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 ∈ H, h1h2 ∈ H. (ii) 1 ∈ H.

(iii) For all h ∈ H, h−1 ∈ 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 ∈ H is the identity for H, and for all h ∈ 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 ∈ 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, {1} ≤ G. We call {1} 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 {0, 1, . . . , n − 1} ⊆ 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 = {0, 1, . . . , n−1} ⊆ 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, {1} × G2 and G1 × {1} 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) = {(0, 0), (1, 0), (0, 1), (1, 1)}.

The subgroups of the form H1 × H2 are the improper subgroup (Z/2Z) × (Z/2Z), the trivial subgroup {(0, 0)} = {0} × {0}, and the subgroups

{0} × Z/2Z = {(0, 0), (0, 1)}, Z/2Z × {0} = {(0, 0), (1, 0)}. However, there is one additional subgroup, the “diagonal subgroup”

H = {(0, 0), (1, 1)} ⊆ (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 ∈ 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 ∈ G. The cyclic subgroup generated by g is the subset

n hgi = {g : n ∈ Z}.

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 = {n · g : n ∈ Z}.

To justify the terminology, we have:

Lemma 2.2. Let G be a group and let g ∈ G.

(i) The cyclic subgroup hgi generated by g is a subgroup of G.

(ii) g ∈ hgi.

(iii) If H ≤ G and g ∈ 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 ∈ hgi, by the rules of exponents

gngm = gn+m ∈ hgi.

Next, 1 = g0 ∈ hgi. Finally, if gn ∈ hgi, then (gn)−1 = g−n ∈ hgi. Hence hgi ≤ G. (ii) Clearly g = g1 ∈ hgi. (iii) If H ≤ G and g ∈ H, then g · g = g2 ∈ H, and by induction

gn+1 = g · gn ∈ H

0 −n n −1 for all n ∈ N. Since 1 = g ∈ H by deﬁnition, and g = (g ) ∈ H for all n n ∈ N, we see that g ∈ H for all n ∈ Z. Thus hgi ≤ H. (iv) Given two elements gn, gm ∈ hgi,

gngm = gn+m = gm+n = gmgn.

Thus hgi is abelian.

Example 2.3. (i) For any group G, h1i = {1} if the operation is multipli- cation, and h0i = {0} 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 ∈ N, hdi = h−di = {nd : n ∈ Z}. Thus hdi (which is often written as dZ) is the subgroup of Z consisting of all multiples of d. (iv) In Z/4Z, h0i = {0}, h1i = h3i = Z/4Z, and h2i = {0, 2} is a nontrivial proper subgroup of Z/4Z. In Z/5Z, h0i = {0}, and hai = Z/5Z for all a 6= 0. In Z/6Z, h0i = {0}, h1i = h5i = Z/6Z, and h2i = h4i = {0, 2, 4} and h3i = {0, 3} are nontrivial proper subgroups of Z/6Z. (v) In R, h2πi = {2nπ : n ∈ Z} is the subset of R consisting of all integral multiples of 2π; it is sometimes denoted by 2πZ. More generally, for any t ∈ R, hti = {nt : n ∈ Z} is the set of all integral multiples of t. ∗ 1 (vi) In Q , h1i = {1} and h−1i = {1, −1}. On the other hand, h 2 i = h2i = n {2 : n ∈ Z}, which is inﬁnite. ∗ 2πi/n (vii) In C , he i = µn. (viii) In (Z/2Z) × (Z/2Z), h(0, 0)i = {(0, 0)}. h(1, 0)i = {(0, 0), (1, 0)}. h(0, 1)i = {(0, 0), (0, 1)}. h(1, 1)i = {(0, 0), (1, 1)}.

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 ∈ G has inﬁnite order. Then hgi = Z. n Proof. Deﬁne f : Z → hgi by f(n) = g . The discussion above shows that f is injective, since f(n) = f(m) ⇐⇒ gn = gm ⇐⇒ n = m. By the deﬁnition of hgi, f is surjective. Thus f is a bijection. Finally, by the rules of exponents, f(n + m) = gn+m = gngm = f(n)f(m). Thus f is an isomorphism.

Case II: g has ﬁnite order n. In this case, we claim that the elements 1 = g0, g = g1, g2, . . . , gn−1 are all diﬀerent. As before, we suppose that ga = gb with 0 ≤ a, b ≤ n − 1. By symmetry we can assume that a ≤ b. Then 1 = gb(ga)−1 = gb−a. But 0 ≤ b − a ≤ b ≤ n − 1 < n. Since the order of g is n, no smaller positive power of g is the identity, so that we must have b − a = 0, i.e. a = b. Thus the powers 1, g, g2, . . . , gn−1 are all diﬀerent. Then gn = 1, gn+1 = gng = g, gn+2 = gng2 = g2,... . In other words, the sequence of powers cycles back over the same values. Moreover, gn−1 = g−1, gn−2 = g−2, and so the negative powers of g look like g−n = 1, g−(n−1) = g−n+1 = g, . . . , g−2 = gn−2, g−1 = gn−1. In other words, the sequence of powers looks the same in the negative direction as well. From this it is easy to see that #(hgi) = n, in other words that the order of an element of ﬁnite order is the same as the order of the cyclic subgroup that it generates, connecting the two diﬀerent meanings of the word order. We will prove all of this more carefully soon, but we will just state the main result now: Proposition 2.5. Suppose that g ∈ G has order n. Then: (i) #(hgi) = n, and hgi = {1, g, g2, . . . , gn−1}.

5 (ii) gN = 1 ⇐⇒ n divides N. ∼ (iii) hgi = Z/nZ. Deﬁnition 2.6. A group G is a cyclic group if there exists a g ∈ G such that G = hgi. In this case, g is called a generator of G.

For example, Z is cyclic; the possible generators are 1 and −1. Z/nZ is 2πi/n cyclic; 1 and −1 are generators, but there may be others. µn is cyclic; e is a generator. However, most groups are not cyclic. For example, we have seen that every cyclic group is abelian, so that a non-abelian group is not cyclic. Also, if G is a cyclic group, then either G is ﬁnite or there is an isomorphism from Z to G, hence a bijection from Z to G. Inﬁnite sets X for which there exists a bijection from Z to X are called countable (because there also exists a bijection from N to X). For example, N, Z, and Q are countable, but R and C are not. Thus, just by a counting argument, R and C are not cyclic groups. It is not hard to show that Q is not cyclic (homework), and a similar method shows that R and C are not cyclic without using counting properties of inﬁnite sets. For an example of a ﬁnite abelian group which is not cyclic, consider (Z/2Z)×(Z/2Z). We have listed the cyclic subgroups of (Z/2Z) × (Z/2Z) corresponding to all possible elements of (Z/2Z) × (Z/2Z), and none of these is equal to (Z/2Z) × (Z/2Z). Thus (Z/2Z) × (Z/2Z) is not cyclic. There is the following easy criterion for when a ﬁnite group is cyclic: Lemma 2.7. Let G be a ﬁnite group with #(G) = n. Then G is cyclic ⇐⇒ there exists a g ∈ G such that the order of g is n. Proof. We use the fact mentioned above, which we shall prove carefully later, that, if the order of g is n, then #(hgi) = n. So, if G is cyclic, say G = hgi, then n = #(G) = #(hgi), and hence the order of g is n. Conversely, if the order of g is n, then #(hgi) = n. Since hgi ⊆ G and they both have the same number of elements, we must have hgi = G, and therefore G is cyclic.

Finally, we say a few words about the subgroups of a group G generated by more than one element. Given a group G and elements g1, . . . , gk ∈ G, we can deﬁne the subgroup generated by g1, . . . , gk to be the smallest subgroup of G containing g1, . . . , gk. We will write this subgroup as hg1, . . . , gki. How- ever, if G is not abelian, then there is no simple description of the elements

6 of hg1, . . . , gki. The most that we can say is that every element looks like a n1 nt product ga1 ··· gat of powers of the elements g1, . . . , gk, where the ai are just some elements in {1, . . . , k} and t can be arbitrarily large in some examples. Here we can take all powers ni to be ±1 by allowing repeats, i.e. by letting gai = gai+1 as needed. In general, we don’t know if any simpliﬁcation is possible. However, if G is abelian, or more generally if the gi commute with each other (i.e. for all i, j with 1 ≤ i, j ≤ k, gigj = gjgi), then it is easy to check that

n1 nk hg1, . . . , gki = {g1 ··· gk : n1, . . . , nk ∈ Z} is a subgroup of G and it is the smallest subgroup of G containing g1, . . . , gk. This should look more familiar if we write the operation on G as +, so that G is abelian by convention. Then

hg1, . . . , gki = {(n1 · g1) + ··· + (nk · gk): n1, . . . , nk ∈ Z}.

Thus, the group generated by g1, . . . , gk is analogous to the span of k vectors in linear algebra.

7