Subgroups and cyclic groups 1 Subgroups In many of the examples of groups we have given, one of the groups is a subset of another, with the same operations. This situation arises very often, and we give it a special name: Definition 1.1. A subgroup H of a group G is a subset H ⊆ G such that (i) For all h1; h2 2 H, h1h2 2 H. (ii) 1 2 H. (iii) For all h 2 H, h−1 2 H. It follows from (i) that the binary operation · on G induces by restriction a binary operation on H. Moreover, (H; ·) is again a group: clearly · remains associative when restricted to elements of H, 1 2 H is the identity for H, and for all h 2 H, the inverse h−1 for h viewed as an element of G is an inverse for h in H. We write H ≤ G to mean that H is a subgroup of G. Somewhat informally, one says that H with the binary operation induced from G is again a group. This assumes the closure property (i). Note that, if H with the induced operation has some identity element, it must automatically be the identity element of G (why?), and if h 2 H has an inverse in H, this inverse must be h−1, the inverse for h in G. Example 1.2. (i) For every group G, G ≤ G. If H ≤ G and H 6= G, we call H a proper subgroup of G. Similarly, for every group G, f1g ≤ G. We call f1g the trivial subgroup of G. Most of the time, we are interested in proper, nontrivial subgroups of a group. (ii) Z ≤ Q ≤ R ≤ C; here the operation is necessarily addition. Similarly, ∗ ∗ ∗ Q ≤ R ≤ C , where the operation is multiplication. Likewise, µn ≤ U(1), ∗ for every n, and U(1) ≤ C . 1 (iii) SOn ≤ On ≤ GLn(R), and SOn ≤ SLn(R) ≤ GLn(R). (iv) The relation ≤ has the following transitivity property: If G is a group, H ≤ G (H is a subgroup of G), and K ≤ H (K is a subgroup of H), then K ≤ G. (A subgroup of a subgroup is a subgroup.) (v) Here are some examples of subsets which are not subgroups. For exam- ∗ ∗ ple, Q is not a subgroup of Q, even though Q is a subset of Q and it is a group. Here, if we don't specify the group operation, the group operation ∗ ∗ on Q is multiplication and the group operation on Q is addition. But Q is not even closed under addition, nor does it contain the identity in Q (i.e. 0). For another example, Z=nZ is not a subgroup of Z. First, as correctly defined, Z=nZ is not even a subset of Z, since the elements of Z=nZ are equivalence classes of integers, not integers. We could try to remedy this by simply defining Z=nZ to be the set f0; 1; : : : ; n − 1g ⊆ Z. But the group operation in Z=nZ would have to be different than the one in Z. For example, if we make the convention that Z=nZ = f0; 1; : : : ; n−1g ⊆ Z, we would have to set 1 + (n − 1) = 0, which is not equal the integer n (in fact n is not even an element of Z=nZ as defined). Another example where subgroups arise naturally is for product groups: For all groups G1 and G2, f1g × G2 and G1 × f1g are subgroups of G1 × G2. More generally, if H1 ≤ G1 and H2 ≤ G2, then H1×H2 ≤ G1×G2. However, not all subgroups of G1 × G2 are of the form H1 × H2 for some subgroups H1 ≤ G1 and H2 ≤ G2. For example, (Z=2Z) × (Z=2Z) is a group with 4 elements: (Z=2Z) × (Z=2Z) = f(0; 0); (1; 0); (0; 1); (1; 1)g: The subgroups of the form H1 × H2 are the improper subgroup (Z=2Z) × (Z=2Z), the trivial subgroup f(0; 0)g = f0g × f0g, and the subgroups f0g × Z=2Z = f(0; 0); (0; 1)g; Z=2Z × f0g = f(0; 0); (1; 0)g: However, there is one additional subgroup, the \diagonal subgroup" H = f(0; 0); (1; 1)g ⊆ (Z=2Z) × (Z=2Z): It is easy to check that H is a subgroup and that H is not of the form H1 × H2 for some subgroups H1 ≤ Z=2Z, H2 ≤ Z=2Z. Lemma 1.3. If H1 and H2 are two subgroups of a group G, then H1 \H2 ≤ G. In other words, the intersection of two subgroups is a subgroup. 2 The proof is an exercise. It is not hard to check that the union of two subgroups of a group G is almost never a subgroup: If H1 and H2 are two subgroups of a group G, then H1 [ H2 ≤ G () either H1 ⊆ H2 or H2 ⊆ H1. Remark 1.4. Later we shall prove Cayley's theorem: If G is a finite group, then there exists an n 2 N such that G is isomorphic to a subgroup of Sn. Thus, the groups Sn contain the information about all finite groups up to isomorphism. Similarly, every finite group is isomorphic to a subgroup of GLn(R) for some n, and in fact every finite group is isomorphic to a subgroup of On for some n. For example, every dihedral group Dn is isomorphic to a subgroup of O2 (homework). 2 Cyclic subgroups In this section, we give a very general construction of subgroups of a group G. Definition 2.1. Let G be a group and let g 2 G. The cyclic subgroup generated by g is the subset n hgi = fg : n 2 Zg: We emphasize that we have written down the definition of hgi when the group operation is multiplication. If the group operation is written as addition, then we write: hgi = fn · g : n 2 Zg: To justify the terminology, we have: Lemma 2.2. Let G be a group and let g 2 G. (i) The cyclic subgroup hgi generated by g is a subgroup of G. (ii) g 2 hgi. (iii) If H ≤ G and g 2 H, then hgi ≤ H. Hence hgi is the smallest subgroup of G containing g. (iv) hgi is abelian. 3 Proof. (i) First, hgi is closed under the group operation: given two elements of hgi, necessarily of the form gn; gm 2 hgi, by the rules of exponents gngm = gn+m 2 hgi: Next, 1 = g0 2 hgi. Finally, if gn 2 hgi, then (gn)−1 = g−n 2 hgi. Hence hgi ≤ G. (ii) Clearly g = g1 2 hgi. (iii) If H ≤ G and g 2 H, then g · g = g2 2 H, and by induction gn+1 = g · gn 2 H 0 −n n −1 for all n 2 N. Since 1 = g 2 H by definition, and g = (g ) 2 H for all n n 2 N, we see that g 2 H for all n 2 Z. Thus hgi ≤ H. (iv) Given two elements gn; gm 2 hgi, gngm = gn+m = gm+n = gmgn: Thus hgi is abelian. Example 2.3. (i) For any group G, h1i = f1g if the operation is multipli- cation, and h0i = f0g if the operation is addition. (ii) Note that hgi = hg−1i, since (g−1)n = g−n. (iii) In Z, h1i = h−1i = Z. For d 2 N, hdi = h−di = fnd : n 2 Zg. Thus hdi (which is often written as dZ) is the subgroup of Z consisting of all multiples of d. (iv) In Z=4Z, h0i = f0g, h1i = h3i = Z=4Z, and h2i = f0; 2g is a nontrivial proper subgroup of Z=4Z. In Z=5Z, h0i = f0g, and hai = Z=5Z for all a 6= 0. In Z=6Z, h0i = f0g, h1i = h5i = Z=6Z, and h2i = h4i = f0; 2; 4g and h3i = f0; 3g are nontrivial proper subgroups of Z=6Z. (v) In R, h2πi = f2nπ : n 2 Zg is the subset of R consisting of all integral multiples of 2π; it is sometimes denoted by 2πZ. More generally, for any t 2 R, hti = fnt : n 2 Zg is the set of all integral multiples of t. ∗ 1 (vi) In Q , h1i = f1g and h−1i = f1; −1g. On the other hand, h 2 i = h2i = n f2 : n 2 Zg, which is infinite. ∗ 2πi=n (vii) In C , he i = µn. (viii) In (Z=2Z) × (Z=2Z), h(0; 0)i = f(0; 0)g. h(1; 0)i = f(0; 0); (1; 0)g. h(0; 1)i = f(0; 0); (0; 1)g. h(1; 1)i = f(0; 0); (1; 1)g. 4 To better get a sense of what the cyclic subgroup hgi looks like in general, we divide up into two cases: Case I: g has infinite order. In this case, we claim that gn = gm () n = m. Clearly, if n = m, then gn = gm. Conversely, suppose that gn = gm. Now either n ≤ m or m ≤ n. By symmetry, we can suppose that m ≤ n. Then, since gn = gm, gn(gm)−1 = 1. But then gn(gm)−1 = gng−m = gn−m = 1. Then n − m ≥ 0, but n − m > 0 is impossible since gk is never 1 for a positive integer k. Thus n − m = 0, i.e. n = m. In particular, we see that hgi is infinite (and hence that G is infinite). There is a more precise statement: ∼ Proposition 2.4. Suppose that g 2 G has infinite order. Then hgi = Z. n Proof.