3 Cyclic Groups

3 Cyclic Groups

3 Cyclic groups Cyclic groups are a very basic class of groups: we have already seen some examples such as Zn. Cyclic groups are nice in that their complete structure can be easily described. The overall approach in this section is to define and classify all cyclic groups and to understand their subgroup structure. 3.1 Definitions and Examples The basic idea of a cyclic group is that it can be generated by a single element. Here are two motivat- ing examples: 1. The group of integers under addition can generated by 1. By this we mean that the element 1 can be combined with itself using only the group operation and inverses to produce the entire set of integers: if n > 0, then n = 1 + 1 + ··· + 1 from which we can easily produce −n and 0 = 1 + (−1). The element −1 could also be used to generate Z. th 2. The group U4 = f1, −1, i, −ig of 4 roots of unity under multiplication and also be generated from a single element i: 2 3 4 U4 = fi, i , i , i g = fi, −1, −i, 1g The same construction can be performed using −i. Now we formalize this idea. Lemma 3.1. Let G be a multiplicative group and let g 2 G. Then the set hgi := fgn : n 2 Zg = f..., g−1, e, g, g2,...g is a subgroup of G. Proof. Non-emptiness Clearly e 2 hgi (similarly g 2 hgi). Closure Every element of hgi has the form gk for some k 2 Z. It suffices to check that gk · gl = gk+l 2 hgi Inverses Since (gk)−1 = g−k 2 hgi, we are done. Definition 3.2. The subgroup hgi defined in Lemma 3.1 is the cyclic subgroup of G generated by g. The order of an element g 2 G is the order jhgij of the subgroup generated by g. G is a cyclic group if 9g 2 G such that G = hgi: we call g a generator of G. We now have two concepts of order. The order of a group is the cardinality of the group viewed as a set. The order of an element is the cardinality of the cyclic group generated by that element. Cyclic groups are the precisely those groups containing elements having the same order as that of the group: these are the generators of the cyclic group. 1 Examples Integers The integers Z form a cyclic group under addition. Z is generated by either 1 or −1. Note that this group is written additively, so that, for example, the subgroup generated by 2 is the group of even numbers under addition: h2i = f2m : m 2 Zg = 2Z Modular Addition For each n 2 N, the group of remainders Zn under addition modulo n is a cyclic group. It is also written additively, and is generated by 1. Typically Zn is also generated by several other elements. For example, Z4 = f0, 1, 2, 3g is generated by both 1 and 3: h1i = f1, 2, 3, 0g h3i = f3, 2, 1, 0g th n−1 Roots of Unity For each n 2 N, the n roots of unity Un = f1, z,..., z g form a cyclic group under multiplication. This group is generated by z, amongst others. The generators of Un are termed the primitive nth roots of unity. We shall see shortly that every cyclic group is isomorphic to either the integers or the modular in- tegers. Regardless, we will still write abstract cyclic groups multiplicatively. This, at least in part, is because the application of cyclic groups very often involves the study of cyclic subgroups of more complex groups, rather than of cyclic groups in their own right. 3.2 Classification of Cyclic Groups Our first goal is to describe all cyclic groups. Lemma 3.3. Every cyclic group is abelian. Proof. Let G = hgi. Then any two elements of G can be written gk, gl for some k, l 2 Z. But then gkgl = gk+l = gl+k = gl gk, and so G is abelian. Note that the converse is false: the Klein 4-group V is abelian but not cyclic. Before going any further, we need to distinguish between finite and infinite groups. To do this, sup- pose that G = hgi is cyclic, and consider the following set of natural numbers1 S = fm 2 N : gm = eg The distinction hinges on whether the set S is empty or not. 1Recall N = f1, 2, 3, . .g. 2 Infinite Cyclic Groups Theorem 3.4. Suppose G = hgi is cyclic and that the set S = fm 2 N : gm = eg is empty. Then G is an infinite group and we have an isomorphism G =∼ Z via m : Z ! G : x 7! gx Proof. We simply check the properties of m: Injectivity WLOG assume that x ≥ y. Then m(x) = m(y) =) gx = gy =) gx−y = e =) x − y = 0 =) x = y since S = Æ. Thus m is 1–1. Surjectivity m is onto by definition: every element of G has the form gx for some x 2 Z. Homomorphism m(x + y) = gx+y = gx gy = m(x)m(y). Finite Cyclic Groups Theorem 3.5. Suppose that G = hgi is cyclic and that S = fm 2 N : gm = eg is non-empty. By the ∼ 2 well-ordering of N, there exists a natural number n = min S. Then G = Zn via the isomorphism x m : Zn ! G : [x] 7! g In particular, n is the order of G. The proof is almost identical to that for infinite groups with two important differences: we need to check well-definition of m since the domain is a set of equivalence classes, and we need to apply the division algorithm to invoke the injectivity argument. Proof. Check the properties of m. Well-definition [x] = [y] =) y = x + ln for some l 2 Z. But then m([x]) = gx = gygln = gy = m([y]) since gn = e. Injectivity Suppose that m([x]) = m([y]). Then gx = gy =) gx−y = e Apply the division algorithm: x − y = ln + r for unique integers l, r with 0 ≤ r < n. But then e = gx−y = glngr = gr =) r = 0 since r < n. But then x − y = ln =) [x] = [y]. Surjectivity This is by definition of m. Homomorphism As in the infinite case, m([x + y]) = gx+y = gx gy = m([x])m([y]). 2 We use the equivalence class notation [x] = fx + ln : l 2 Zg to be clear regarding the nature of the elements of Zn. In particular, this stresses the fact that we need to check well-definition of m. 3 Examples th 1. The group of 7 roots of unity (U7, ·) is isomorphic to (Z7, +7) via the isomorphism k f : Z7 ! U7 : k 7! z7 2. The group 5Z = h5i is an infinite cyclic group. It is isomorphic to the integers via f : (Z, +) =∼ (5Z, +) : z 7! 5z 3. The real numbers R form an infinite group under addition. This cannot be cyclic because its @ cardinality 2 0 is larger than the cardinality @0 of the integers. The cyclic group of order n? Given the Theorem, we see that all cyclic groups of order n are isomorphic. This is usually summa- rized by saying that there is exactly one cyclic group of order n (up to isomorphism). We label this abstract group Cn. The fact that Cn is somewhat nebulous turns out to be convenient. For example: in some texts you may find Z6 referred to as the cyclic group of order 6. What, then, should we call the subgroup h2i = f2, 4, 6, 8, 10, 0g ≤ (Z12, +)? Certainly h2i is isomorphic to Z6 via f : 2z 7! z. It would be erroneous however to say that h2i equals Z6, since Z6 = f0, 1, 2, 3, 4, 5g has different elements. The notation C6 is neutral: it’s elements have no explicit interpretation. It is therefore a little safer to ∼ write h2i = C6. 3.3 Subgroups of cyclic groups We can very straightforwardly classify all the subgroups of a cyclic group. Theorem 3.6. All subgroups of a cyclic group are themselves cyclic. Proof. Let G = hgi and let H ≤ G. If H = feg is trivial, we are done. Otherwise, since all elements of H are in G, there must exist3 a smallest natural number s such that gs 2 H. We claim that H = hgsi. Let gm 2 H be a general element of H. The division algorithm says that there exist unique integers q, r such that m = qs + r and 0 ≤ r < s Therefore gm = gqs+r = (gs)qgr =) gr = (gs)−qgm 2 H since H is closed under · and inverses. But this forces r = 0, since r < s. In summary gm = (gs)q 2 hgsi and we are done. 3Well-ordering again. 4 Subgroups of infinite cyclic groups If G is an infinite cyclic group, then any subgroup is itself cyclic and thus generated by some element. It is easiest to think about this for G = Z. There are two cases: • The trivial subgroup: h0i = f0g ≤ Z. • Every other subgroup: hsi = sZ ≤ Z for s 6= 0. All these subgroups are isomorphic to Z via the isomorphism m : Z ! sZ : x 7! sx. It should also be clear that nZ ≤ mZ () n j m. In the language of an abstract cyclic group G = hgi the non-trivial subgroups have the form hgsi =∼ G via m : G ! hgsi : gx 7! gsx You should be comfortable comparing notations. Subgroups of finite cyclic groups As above, it is easier to consider the case where G = Zn first.

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