Chapter 8 Cosets In this chapter, we develop new tools which will allow us to extend to every finite group some of the results we already know for cyclic groups. More specifically, we will be able to generalize results regarding the order of a subgroup. We know that in the case of cyclic groups, the order of every element and hence of every subgroup must divide the order of the group. The tools we develop here will allow us to generalize this result to every finite group. 8.1 Definitions and Examples We begin with some definitions which apply to sets in general, not just groups. Definition 295 Let G be a group and H a subset of G (not necessarily a subgroup). Let a G. 1. The set denoted ah is defined to be: 2. The set denoted Ha is defined to be: ah = {ah h H} Ha = {ha h H} 3. The set denoted aha 1 is defined to be: aha 1 = { aha 1 h H } Remark 296 If the operation on G is written additively, then instead of ah we write a + H. Instead of Ha we write H + a. For the rest of this chapter, H will in fact be a subgroup of G. In this case, we have the following definitions: 89
90 CHAPTER 8. COSETS Definition 297 Let G be a group and H a subgroup of G. 1. ah is called the left coset of H containing a. 2. Ha is called the right coset of H containing a. 3. ah denotes the number of elements of ah. Ha denotes the number of elements of Ha. Example 298 Let G = { e, a, a 2, a 3} = a where a = 4. Let H = { e, a 2} = a 2. Then, Ha = { a, a 3}. ah = { a, a 3}. Ha 2 = { a 2, a 4} = { e, a 2} = H since a = 4. Similarly, a 2 H = H. a 3 H = { a 3, a } = ah. Similarly, Ha 3 = H. So, we see that the right cosets of H are H and Ha 2 and the left cosets are H and a 2 H. Let us make a few remarks about this example. Remark 299 The right and left cosets in the above example were the same. This is not usually the case. Only when the group G is Abelian will the left and right cosets be the same for sure. In the example above, G is Abelian since it is cyclic. Remark 300 The right and left cosets of H were equal to H when they were built using an element which was already in H. We will prove this is true as long as H is a subgroup of G. Example 301 Let G = D 4 = {R 0, R 90, R 180, R 270, H, V, D, D } and K = {R 0, R180} (we used K for the name of the subgroup because H is an element of D 4 ). Recall the operation table for D 4 was R 0 R 90 R 180 R 270 H V D D R 0 R 0 R 90 R 180 R 270 H V D D R 90 R 90 R 180 R 270 R 0 D D H V R 180 R 180 R 270 R 0 R 90 V H D D R 270 R 270 R 0 R 90 R 180 D D V H H H D V D R 0 R 180 R 90 R 270 V V D H D R 180 R 0 R 270 R 90 D D V D H R 270 R 90 R 0 R 180 D D H D V R 90 R 270 R 180 R 0 Find the left cosets of K. Using the above table, we have: R 0 K = K = R 180 K R 90 K = {R 90, R 270 } = R 270 K V K = {V, H} = HK DK = {D, D } = D K So, there are four distinct left cosets.
8.2. PROPERTIES OF COSETS 91 Example 302 Let G = (Z, +) and H = {6k k Z} = {0, ±6, ±12,...}. Find the left cosets of H. Here, the operation is addition. The left cosets are: 0 + H = H 1 + H = {6k + 1 k Z} 2 + H = {6k + 2 k Z} 3 + H = {6k + 3 k Z} 4 + H = {6k + 4 k Z} 5 + H = {6k + 5 k Z} 6 + H = {6k + 6 k Z} = H There are no more. The left cosets are H, 1 + H, 2 + H, 3 + H, 4 + H, 5 + H. In the above examples, we noticed that sometimes two different elements produce the same cosets, but not always. When do we know? Here are a few questions we try to address in the next section. 1. When is ah = bh or when is Ha = Hb? 2. Are all cosets subgroups? If not, which ones are? 3. Do different cosets of the same subgroup have the same number of elements in other words is ah = bh? 4. Can different cosets have elements in common? 5. Is ah = Ha? 8.2 Properties of Cosets We give the properties in the form of a lemma. The properties are given for left cosets. Similar properties hold for right cosets. Lemma 303 Let G be a group, H a subgroup of G and a and b elements of G. The following is true: 1. a ah 2. ah = H a H 3. ah = bh a bh 4. Either ah = bh or ah bh =
92 CHAPTER 8. COSETS 5. ah = bh ab 1 H 6. ah = bh 7. ah = Ha 8. ah = Ha H = aha 1 9. ah is a subgroup of G if and only if a H Proof. We prove some of these, the rest is left as homework. 1. Since H is a subgroup of G, e H. a = ae ah. 2. We need to prove both directions. ( = ) Suppose ah = H, prove a H. From #1, a ah but ah = H hence a H. ( = ) Suppose a H, prove ah = H by showing ah H and H ah. First, we prove that ah H. Let x ah then x = ah for some h H. Since H is a group, it is closed under the operation so if both a and h belong to H, x = ah H. hence, ah H. Next, we show that H ah that is if x H, then x ah that is x = ah for some h H. Since a H, a 1 H. Now, x = ex = aa 1 x = a ( a 1 x ). Both a 1 and x are in H, thus a 1 x H. Since x = a ( a 1 x ), this proves x ah. 3. We need to prove both directions. ( = ) Suppose ah = bh, prove a bh. a = ae ah = bh. ( = ) Suppose a bh, prove ah = bh. For this, we prove both inclusions. First, ah bh. Let x ah, then x = ah 1 for h 1 H but a bh, so a = bh 2 for h 2 H. Thus, x = bh 2 h 1 bh since h 2 h 1 H (H is closed under the operation). Next, bh ah. Let x bh then x = bh 1 for some h 1 H. but a bh, so a = bh 2 for h 2 H and therefore b = ah 1 2. It follows that x = ah 1 2 h 1 ah. 4. Either ah bh = or they have at least one element in common, say c. Thus c ah bh. From property 3, this means that ch = ah and ch = bh thus ah = bh. 5. From property 3, 6. See problems. ah = bh a bh a = bh for some h H ab 1 = h ab 1 H 7. From property 6, ah = eh = H = He = Ha
8.3. PROBLEMS 93 8. See problems 9. We need to prove both directions. ( = ) Suppose ah is a subgroup of G, show that a H. With the assumptions, e ah, so e = ah for some h H. Then, a = h 1 H. ( = ) Suppose a H, show that ah is a subgroup of G. From property 2, ah = H which is a subgroup of G. Remark 304 Properties 1, 4 and 6 of the lemma tell us that the cosets (right or left) of the subgroup H of a group G partition G into blocks of equal size. This will be useful in the next chapter. 8.3 Problems Do the following problems. 1. This problem is another approach to cosets. To answer the questions, you cannot use lemma 303. Let G be a group and H a subgroup of G. Define a relation on G, denoted H by a H b ab 1 H. (a) Prove H is an equivalence relation on G. (b) If we denote [a] the equivalence class of a under H, show that [a] = Ha. (c) Explain how parts a and b can be used to quickly prove parts 1-5 of lemma 303. 2. Let H be a subgroup of a finite group G. Prove that Ha = Hb by proving that there exists a bijection f : Ha Hb. Find this bijection. In particular you must prove that your candidate is indeed a bijection. 3. Prove part 8 of lemma 303. 4. Do # 1, 3, 4, 5, 6, 7, 8, 9 on page 149 of the book.