# Examples: The (left or right) cosets of the subgroup H = 11 in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are

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1 Cosets Let H be a subset of the group G. (Usually, H is chosen to be a subgroup of G.) If a G, then we denote by ah the subset {ah h H}, the left coset of H containing a. Similarly, Ha = {ha h H} is the right coset of H containing a. In both situations, we call a the coset representative. Examples: The (left or right) cosets of the subgroup H = 11 in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are 1H = {1, 11} = 11H, 7H = {7, 17} = 17H, 13H = {13, 23} = 23H, 19H = {19, 29} = 29H. The (left or right) cosets of the subgroup 4Z in the additive group Z are 4Z = 0 + 4Z = ±4 + 4Z = ±8 + 4Z = ±12 + 4Z = L, L = 3 + 4Z = 1 + 4Z = 5 + 4Z = 9 + 4Z = L, L = 2 + 4Z = 2 + 4Z = 6 + 4Z = Z = L, L = 1 + 4Z = 3 + 4Z = 7 + 4Z = Z = L.

2 The left cosets of the subgroup {R 0,H} in D 4 are R 0 { R 0,H} = {R 0,H} = H{ R 0,H}, R 90 { R 0,H} = {R 90, D } = D {R 0,H}, R 180 { R 0,H} = {R 180,V} = V {R 0,H}, R 270 { R 0,H} = {R 270, D} = D{R 0,H}. There are a number of patterns evidenced in these examples that are true more generally. Theorem Let H be a subgroup of G and suppose a,b G. Then the left cosets of H in G satisfy the following properties (with correspondingly similar results for the right cosets of H): 1. a ah, 2. ah = H a H, 3. either ah = bh or ah bh =, 4. ah = bh a 1 b H, 5. ah = bh, 6. ah = Ha H = aha 1, 7. ah is a subgroup of G a H. Proof 1. Trivial. 2. ( ) a = ae ah = H. ( ) a H a 1 H and for every h H, ah H by closure in H, whence ah H ; and h = a(a 1 h) ah, whence ah H. Thus, ah = H.

3 3. Suppose ah bh. Then there is some x that lies in both cosets; that is, there are elements y,z H for which x = ay = bz. Thus, ah = (bzy 1 )H = b(zy 1 H ) = bh (by #2 above, since zy 1 H ). 4. ah = bh H = a 1 bh a 1 b H by #2 above. 5. The function from the set ah to the set bh given by ah a bh is clearly onto. It is one-to-one since bh 1 = bh 2 h 1 = h 2 ah 1 = ah 2. So ah = bh. 6. ah = Ha aha 1 = (ah )a 1 = (Ha)a 1 = He = H. 7. ah G e ah ah H ah = H a H (using #3 and #2 above). Conversely, by #2, a H ah = H G. // Properties #1, #3, and #5 above combine to assert that G is partitioned by the left cosets of H into subsets of the same size. In fact, we can define a relation on the elements of G by a ~ b when a and b lie in the same coset, i.e., ah = bh. This relation is reflexive (#1), symmetric, and transitive (by #3), so it is an equivalence relation on G. (See pp ) Examples: Let Π be a plane through the origin in R 3. Then Π is an additive subgroup of R 3. The cosets of Π represent all the planes in R 3 parallel to Π. The cosets of SL(2, R) partition GL(2, R) into subsets consisting of all matrices with a given determinant.

4 Lagrange s Theorem One of the most versatile results in all of group theory is Lagrange s Theorem Let G be a finite group and suppose that H < G. Then H divides G. Moreover, the number of distinct left (or right) cosets of H in G, called the index of H in G and denoted G:H, equals G / H. Proof As we mentioned in the comment following the previous theorem, properties #1, #3 and #5 imply that G is the disjoint union of the distinct cosets of H, all of which have the same number of elements. Since H is one of these cosets, it follows that G is the disjoint union of G:H cosets, all of which have size H. That is, G = G:H H, from which the final claim of the theorem follows. // Corollary The order of any element of a finite group divides the order of the group. Proof For any element a of a group, a = a. Corollary Groups of prime order must be cyclic. Proof Let a be an element of a group G of prime order p which is not the identity element. Then a divides p but a 1. So a = p and G = a. //

5 Corollary If G is a finite group and a G, then a G = e. Proof a must divide G. // Corollary [Fermat s Little Theorem (F lt)] If p is a prime number and a is any integer, then a p mod p = amod p. Proof If a is a multiple of p, then both amod p and a p mod p equal 0. If a is not a multiple of p, then both these numbers are elements of the multiplicative group U(p), which has order p 1. Therefore, a p 1 mod p = 1. Multiplication by a yields a p mod p = amod p. //

6 Lagrange s Theorem says that a group of order n can only have subgroups whose orders are divisors of n. Note however that the converse of this theorem need not be true: Proposition A 4 = 12, but A 4 has no subgroup of order 6. Proof Suppose H were a subgroup of order 6. Then since A 4 :H = 2, we can conclude that H has exactly two distinct cosets in A 4. Let α represent any one of the eight elements of order 3 in A 4 ( α 5,α 6,K,α 12 in the Cayley table of A 4 on p. 105). Then at least one pair of the three cosets H,αH,α 2 H must coincide and regardless which pair it is, we must then have that α H. But then H must contain all eight elements of order 3, which is impossible. //

7 Lagrange s Theorem gives us a powerful tool for investigating the nature of finite groups. For example: Theorem A group whose order is twice an odd prime p must be isomorphic to either Z 2 p or D p. Proof If p is an odd prime and G has order 2p, we have two cases to consider: (1) G has an element of order 2p. Then clearly G Z 2 p. (2) G has no element of order 2p. By Lagrange s Theorem, any element different from e must have order 2 or order p. If every element had order 2, then every element would be its own inverse and we could write xy = (xy) 1 = y 1 x 1 = yx, showing that G is Abelian; moreover, this would allow any two such elements to generate a subgroup {e,x, y,xy} of order 4. But 4 does not divide 2p, so this is not possible. Therefore, not every element has order 2 and there must be some element a of order p. Now let b be an element not in a. Then the coset b a is distinct from a, and since G: a = 2, these are the only two distinct cosets. So b 2 a must conicide with one of them. Since b 2 a = b a implies b a = a, which is false, we must have b 2 a = a b 2 a. So b 2 divides a = p. It

8 follows that b 2 = 1, for if b 2 = p, the fact that b 2p together with b 2 b <G means that b = p, and since b 2 a, b = b p+1 = (b 2 ) p +1 2 = (a k ) p +1 2 a, which we know to be false. Thus b 2 = 1, and b must have order 2. Next, ab a (since ab a b a ), so the same argument we used in the last paragraph will show that ab = 2. Also, ab = (ab) 1 = b 1 a 1 = ba 1. Therefore, G = {e,a,a 2,,a p 1,b,ba,ba 2,,ba p 1 } and the group operation is determined by the relations between a and b. In particular, a k b = a k 1 (ab) = a k 1 (ba 1 ) = a k 1 ba 1 = a k 2 (ab)a 1 = a k 2 (ba 1 )a 1 = a k 2 ba 2 =L = ba k so the product of any two elements in G is completely determined, as follows:

9 a k a l = a k+l ; (ba k )a l = ba k+l ; a k (ba l ) = ba l k ; (ba k )(ba l ) = a l k. In other words, every group of order 2p without an element of order 2p must have this structure: all such groups are isomorphic. As D p is a group of this form, all such groups are isomorphic to D p. // Corollary S 3 D 3. //

10 Orbits and Stabilizers We saw from Cayley s Theorem how every group is a group of permutations on some set. Suppose that G is a group of permutations of the objects in the set S. Then for each s S, the set orb G (s) = {ϕ(s) S ϕ G}, called the orbit of s under G, is the subset of elements in S to which s is moved under the action of G. We also define stab G (s) = {ϕ G ϕ(s) = s}, called the stabilizer of s in G, to be the subset of G consisting of those elements that fix s. As this (nonempty) subset of G is clearly closed under composition and under the taking of inverses, it is a subgroup of G.

11 Theorem Let G be a finite group of permutations of the set S. Then for any s S, G = orb G (s) stab G (s). Proof By Lagrange s Theorem, G / stab G (s) is the number of left cosets of stab G (s) in G. So we are done if we can establish a bijection between the left cosets of stab G (s) and the orbit of s under G. To this end, define the function f as ϕstab G (s) aϕ(s). Before we continue, we must first show that this is well-defined; that is, the coset ϕstab G (s) may be equal to the coset ψ stab G (s), so we must ensure that ϕ(s) = ψ (s). But this is always true, for ϕstab G (s) = ψ stab G (s) ϕ 1 ψ stab G (s), which means that ϕ 1 ψ (s) = s, or ϕ(s) = ψ (s). The function f is one-to-one because we can reverse this last argument, and it is onto because if t is in the orbit of s, then t =ϕ(s) for some ϕ G, whence f maps the coset ϕstab G (s) to t =ϕ(s). //

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