Chapter 7 Cosets and Lagrange's Theorem Properties of Cosets Lagrange's Theorem and Consequences An Application of Cosets to Permutation Groups Ahmed EL-Mabhouh Abstract Algebra I Properties of Cosets Definition (Coset of H in G) Let G be a group and let H be a subset of G: For any a 2 G; we define the following sets: aH = fahjh 2 Hg; Ha = fhajh 2 Hg; aHa−1 = aha−1jh 2 H : When H is a subgroup of G; aH is called the left coset of H in G containing a. Ha is called the right coset of H in G containing a. the element a is called the coset representative of aH (or Ha). jaHj; (jHaj) denotes the number of elements in the set aH; (Ha) respectively. Note that if G is Abelian, then aH = Ha Ahmed EL-Mabhouh Abstract Algebra I Properties of Cosets Example (1) Let G = S3 and H = f(1); (13)g: Then the left cosets of H in G are: (1)H = H (12)H = f(12); (12)(13)g = f(12); (132)g = (132)H (13)H = f(13); (1)g = H (23)H = f(23); (23)(13)g = f(23); (123)g = (123)H Example (2) Let K = fR0; R180g and G = D4: Find the distinct left cosets of K in D4: Proof. Ahmed EL-Mabhouh Abstract Algebra I Properties of Cosets Example (3) Let H = f0; 3; 6g in Z9 under addition. Then the distinct left cosets of H in Z9 are: 0 + H = f0; 3; 6g = 3 + H = 6 + H 1 + H = f1; 4; 7g = 4 + H = 7 + H 2 + H = f2; 5; 8g = 5 + H = 8 + H Notes From the three examples above, we notice that: 1 aH is not necessarily a subgroup of G. 2 aH = bH ; a = b 3 aH 6= Ha in general. The following lemma summarise the properties of left (right) cosets. Ahmed EL-Mabhouh Abstract Algebra I Properties of Cosets Lemma (Properties of Cosets) Let H be a subgroup of G; and let a; b 2 G: Then 1 a 2 aH 2 aH = H if and only if a 2 H 3 aH = bH if and only if a 2 bH 4 aH = bH or aH \ bH = ? −1 5 aH = bH if and only if a b 2 H 6 jaHj = jbHj −1 7 aH = Ha if and only if H = aHa 8 aH is a subgroup of G if and only if a 2 H: Ahmed EL-Mabhouh Abstract Algebra I Properties of Permutations Proof. Ahmed EL-Mabhouh Abstract Algebra I Properties of Permutations Note that properties 1, 4, and 6 of the lemma guarantee that the left cosets of a subgroup H of G partition G into blocks of equal size. Example (4) 3 Let G = R and H any plane through the origin. Any left coset of H in G is of the form (a; b; c) + H which is the plane passing through the point (a; b; c) and parallel to H: Thus, the left cosets of H partition the 3-space into planes parallel to H. Example (5) If G = GL(2; R) and H = SL(2; R) and A is any matrix in G; then the coset AH is the set of all 2 × 2 matrices with the same determinant as A. Thus, 2 0 H is the set of all 2 × 2 matrices of determinant 2 0 1 Ahmed EL-Mabhouh Abstract Algebra I Properties of Permutations Example (6) Find the distict cosets of H = f1; 15g in G = U(32) = f1; 3; 5; 7; 9; 11; 13; 15; 17; 19; 21; 23; 25; 27; 29; 31g: Proof. We begin with H, so 1:H = H = f1; 15g Choose an element not in H; say 3. A new left coset is 3H = f3; 13g Choose an element not in H and not in 3H, say 5, another left coset is 5H = f5; 11g and so on. We have 8 distinct left cosets of H in G write them all. Ahmed EL-Mabhouh Abstract Algebra I Lagrange's Theorem and Consequences Theorem (1, Lagrange's Theorem) If G is a finite group and H is a subgroup of G; then jHj divides jGj Moreover, the number of distinct left (right) cosets of H in G is jGj=jHj Proof. Let a1H; a2H;:::; ar H denote the distinct left cosets of H in G: We show r = jGj=jHj; or jGj = rjHj: Consequently, jHj j jGj: Let a 2 G: Then aH = ai H for some i = 1; 2; :::; r: By Lemma part (1), a 2 aH = ai H: Hence, G = a1H [···[ ar H Since ai H \ aj H = φ for all i 6= j; we have jGj = ja1Hj + ja2Hj + ··· + jar Hj By part (6), jHj = jai Hj and so jGj = jHj + jHj + ··· + jHj = r jHj Ahmed EL-Mabhouh Abstract Algebra I Consequences of Lagrange's Theorem Definition (Index of H in G) The index of a subgroup H in G is the number of distinct left cosets of H in G and is denoted by jG : Hj: Corollary (1) If G is a finite group and H is a subgroup of G; then jG : Hj = jGj=Hj Corollary (2) In a finite group, the order of each element of the group divides the order of the group. Corollary (3) A group of prime order is cyclic. Ahmed EL-Mabhouh Abstract Algebra I Consequences of Lagrange's Theorem Corollary (4) Let G be a finite group, and let a 2 G: Then, ajGj = e Proof. By Corollary 2, jaj j jGj; so jGj = jajk for some positive integer k: Then ajGj = ajajk = (ajaj)k = ek = e Corollary (5, Fermat's Little Theorem) For every integer a and every prime p; ap mod p = a mod p Proof. By the division algorithm, a = pm + r; where 0 ≤ r < p: Therefore, a mod p = r Hence, ap mod p = r p mod p: So, we only show that r p mod p = r: Since r 2 U(p) and U(p) is a group of order p − 1; by corollary (4), r p−1 mod p = 1 and, therefore, r p mod p = r Ahmed EL-Mabhouh Abstract Algebra I Consequences of Lagrange's Theorem Example (7, The Converse of Lagrange's Theorem Is False) Give a counter example to show that the Converse of Lagrange's Theorem Is False. That is if k j jGj then G does not have a subgroup of order k. Proof. The group G = A4 has order 12 and 6j12 but A4 has no subgroup of order 6. to prove this; Note that G = A4 has 8 elements of order 3, namely, (α5; :::; α12) Suppose H is a subgroup of G = A4 of order 8, let a be any element of order 3 in A4: 12 Since j A4 : H j= 6 = 2; at most two of the cosets H; aH; and a2H are distinct. Ahmed EL-Mabhouh Abstract Algebra I Consequences of Lagrange's Theorem Proof. If H = aH; then a 2 H: This is true for any element a of order 3, and so H contains at least 8 elements, a contradiction. If aH = a2H; multiply by a−1 we have H = aH; a contradiction. If H = a2H; then by multiplying by a we get aH = a3H = eH = H; a contradiction. Theorem Classification of Groups of Order 2p Let G be a group of order 2p; where p is a prime greater than 2: Then G is isomorphic to Z2p or Dp . Note that if G is a non Abelian group of order 2p, then G ≈ Dp: if G is an Abelian group of order 2p, then G is cyclic and G ≈ Z2p: Ahmed EL-Mabhouh Abstract Algebra I An Application of Cosets to Permutation Groups Definition (Stabilizer of a Point) Let G be a group of permutations of a set S: For each i 2 S; let stabG (i) = fφ 2 Gjφ(i) = ig: We call stabG (i) the stabilizer of i in G . Exercise: Show that stabG (i) is a subgroup of G: Definition (Orbit of a Point) Let G be a group of permutations of a set S: For each s 2 S; let orbG (s) = fφ(s)jφ 2 Gg: The set orbG (s) is a subset of S called the orbit of s under G . We use j orbG (s)j to denote the number of elements in orbG (s): Ahmed EL-Mabhouh Abstract Algebra I An Application of Cosets to Permutation Groups Example (8) Let G = f(1);(132)(465)(78); (132)(465); (123)(456) (123)(456)(78); (78)g be a group of permutations on the set f1; 2; 3; 4; 5; 6; 7; 8g: Then orbG (1) = f1; 3; 2g stabG (1) = f(1); (78)g orbG (2) = f2; 1; 3g; stabG (2) = f(1); (78)g orbG (4) = f4; 6; 5g; stabG (4) = f(1); (78)g orbG (7) = f7; 8g; stabG (7) = f(1); (132)(465); (123)(456)g Ahmed EL-Mabhouh Abstract Algebra I An Application of Cosets to Permutation Groups Example (9) We may view D4 as a group of permutations of a square region. Let p and q be two points inside the square as shown in the figure below. What is the orbG (p)? orbG (q)? What is the stabG (p)? stabG (q)? Ahmed EL-Mabhouh Abstract Algebra I An Application of Cosets to Permutation Groups Theorem (Orbit-Stabilizer Theorem) Let G be a finite group of permutations of a set S: Then, for any i 2 S; jGj = j orbG (i)jj stabG (i)j: Proof.