Partitions of Groups

Transcription

1 Partitions of Groups S. F. Ellermeyer November 15, 2006 Suppose that G is a group and suppose that H is a subgroup of G. For any given element a 2 G, we may consider the left coset ah = fah j h 2 Hg and the right coset Ha = fha j h 2 Hg. If G is an abelian group and a is any member of H, then obviously the sets ah and Ha are the same. If additive notation (rather than multiplicative notation) is in order, then we write a + H instead of ah and we write H + a instead of Ha. Example 1 Let G be the group of integers under addition and let H be the group of even integers under addition. Thus and G = Z = f: : : ; 3; 2; 1; 0; 1; 2; 3; : : :g H = 2Z = f: : : ; 6; 4; 2; 0; 2; 4; 6; : : :g. We are well familiar with the fact that H is a subgroup of G. Let us nd all of the left cosets of H in G. These are all of the sets of the form a + H where a 2 G. We see that 0 + H = f: : : ; 6; 4; 2; 0; 2; 4; 6; : : :g 1 + H = f: : : ; 5; 3; 1; 1; 3; 5; 7; : : :g 2 + H = f: : : ; 4; 2; 0; 2; 4; 6; 8; : : :g 3 + H = f: : : ; 3; 1; 1; 3; 5; 7; 9; : : :g. etc. It can thus be seen that if a is any even integer, then a + H = f: : : ; 6; 4; 2; 0; 2; 4; 6; : : :g = H 1

2 and if a is any odd integer, then a + H = f: : : ; 5; 3; 1; 1; 3; 5; 7; : : :g. Thus H has only two distinct left cosets in G. (H also has only two distinct right cosets in G and they are the same as the left cosets because G is abelian). The only cosets of H in G are 0 + H and 1 + H. Note that 0 + H = H is, of course a subgroup of G: However, 1 + H is not a subgroup of G. In fact 1 + H is simply the set of all odd integers. Another important observation is that f0 + H; 1 + Hg is a partition of G. This means that and (0 + H) [ (1 + H) = G (0 + H) \ (1 + H) = ;. As will be proved, it is always true (for any group G and any subgroup, H, of G) that the set of distinct left cosets of H form a partition of G (and likewise with right cosets, although left and right cosets may yield di erent partitions if G is not abelian). Example 2 Let G = U 8 = f1; 3; 5; 7g under multiplication and let H = f1; 5g. Then H is a subgroup of G. The left cosets of H in G are 1H = 1 f1; 5g = f1; 5g = H 3H = 3 f1; 5g = f3; 7g 5H = 5 f1; 5g = f5; 1g = f1; 5g = H 7H = 7 f1; 5g = f7; 3g = f3; 7g. Thus 1H = f1; 5g and 3H = f3; 7g are the only two left cosets of H in G (because 5H = 1H and 7H = 3H). Observe that 1H and 3H form a partition of G. Also observe that 3H is not a subgroup of G. Example 3 Let G = Z 6 under addition and let H = f0; 2; 4g. Then H is a subgroup of G and the left cosets of H in G are 0 + H = f0; 2; 4g = H 1 + H = 1 + f0; 2; 4g = f1; 3; 5g 2 + H = 2 + f0; 2; 4g = f2; 4; 0g = H 3 + H = 3 + f0; 2; 4g = f3; 5; 1g = f1; 3; 5g 4 + H = 4 + f0; 2; 4g = f4; 0; 2g = H 5 + H = 5 + f0; 2; 4g = f5; 1; 3g = f1; 3; 5g. 2

3 Thus the only distinct left cosets of H in G are 0 + H and 1 + H. These sets form a partition of G. Example 4 We now consider an example that involves a non abelian group. In particular we take G = D 3 (whose Cayley table is shown below) and H = fr 0 ; F a g. Note that H is a subgroup of G. The left cosets of H in G are * R 0 R 1 R 2 F a F b F c R 0 R 0 R 1 R 2 F a F b F c R 1 R 1 R 2 R 0 F c F a F b R 2 R 2 R 0 R 1 F b F c F a F a F a F b F c R 0 R 1 R 2 F b F b F c F a R 2 R 0 R 1 F c F c F a F b R 1 R 2 R 0 R 0 H = R 0 fr 0 ; F a g = fr 0 ; F a g = H R 1 H = R 1 fr 0 ; F a g = fr 1 ; F c g R 2 H = R 2 fr 0 ; F a g = fr 2 ; F b g F a H = F a fr 0 ; F a g = ff a ; R 0 g = fr 0 ; F a g = H F b H = F b fr 0 ; F a g = ff b ; R 2 g = fr 2 ; F b g F c H = F c fr 0 ; F a g = ff c ; R 1 g = fr 1 ; F c g. Thus H has three distinct left cosets in G. They are R 0 H = fr 0 ; F a g, R 1 H = fr 1 ; F c g, and R 2 H = fr 2 ; F b g. These three cosets form a partition of G. Exercise 5 For the group D 3, nd all of the right cosets of the subgroup H = fr 0 ; F a g. You will see that they are di erent from the corresponding left cosets (although they still form a partition of D 3 ). Exercise 6 For the group D 3, nd all of the left and right cosets of the subgroup H = fr 0 ; R 1 ; R 2 g. 1 Basic Facts About Cosets Having considered several examples of subgroups and their cosets, we will now prove that the various things that those examples were observed to have 3

4 in common are true in general. We will limit our attention to groups of nite order. Speci cally, we will prove that if G is a group of nite order and H is a subgroup of G, then the collection of distinct left cosets of H partition G into sets which all have the same number of members. (The same is true if we consider right cosets instead of left cosets, although the partition obtained might be di erent if the group G is not abelian.) Recall that a partition of a set, A, is a collection of non empty subsets of A such that each member of A belongs to exactly one of these subsets. Theorem 7 Let G be a group of nite order and let H be a subgroup of G. Then each coset of H contains the same number of members as H and the collection of distinct cosets of H form a partition of G. Proof. First we will consider three simple cases: the case that jgj = 1, the case that jgj > 1 and jhj = 1, and the case that jgj > 1 and jhj = jgj. Suppose that jgj = 1. Then G is the trivial group and the only subgroup of G is G. In addition, G is the only coset of G and fgg is obviously a partition of G. Suppose that jgj > 1 and jhj = 1. Then H is the trivial subgroup of G and the cosets of H are all of the singleton sets fag where a is any member of G. The collection ffag j a 2 Gg is obviously a partition of G. Now suppose that jgj > 1 and jhj = jgj. Then H = G and the only coset of H is G. It is obvious that fgg is a partition of of G. Having disposed of the simple cases, we may now assume that jgj > 1 and that 1 < jhj < jgj. Speci cally, let us suppose that jgj = n (where n > 1) and that jhj = m (where 1 < m < n). We can list the members of H as H = fa 0 ; a 1 ; a 2 ; : : : ; a m 1 g where a 0 = e. (Note that e 2 H because H is a subgroup of G). If a is any member of G, then the coset ah is ah = faa 0 ; aa 1 ; aa 2 ; : : : ; aa m 1 g. It is clear that ah cannot contain more than m members. Also, since aa i = aa j implies that a 1 (aa i ) = a 1 (aa j ) which implies that a i = a j, we see that ah cannot contain fewer than m members. Therefore ah contains exactly m members. We have now proved that every coset of H contains exactly m members (the same number of members that H contains). We 4

5 will complete the proof by showing that each member of G is a member of exactly one of the cosets of H. Let a 2 G. Then, since e 2 H, it is clear that a is a member of the coset ah. Suppose now that a is a member of some coset bh. Then there exists some h 2 H such that a = bh. We will show that the cosets ah and bh are actually the same. To do this, we must show that ah bh and that bh ah. First we will show that ah bh. To do this, let x 2 ah. Then there exists some h 1 2 H such that x = ah 1. Since a = bh, we obtain x = ah 1 = bhh 1 and since hh 1 2 H (because H is a subgroup) we see that x 2 bh. Therefore ah bh. We will now show that bh ah. To do this, suppose that x 2 bh. Then there exists some h 1 2 H such that x = bh 1. Since a = bh, we see that b = ah 1 and hence that x = bh 1 = ah 1 h 1. Since h 1 h 1 2 H, we see that x 2 ah. Therefore bh ah. We have now shown that each member of G belongs to exactly one of the distinct cosets of H. 2 Corollaries We now give some corollaries that result from Theorem 7. The rst of these corollaries is LaGrange s Theorem. 5

6 Joseph Louis LaGrange ( ) Corollary 8 (LaGrange s Theorem) If G is a group of nite order and H is a subgroup of G, then jgj is divisible by jhj. Proof. We know that the distinct cosets of H all have the same number of members as H, that these cosets are pairwise disjoint (meaning that any two distinct cosets do not intersect each other), and that the union of these cosets if all of G. This implies that jgj must be divisible by jhj. Example 9 Suppose that G is a group of order 20. Is it possible that G has a subgroup of order 12? The answer is no because LaGrange s Theorem tells us that the only possible orders of subgroups of G are 1, 2; 4, 5, 10, and 20. 6

7 Corollary 10 Suppose that G is group and that jgj = p where p is a prime number. Then G is cyclic and every non identity element of G generates G. Proof. Let a be any member of G with a 6= e. The cyclic subgroup hai must contain at least two elements because a 6= e. Also, by LaGrange s Theorem, p must be divisible by jhaij. Since p is a prime number and jhaij 6= 1, it must be the case that jhaij = p. Therefore hai = G. Example 11 Since 5 is a prime number, any group of order 5 is cyclic and is generated by each of its non identity elements. This means that all groups of order 5 are isomorphic to each other. (Hence there is essentially only one group of order 5.) Exercise 12 Is it possible for a group of order 111 to contain a subgroup of order 37? If so, can you give a speci c example of such a group? Corollary 13 Suppose that G is a group of nite order and suppose that a is any element in G. Then a jgj = e. Proof. If a = e, then it is obvious that a jgj = e. Thus suppose that a 6= e. Then hai is a subgroup of G that contains at least two members. In addition, we know that a jhaij = e. By LaGrange s Theorem, jgj is divisible by jhaij. This means that jgj = q jhaij for some positive integer q. It follows that a jgj = a qjhaij = a jhaij q = e q = e. Exercise 14 Verify by direct computation that every member, a, of the multiplicative group U 10 = f1; 3; 7; 9g satis es a 10 = 1. 7