Cosets and Lagrange’s Theoremfacstaff.cbu.edu/wschrein/media/M402 Notes/M402C7.pdf7. COSETS AND LAGRANGE’S THEOREM 93 When the group operation is addition, we use a+H and H +a

CHAPTER 7

Cosets and Lagrange’s Theorem

Properties of Cosets

Definition (Coset of H in G).

Let G be a group and H ✓ G. For all a 2 G, the set {ah|h 2 H} is denotedby aH. Analagously, Ha = {ha|h 2 H} and aHa�1 = {aha�1|h 2 H}.When H G, aH is called the left coset of H in G containing a, and Hais called the right coset of H in G containing a. In this case the element a iscalled the coset representative of aH or Ha. |aH| and |Ha| are used to denotethe number of elements in aH and Ha, respectively.

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92 7. COSETS AND LAGRANGE’S THEOREM

Example. Consider the left cosets of

H = {(1), (1 2)(34), (1 3)(2 4), (1 4)(2 3)} = {↵1,↵2,↵3,↵4} A4

from the table below:

H = 1H = ↵1H = {↵1,↵2,↵3,↵4} = ↵2H = ↵3H = ↵4H

↵5H = {↵5,↵6,↵7,↵8} = ↵6H = ↵7H = ↵8H.

↵9H = {↵9,↵10,↵11,↵12} = ↵10H,↵11H,↵12H.

Also, for k = 1, 2, . . . , 12,↵kH = H↵k.

7. COSETS AND LAGRANGE’S THEOREM 93

When the group operation is addition, we use a + H and H + a instead of aHand Ha.

Example. Let G be the group of vectors in the plane with addition. LetH be a subgroup which is a line through the origin, i.e.,

H = {tx|t 2 R and kxk = 1}.

Then the left coset v + H = {v + x|x 2 H} and the right coset

H + v = {x + v|x 2 H} are the same line, and is parallel to H.

Lemma (Properties of Cosets). Let H G, and let a, b 2 G. Then

(1) a 2 aH.

Proof. a = ae 2 aH. ⇤

(2) aH = H () a 2 H.Proof.

(=)) Suppose aH = H. Then a = ae 2 aH = H.


((=) Now assume a 2 H. Since H is closed, aH ✓ H. Next assume h 2 Halso, so a�1h 2 H since H G. Then

h = eh = (aa�1)h = a(a�1h) 2 aH,

so H ✓ aH. By mutual inclusion, aH = H. ⇤

(3) (ab)H = a(bH) and H(ab) = (Ha)b.

Proof.

Follows from the associative property of group multiplication. ⇤

(4) aH = bH () a 2 bH.

Proof.

(=)) aH = bH =) a = ae 2 aH = bH.

((=)a 2 bH =) a = bh where h 2 H =) aH = (bh)H = b(hH) = bH ⇤

(5) aH = bH or aH \ bH = ;.Proof.

Suppose aH \ bH 6= ;. Then 9 x 2 aH \ bH =) 9 h1, h2 2 H 3��x = ah1 and x = bh2. Thus

a = xh�11 = bh2h

�11 and aH = bh2h

�11 H = b(h2h

�11 H) = bH

by (2). ⇤

(6) aH = bH () a�1b 2 H

Proof.

aH = bH () H = a�1bH(2)() a�1b 2 H. ⇤


(7) |aH| = |bH|.Proof.

[Find a map ↵ : aH ! bH that is 1–1 and onto]

Consider ↵ : aH ! bH defined by ↵(ah) = bh. This is clearly onto bH.Suppose ↵(ah1) = ↵(ah2). Then bh1 = bh2 =) h1 = h2 by left cancellation=) ah1 = ah2, so ↵ is 1–1. Since ↵ provides a 1-0-1 correspondence betweenaH and bH, |aH| = |bH|. ⇤

(8) aH = Ha () H = aHa�1.

Proof.

aH = Ha () (aH)a�1 = (Ha)a�1 = H(aa�1) = H () aHa�1 =H. ⇤

(9) aH G () a 2 H.

Proof.

(=)) If aH G, e 2 ah =) aH \ eH 6= ; (5)=) aH = eH = H

(2)=) a 2 H.

((=) If a 2 H, aH = H by (2), so aH G since H G. ⇤

Note. Analagous results hold for right cosets.

Note. From(1), (5), and (7), the left (right) cosets of H partition G intoequivalence classes under the relation

a ⇠ b () aH = bH ( or Ha = Hb).


Lagrange’s Theorem and Consequences

Theorem (7.1 — Lagrange’s Theorem: |H| Divides |G|). If G is a finitegroup and H G, then |H|

��|G|. Moreover, the number of distinct left

(right) cosets of H in G is|G||H|.

Proof.

Let a1H, a2H, . . . , arH denote the distinct left cosets of H in G. Then, for alla 2 G, aH = aiH for some i = 1, 2, . . . , r. By (1) of the Lemma, a 2 aH.Thus

G = a1H [ a2H [ · · · [ arH.

By (5) of the Lemma, this union is disjoint, so

|G| = |a1H| + |a2H| + · · · + |arH| = r|H|since |aiH| = |aH| for i = 1, 2, . . . r. ⇤

Definition. The index of a subgroup H in G is the number of distinct leftcosets of H in G, and is denoted by |G : H|.

Corollary (1 — |G : H| =|G||H|.). If G is a finite group and H G,

then |G : H| =|G||H|.

Proof. Immediate consequence of Lagrange’s Theorem. ⇤

Corollary (2— |a| Divides |G|). In a finite group, the order of eachelement divides the order of the group.

Proof. For a 2 G, |a| = |hai|, so |a|��|G|. ⇤


Corollary (3 — Groups of Prime Order are Cyclic). A group of primeorder is cyclic.

Proof.

Suppose G has prime order. Let a 2 G, a 6= e. The |hai|��|G| and |hai| 6= 1,

so |hai| = |G| =) hai = G. ⇤

Corollary (4 — a|G| = e.). Let G be a finite group and a 2 G. Thena|G| = e.

Proof.

By Corollary 2, |G| = |a|k for some k 2 N. Then

a|G| = a|a|k = (a|a|)k = ek = e.

⇤

Corollary (5 — Fermat’s Little Theorem.). For all a 2 Z and everyprime p,

ap mod p = a mod p.Proof.

By the division algorithm, a = pm + r where 0 r < p. Thus a mod p = r,so we need only show rp mod p = r. if r = 0, the result is clear, so

r 2 U(p) = {1, 2, . . . , p� 1}.Then, by Corollary 4, rp�1 mod p = 1 =) rp mod p = r. ⇤


Example.

In A4, there are 8 elements of order 3 (↵5–↵12). Suppose H A4 with |H| = 6.Let a 2 A4 with |a| = 3. Since |A4 : H| = 2, at most two of H, aH, and a2Hare distinct. If a 2 H,H = aH = a2H, a contradiction.

If a 62 H, then A4 = H [ aH =) a2 2 H or a2 2 aH.

Now a2 2 H =) (a2)2 = a4 = a 2 H, again a contradiction.

If a2 2 aH, a2 = ah for some h 2 H =) a 2 H. Thus all 8 elemts of order 3are in H, a group of order 6, a contradiction.

Note. This example means the converse of Lagrange’s Theorem is not true.


Theorem (7.2 — |HK| =|H||K||H \K|). For two finite subgroups H and

K of a group G, define the set HK = {hk | h 2 H, k 2 K}. Then

|HK| =|H||K||H \K|.

Proof.

Although the set HK has |H||K| products, all not need be distinct. That is,we may have hk = h0k0 where h 6= h0 and k 6= k0. To determine |HK|, weneed to determine the extent to which this happens. For every t 2 H \ K,hk = (ht)(t�1k), so each group element in HK is represented by at least|H \K| products in HK. But hk = h0k0 =) t = h�1h0 = kk0�1 2 H \K,so h0 = ht and k0 = t�1k. Thus each element in HK is represented by exactly

|H \K| products, and so |HK| =|H||K||H \K|. ⇤

Problem (Page 158 # 41). Let G be a group of order 100 that has asubgroup H of order 25. Prove that every element of order 5 in G is in H.

Solution.

Let a 2 G and |a| = 5. Then by Theorem 7.2 the set haiH has exactly5 · |H|

|hai \H| elements and |hai \H| divides |hai| = 5 =) |hai \H| = 5 =)

hai \H = hai =) a 2 H. ⇤


Theorem (7.3 — Classification of Groups of Order 2p.). Let G be a groupof order 2p where p is a prime greater than 2. Then G ⇡ Z2p or G ⇡ Dp.

Proof.

Assume G has no element of order 2p. [To show G ⇡ Dp.]

[Show G has an element of order p.] By Lagrange’s Theorem, every nonidentityelement must have order 2 or p. So assume every nonidentity element has order2. Then, for all a, b 2 G,

(ab) = (ab)�1 = b�1a�1 = ba,

so G is Abelian. Thus, for a, b 2 G, a 6= e, b 6= e, {e, a, b, ab} is closed and sois a subgroup of G of order 4, contradicting Lagrange’s Theorem. Thus G hasan element of order p. Call it a.

[To show any element not in hai has order 2.] Suppose b 2 G, b 62 hai. ByLagrange’s theorem and our assumption that G does not have an element oforder of 2p, |b| = 2 or |b| = p. Because |hai \ hbi| divides |hai| = p andhai 6= hbi, |hai \ hbi| = 1. Now suppose |b| 6= 2. Then by Theorem 7.2|haihbi| = |hai||hbi| = p2 > 2p = |G|, which is impossible. Thus |b| = 2.

Now ab 62 hai =) |ab| = 2 by the same argument as above. Then

ab = (ab)�1 = b�1a�1 = ba�1.

This relation completely determines the multiplication table for G. [For exam-ple,

a3ba4 = a2(ab)a4 = a2(ba�1)a4 = a(ab)a3 =

a(ba�1)a3 = (ab)a2 = (ba�1)a2 = ba.]

Thus all noncyclic groups of order 2p must be isomorphic to each other, andDp is one such group. ⇤

Example. S3 ⇡ D3.

Proof. |S3| = 6 = 2 · 3 and S3 is not cyclic. ⇤


An Application of Cosets to Permutation Groups

Definition (Stabilizer of a Point). Let G be a group of permutations of aset S. For each i 2 S, let stabG(i) = {� 2 G|�(i) = i}. We call stabG(i) thestabilizer of i in G.

Corollary. stabG(i) is a subgroup of G.

Proof.

[Page 120 # 35] Let ↵,� 2 stabG(i). Then ↵(i) = i and �(i) = i, so

(↵�)(i) = ↵��(i)

�= ↵(i) = i,

so ↵� 2 stabG(i). Also,

↵�1�a(i)

�= ↵�1(i) =) i = ↵�1(i),

so ↵�1 2 stabG(i).

Thus stabG(i) G by the two-step test. ⇤

Definition (Orbit of a Point). Let G be a group of permutations of a setS. For each i 2 S, let orbG(s) = {�(s)|� 2 G}. The set orbG(s) is a subsetof S called the orbit of s under G. We use |orbG(s)| to denote the number ofelements in orbG(s).


Problem (Page 158 # 45).

Let

G = {(1), (1 2)(3 4), (1 2 3 4)(5 6), (1 3)(2 4), (1 4 3 2)(5 6),

(5 6)(1 3), (1 4)(2 3), (2 4)(5 6)}Solution.

stabG(1) = {(1), (2 4)(5 6)}.

stabG(3) = {(1), (2 4)(5 6)}.

stabG(5) = {(1), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}.

orbG(1) = {1, 2, 3, 4}.

orbG(3) = {1, 2, 3, 4}.

orbG(5) = {5, 6}.

⇤


Theorem (7.4 — Orbit-Stabilizer Theorem). Let G be a group of permu-tations of a set S. Then, for any i 2 S, |G| = |orbG(i)| · |stabG(i)|.

Proof.

By Lagange’s Theorem,|G|

|stabG(i)| is the number of distinct left cosets of

stabG(i) in G. [To show a 1–1 correspondence between these cosets and theelements of orbG(i).]

Define

T : {� stabG(i)|� 2 G} ! orbG(i) by T (� stabG(i)) = �(i).

[To show T is well-defined, i.e., that ↵ stabG(i) = � stabG(i) =) ↵(i) = �(i)or that the result of the mapping depends on the coset and not on a particularrepresentation of it.] Now,

↵ stabG(i) = � stabG(i) () stabG(i) = ↵�1� stabG(i) ()↵�1� 2 stabG(i) () ↵�1�(i) = i () �(i) = ↵(i)

so T is well-defined and, from the reverse implications, T is 1–1.

[To show T is onto.] Let j 2 orbG(i). Then ↵(i) = j for some ↵ 2 G, andT (↵ stabG(i)) = ↵(i) = j, so T is onto.

Thus T is a 1–1 correspondence. ⇤

The Rotation Group of a Cube

Let G be the rotation group of a cube. Label the faces 1–6. Since each rotationmust carry each face to exactly one other face, G is a group of permutationson {1, 2, 3, 4, 5, 6}.There is a central horizontal or vertical permutation thatcarries face 1 to any other face, so |orbG(1)| = 6. Also, there are 4 rotations(0�, 90�, 180�, 270� about the line ? to the face and passing through its center),so |stabG(1)| = 4. By Theorem 7.4,

|G| = |orbG(1)| · |stabG(1)| = 6 · 4 = 24.


Theorem (The rotation Group of the Cube). The group of rotations ofa cube is isomorphic to S4.

Proof.

Since |G| = |S4|, we need only show that G is isomorphic to a subgroup of S4.Now a cube has 4 diagonals and any rotation induces a permutation of thesediagonals. But we cannot just assume that di↵erent rotations correspond todi↵erent rotations.

↵ = (1 2 3 4) � = (1 4 3 2)

We need to show all 24 permutations of the diagonals come from rotations.Numbering the diagonals as above, we see two perpendicular axes where 90�

rotations give the permutations ↵ = (1 2 3 4) and � = (1 4 3 2). These inducean 8 element subgroup {",↵,↵2,↵3,�2,�2↵,�2↵2,�2↵3} and the 3 elementsubgroup {",↵�, (↵�)2}. Thus the rotations induce all 24 permutations since24 = lcm(8, 3). ⇤


Problem (Page 158 # 46). Prove that a group G of order 12 must havean element of order 2.

Proof.

Let a 2 G, a 6= e. By Lagrange’s theorem, |a| = 12, 6, 4, 3, or 2.

If |a| = 12, |a6| = 2. If |a| = 6, |a3| = 2. If |a| = 4, |a2| = 2.

Suppose all non-identity elements of G have order 3. But such elements comein pairs, e.g., if |a| = 3, |a2| = 3. But there are 11 non-identity elements, acontradiction. Thus G has an element of order 2. ⇤

Problem (Page 158 # 47). Show that in a group G of odd order, theequation x2 = a has a unique solution.

Proof.

Suppose y is a solution also, i.e., y2 = a =) x2 = y2. Let |G| = 2k + 1. Then

x = xe = xx2k+1 = x2k+2 = (x2)k+1 = (y2)k+1 = y2k+2 = yy2k+1 = ye = y.

⇤

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Cosets and Lagrange’s Theoremfacstaff.cbu.edu/wschrein/media/M402 Notes/M402C7.pdf7. COSETS AND LAGRANGE’S THEOREM 93 When the group operation is addition, we use a+H and H +a