Cosets and Lagrange's Theorem

Home , Index of a subgroup

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 ∈ G, we deﬁne the following sets:

aH = {ah|h ∈ H}, Ha = {ha|h ∈ H}, aHa−1 = aha−1|h ∈ 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). |aH|, (|Ha|) 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 = {(1), (13)}. Then the left cosets of H in G are: (1)H = H (12)H = {(12), (12)(13)} = {(12), (132)} = (132)H (13)H = {(13), (1)} = H (23)H = {(23), (23)(13)} = {(23), (123)} = (123)H

Example (2)

Let K = {R0, R180} 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 = {0, 3, 6} in Z9 under addition. Then the distinct left cosets of H in Z9 are: 0 + H = {0, 3, 6} = 3 + H = 6 + H 1 + H = {1, 4, 7} = 4 + H = 7 + H 2 + H = {2, 5, 8} = 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 ∈ G. Then 1 a ∈ aH 2 aH = H if and only if a ∈ H 3 aH = bH if and only if a ∈ bH 4 aH = bH or aH ∩ bH = ∅ −1 5 aH = bH if and only if a b ∈ H 6 |aH| = |bH| −1 7 aH = Ha if and only if H = aHa 8 aH is a subgroup of G if and only if a ∈ 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 = {1, 15} in G = U(32) = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31}.

Proof. We begin with H, so 1.H = H = {1, 15} Choose an element not in H, say 3. A new left coset is 3H = {3, 13} Choose an element not in H and not in 3H, say 5, another left coset is 5H = {5, 11} 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 |H| divides |G| Moreover, the number of distinct left (right) cosets of H in G is |G|/|H|

Proof.

Let a1H, a2H,..., ar H denote the distinct left cosets of H in G. We show r = |G|/|H|, or |G| = r|H|. Consequently, |H| | |G|.

Let a ∈ G. Then aH = ai H for some i = 1, 2, ..., r. By Lemma part (1), a ∈ aH = ai H. Hence, G = a1H ∪ · · · ∪ ar H Since ai H ∩ aj H = φ for all i 6= j, we have

|G| = |a1H| + |a2H| + ··· + |ar H|

By part (6), |H| = |ai H| and so

|G| = |H| + |H| + ··· + |H| = r |H|

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 |G : H|.

Corollary (1) If G is a ﬁnite group and H is a subgroup of G, then |G : H| = |G|/H|

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 ∈ G. Then, a|G| = e

Proof. By Corollary 2, |a| | |G|, so |G| = |a|k for some positive integer k. Then a|G| = a|a|k = (a|a|)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 ∈ 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 | |G| then G does not have a subgroup of order k.

Proof.

The group G = A4 has order 12 and 6|12 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 | A4 : H |= 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 ∈ 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 ∈ S, let

stabG (i) = {φ ∈ G|φ(i) = i}.

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 ∈ S, let

orbG (s) = {φ(s)|φ ∈ G}.

The set orbG (s) is a subset of S called the orbit of s under G . We use | orbG (s)| 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 = {(1),(132)(465)(78), (132)(465), (123)(456) (123)(456)(78), (78)} be a group of permutations on the set {1, 2, 3, 4, 5, 6, 7, 8}. Then

orbG (1) = {1, 3, 2} stabG (1) = {(1), (78)} orbG (2) = {2, 1, 3}, stabG (2) = {(1), (78)} orbG (4) = {4, 6, 5}, stabG (4) = {(1), (78)} orbG (7) = {7, 8}, stabG (7) = {(1), (132)(465), (123)(456)}

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 ∈ S, |G| = | orbG (i)|| stabG (i)|.

Proof.

Ahmed EL-Mabhouh Abstract Algebra I An Application of Cosets to Permutation Groups

Proof.

T is well deﬁned: Let α tabG (i) = β stabG (i), then −1 α β ∈ stabG (i), so α−1β (i) = i, therfore β(i) = α(i) T is 1-1: Reversing the arguments above from the last step to the ﬁrst step shows that T is also one-to-one.

T is onto: Let j ∈ orbG (i). Then α(i) = j for some α ∈ G Therefore, T (α stabG (i)) = α(i) = j, so that T is onto.

Ahmed EL-Mabhouh Abstract Algebra I