order n is φ (n).
Proof. This follows at once from the Propositions 2.69 and 1.39. •
Proposition 2.71. Every subgroup S of a cyclic group G= hai is itself cyclic.
In fact, am is a generator of S, where m is the smallest positive integer with am ∈ S.
Proof. We may assume that S is nontrivial; that is, S6= {1}, for the proposition is obviously true when S = {1}. Let I = {m ∈ : am ∈ S}; it is easy to check that I satisfies the conditions in Corollary 1.34. (i): 0 ∈ I , for a0 = 1 ∈ S.
(ii): If m, n∈ I , then am,an∈ S, and so ama−n= am−n ∈ S; hence, m − n ∈ I . (iii): If m ∈ I and i ∈ , then am ∈ S, and so (am)i = ai m ∈ S; hence, i m ∈ I . Since S 6= {1}, there is some 1 6= aq ∈ S; thus, q ∈ I and I 6= {0}.
By Corollary 1.34, if k is the smallest positive integer in I , then k| m for every mi n I . We claim thathaki = S. Clearly, haki ≤ S. For the reverse inclusion, take s ∈ S. Now s = am for some m, so that m ∈ I and m = k` for some `.
Therefore, s= am = ak`∈ haki. •
Proposition 2.78 will give a number-theoretic interpretation of this last result.
Proposition 2.72. Let G be a finite group and let a ∈ G. Then the order of a is the number of elements inhai.
Proof. We will use the idea in Lemma 2.23. Since G is finite, there is an integer k ≥ 1 with 1, a, a2, . . . ,ak−1 consisting of k distinct elements, while 1, a, a2, . . . ,ak has a repetition; hence ak ∈ {1, a, a2, . . . ,ak−1}; that is, ak = ai for some i with 0≤ i < k. If i ≥ 1, then ak−i = 1, contradicting the original list having no repetitions. Therefore, ak = a0= 1, and k is the order of a (being the smallest positive such k).
If H = {1, a, a2, . . . ,ak−1}, then |H| = k; it suffices to show that H = hai.
Clearly, H ⊆ hai. For the reverse inclusion, take ai ∈ hai. By the division algorithm, i = qk + r, where 0 ≤ r < k. Hence ai = aqk+r = aqkar = (ak)qar = ar ∈ H; this gives hai ⊆ H, and so hai = H. •
Definition. If G is a finite group, then the number of elements in G, denoted by|G|, is called the order of G.
The word “order” is used in two senses: the order of an element a ∈ G and the order|G| of a group G. Proposition 2.72 shows that the order of a group element a is equal to|hai|.
The following characterization of finite cyclic groups will be used to prove Theorem 3.122 showing that the multiplicative group of a finite field is cyclic.
Proposition 2.73. A group G of order n is cyclic if and only if, for each divisor d of n, there is at most one cyclic subgroup of order d.
Proof. Suppose that G= hai is a cyclic group of order n. We claim that han/di has order d. Clearly, (an/d)d = an = 1, and it suffices to show that d is the smallest such positive integer. If (an/d)r = 1, then n | (n/d)r, by Lemma 2.53.
Hence, there is some integer s with (n/d)r = ns, so that r = ds and r ≥ d.
To prove uniqueness, let C be a subgroup of G of order d; by Proposi-tion 2.71, the subgroup C is cyclic, say, C = hxi. Now x = am has order d, so that 1= (xm)d. Hence, n | md, by Lemma 2.53, and so md = nk for some integer k. Therefore, x = am = (an/d)k, so that C = hxi ⊆ han/di. Since both subgroups have the same order, however, it follows that C = han/di.
Conversely, define a relation on a group G by a ≡ b if hai = hbi. It is easy to see that this is an equivalence relation and that the equivalence class[a] of a∈ G consists of all the generators of C = hai. Thus, we denote [a] by gen(C),
C|gen(C)|, where the sum is over all the cyclic subgroups of G. But Corollary 2.70 gives|gen(C)| = φ(|C|). By hypothesis, G has at most one (cyclic) subgroup of any order, so that
n=X
C
|gen(C)| ≤X
d|n
φ (d)= n,
the last equality being Corollary 1.28. Therefore, for each divisor d of n, there must be a cyclic subgroup C of order d contributing φ (d) toP
C|gen(C)|. In particular, there must be a cyclic subgroup C of order n, and so G is cyclic. •
Here is a way of constructing a new subgroup from given ones.
Proposition 2.74. The intersection T
i∈I Hi of any family of subgroups of a group G is again a subgroup of G. In particular, if H and K are subgroups of G, then H∩ K is a subgroup of G.
Corollary 2.75. If X is a subset of a group G, then there is a subgrouphXi of G containing X that is smallest in the sense thathXi ≤ H for every subgroup H of G which contains X .
Proof. First of all, note that there exist subgroups of G which contain X ; for example, G itself contains X . DefinehXi = T
X⊆HH , the intersection of all the subgroups H of G which contain X . By Proposition 2.74,hXi is a subgroup of G; of course,hXi contains X because every H contains X. Finally, if H is any subgroup containing X , then H is one of the subgroups whose intersection ishXi; that is, hXi ≤ H. •
Note that there is no restriction on the subset X in the last corollary; in particular, X = is allowed. Since the empty set is a subset of every set, we have ⊆ H for every subgroup H of G. Thus, h i is the intersection of all the subgroups of G, one of which is{1}, and so h i = {1}.
Definition. If X is a subset of a group G, thenhXi is called the subgroup generated by X .
Example 2.76.
(i) If G = hai is a cyclic group with generator a, then G is generated by the subset X = {a}.
(ii) The symmetry group 6(πn)of a regular n-gon πn is generated by a, b, where a is a rotation about the origin by (360/n)◦and b is a reflection (see Theorem 2.63). These generators satisfy the conditions an = 1, b2 = 1, and bab= a−1, and 6(πn)is a dihedral group D2n.
The next proposition gives a more concrete description of the subgroup gen-erated by a subset.
Definition. Let X be a nonempty subset of a group G. Then a word on X is either the identity element or an element of G of the form w = x1e1xe22· · · xnen, where n≥ 1, xi ∈ X for all i, and ei = ±1 for all i.
Proposition 2.77. If X is a nonempty subset of a group G, thenhXi is the set of all the words on X .
Proof. We begin by showing that W , the set of all words on X , is a sub-group of G. By definition, 1 ∈ W . If w, w0 ∈ W , then w = x1e1x2e2· · · xnen
and w0 = y1f1y2f2· · · ymfm, where yj ∈ X and fj = ±1. Hence, ww0 = x1e1x2e2· · · xneny1f1y2f2· · · ymfm, which is a word on X , and so ww0 ∈ W . Finally, (w)−1 = x−en nxn−e−1n−1· · · x1−e1 ∈ W . Thus, X is a subgroup of G, and it clearly contains every element of X . We conclude thathXi ≤ W . For the reverse in-equality, we show that if S is any subgroup of G containing X , then S contains
every word on X . But this is obvious: since S is it a subgroup, it contain xe whenever x ∈ X and e = ±1, and it contains all possible products of such elements. Therefore, W ≤ S for all such S, and so W ≤T
S= hXi. • Proposition 2.78. Let a and b be integers and let A= hai and B = hbi be the cyclic subgroups of they generate.
(i) If A+ B is defined to be {a + b : a ∈ A and b ∈ B}, then A + B = hdi, where d = gcd(a, b).
(ii) A∩ B = hmi, where m = lcm(a, b).
Proof.
(i) It is straightforward to check that A+ B is a subgroup of (in fact, A+ B is precisely the set of all the linear combinations of a and b). By Proposition 2.71, the subgroup A+ B is cyclic: A + B = hdi, where d can be chosen to be the smallest non-negative number in A+ B. Thus, d is a common divisor of a and b, and it is the smallest such; that is, d= gcd(a, b).
(ii) If c ∈ A ∩ B, then c ∈ A and a | c; similarly, if c ∈ A ∩ B, then c ∈ B and b | c. Thus, every element in A ∩ B is a common multiple of a and b.
Conversely, every common multiple lies in the intersection. By Proposition 2.71, the subgroup A∩ B is cyclic: A ∩ B = hmi, where m can be chosen to be the smallest non-negative number in A∩ B. Therefore, m is the smallest common multiple; that is, m= lcm(a, b). •
Perhaps the most fundamental fact about subgroups H of a finite group G is that their orders are constrained. Certainly, we have|H| ≤ |G|, but it turns out that|H| must be a divisor of |G|. To prove this, we introduce the notion of coset.
Definition. If H is a subgroup of a group G and a∈ G, then the coset14a H is the subset a H of G, where
a H = {ah : h ∈ H}.
Of course, a = a1 ∈ a H. Cosets are usually not subgroups. For example, if a /∈ H, then 1 /∈ a H (otherwise 1 = ah for some h ∈ H, and this gives the contradiction a= h−1∈ H).
If we use the∗ notation for the operation in a group G, then we denote the coset a H by a∗ H, where
a∗ H = {a ∗ h : h ∈ H.}
14The cosets just defined are often called left cosets; there are also right cosets of H , namely, subsets of the form H a = {ha : h ∈ H}; these arise in further study of groups, but we shall work almost exclusively with (left) cosets.
In particular, if the operation is addition, then the coset is denoted by a+ H = {a + h : h ∈ H}.
Example 2.79.
(i) Consider the plane 2 as an (additive) abelian group and let ` be a line through the origin O (see Figure 2.17 on page 152); as in Example 2.65(iii), the line ` is a subgroup of 2. If β ∈ 2, then the coset β+ ` is the line
`0containing β which is parallel to `, for if r α∈ `, then the parallelogram law gives β+ rα ∈ `0.
r
+ r +
O
Figure 2.17 The Coset β+ `
(ii) If G= S3and H = h(1 2)i, there are exactly three cosets of H, namely:
H = {(1), (1 2)} = (1 2)H, (1 3)H = {(1 3), (1 2 3)} = (1 2 3)H, (2 3)H = {(2 3), (1 3 2)} = (1 3 2)H, each of which has size 2.
Observe, in our examples, that different cosets of a given subgroup do not overlap.
If H is a subgroup of a group G, then the relation on G, defined by a≡ b if a−1b∈ H,
is an equivalence relation on G. If a ∈ G, then a−1a = 1 ∈ H, and a ≡ a;
hence,≡ is reflexive. If a ≡ b, then a−1b∈ H; since subgroups are closed under inverses, (a−1b)−1= b−1a∈ H and b ≡ a; hence ≡ is symmetric. If a ≡ b and b≡ c, then a−1b, b−1c ∈ H; since subgroups are closed under multiplication, (a−1b)(b−1c)= a−1c∈ H, and a ≡ c. Therefore, ≡ is transitive, and hence it is an equivalence relation.
We claim that the equivalence class of a∈ H is the coset a H. If x ≡ a, then x−1a ∈ H, and Lemma 2.80 gives x ∈ x H = a H. Thus, [a] ⊆ a H. For the reverse inclusion, it is easy to see that if x = ah ∈ a H, then x−1a= (ah)−1a= h−1a−1a = h ∈ H, so that x ≡ a and x ∈ [a]. Hence, a H ⊆ [a], and so [a] = a H.
Lemma 2.80. Let H be a subgroup of a group G, and let a, b∈ G.
(i) a H = bH if and only if b−1a ∈ H. In particular, a H = H if and only if a∈ H.
(ii) If a H∩ bH 6= , then a H = bH.
(iii) |a H| = |H| for all a ∈ G.
Proof.
(i) This is a special case of Lemma 2.19, for cosets are equivalence classes. The second statement follows because H = 1H.
(ii) This is a special case of Proposition 2.20, for the equivalence classes com-prise a partition of X .
(iii) The function f: H → a H, given by f (h) = ah, is easily seen to be a bijection [its inverse a H → H is given by ah 7→ a−1(ah)= h]. Therefore, H and a H have the same number of elements, by Exercise 2.11 on page 101. • Theorem 2.81 (Lagrange’s Theorem). If H is a subgroup of a finite group G, then|H| is a divisor of |G|.
Proof. Let{a1H, a2H, . . . , atH} be the family of all the distinct cosets of H in G. Then
G= a1H∪ a2H∪ · · · ∪ atH,
because each g∈ G lies in the coset g H, and g H = aiH for some i . Moreover, Lemma 2.80(ii) shows that distinct cosets aiH and ajH are disjoint. It follows that
|G| = |a1H| + |a2H| + · · · + |atH|.
But|aiH| = |H| for all i, by Lemma 2.80(iii), so that |G| = t|H|. •
Definition. The index of a subgroup H in G, denoted by[G : H], is the number of cosets of H in G.
When G is finite, the index[G : H] is the number t in the formula |G| = t|H| in the proof of Lagrange’s theorem, so that
|G| = [G : H]|H|.
This formula shows that the index[G : H] is also a divisor of |G|.
Corollary 2.82. If H is a subgroup of a finite group G, then [G : H] = |G|/|H|.
Proof. This follows at once from Lagrange’s theorem. •
Recall Theorem 2.63: the symmetry group 6(πn)of a regular n-gon is a dihedral group of order 2n. It contains a cyclic subgroup of order n, generated by a rotation a, and the subgrouphai has index [6(πn): hai] = 2. Thus, there are two cosets:hai and bhai, where b is any symmetry outside of hai.
We now see why the orders of elements in S5, displayed in Table 2.3 on page 134, are divisors of 120. Corollary 2.144 explains why the number of permutations in S5of any given cycle structure is a divisor of 120.
Corollary 2.83. If G is a finite group and a∈ G, then the order of a is a divisor of|G|.
Proof. By Proposition 2.72, the order of the element a is equal to the order of the subgroup H = hai. •
Corollary 2.84. If a finite group G has order m, then am = 1 for all a ∈ G.
Proof. By Corollary 2.83, a has order d, where d | m; that is, m = dk for some integer k. Thus, am= adk = (ad)k = 1. •
Corollary 2.85. If p is a prime, then every group G of order p is cyclic.
Proof. Choose a ∈ G with a 6= 1, and let H = hai be the cyclic subgroup generated by a. By Lagrange’s theorem,|H| is a divisor of |G| = p. Since p is a prime and|H| > 1, it follows that |H| = p = |G|, and so H = G. •
Lagrange’s theorem says that the order of a subgroup of a finite group G is a divisor of|G|. Is the “converse” of Lagrange’s theorem true? That is, if d is a divisor of|G|, must there exist a subgroup of G having order d? The answer is
“no;” Proposition 2.97 shows that the alternating group A4is a group of order 12 which has no subgroup of order 6.
E
XERCISES2.46 (i) Define the special linear group by
SL(2, )= {A ∈ GL(2, ): det(A) = 1}.
Prove that SL(2, )is a subgroup of GL(2, ).
(ii) Prove that GL(2, ) is a subgroup of GL(2, ).
*2.47 Give an example of two subgroups H and K of a group G whose union H∪ K is not a subgroup of G.
*2.48 Let G be a finite group with subgroups H and K . If H ≤ K , prove that [G : H] = [G : K ][K : H].
2.49 If H and K are subgroups of a group G and if|H| and |K | are relatively prime, prove that H∩ K = {1}.