Subgroups and cyclic groups - Columbia University

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 = 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, ?n U (1), for every n, and U (1) C.

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(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 example, 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.

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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

g = {gn : n Z}.

We emphasize that we have written down the definition of g when the group operation is multiplication. If the group operation is written as addition, then we write:

g = {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 g generated by g is a subgroup of G.

(ii) g g .

(iii) If H G and g H, then g H. Hence g is the smallest subgroup of G containing g.

(iv) g is abelian.

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Proof. (i) First, g is closed under the group operation: given two elements of g , necessarily of the form gn, gm g , by the rules of exponents

gngm = gn+m g .

Next, 1 = g0 g . Finally, if gn g , then (gn)-1 = g-n g . Hence g G. (ii) Clearly g = g1 g . (iii) If H G and g H, then g ? g = g2 H, and by induction

gn+1 = g ? gn H

for all n N. Since 1 = g0 H by definition, and g-n = (gn)-1 H for all n N, we see that gn H for all n Z. Thus g H.

(iv) Given two elements gn, gm g ,

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

Thus g is abelian.

Example 2.3. (i) For any group G, 1 = {1} if the operation is multiplication, and 0 = {0} if the operation is addition.

(ii) Note that g = g-1 , since (g-1)n = g-n.

(iii) In Z, 1 = -1 = Z. For d N, d = -d = {nd : n Z}. Thus d (which is often written as dZ) is the subgroup of Z consisting of all multiples of d.

(iv) In Z/4Z, 0 = {0}, 1 = 3 = Z/4Z, and 2 = {0, 2} is a nontrivial proper subgroup of Z/4Z. In Z/5Z, 0 = {0}, and a = Z/5Z for all a = 0. In Z/6Z, 0 = {0}, 1 = 5 = Z/6Z, and 2 = 4 = {0, 2, 4} and 3 = {0, 3} are nontrivial proper subgroups of Z/6Z.

(v) In R, 2 = {2n : n Z} is the subset of R consisting of all integral

multiples of 2; it is sometimes denoted by 2Z. More generally, for any

t R, t = {nt : n Z} is the set of all integral multiples of t.

(vi) In Q,

1

= {1} and

-1 = {1, -1}. On the other hand,

1 2

=

2

=

{2n : n Z}, which is infinite.

(vii) In C, e2i/n = ?n.

(viii) In (Z/2Z) ? (Z/2Z), (0, 0) = {(0, 0)}. (1, 0) = {(0, 0), (1, 0)}.

(0, 1) = {(0, 0), (0, 1)}. (1, 1) = {(0, 0), (1, 1)}.

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To better get a sense of what the cyclic subgroup g 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 g 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 g = Z.

Proof. Define f : Z g by f (n) = gn. The discussion above shows that f is injective, since f (n) = f (m) gn = gm n = m. By the

definition of g , 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 #( g ) = 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) #( g ) = n, and g = {1, g, g2, . . . , gn-1}.

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