# 6.3.2 Cyclic groups - PowerPoint PPT Presentation

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6.3.2 Cyclic groups. 1.Order of an element Definition 13: Let G be a group with an identity element e. We say that a is of order n if a n =e, and for any 0<m<n, a m  e. We say that the order of a is infinite if a n  e for any positive integer n. Example:group[{1,-1,i.-i};  ],

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6.3.2 Cyclic groups

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### 6.3.2 Cyclic groups

• 1.Order of an element

• Definition 13: Let G be a group with an identity element e. We say that a is of order n if an =e, and for any 0<m<n, ame. We say that the order of a is infinite if ane for any positive integer n.

• Example:group[{1,-1,i.-i};],

• i2=-1,i3=-i, i4=1

• (-i)2=-1, (-i)3=i, (-i)4=1

• Theorem 6.14: Let a is an element of the group G, and let its order be n. Then am=e for mZ iff n|m.

• Example: Let the order of the element a of a group G be n. Then the order of ar is n/d, where d=(r,n) is maximum common factor of r and n.

• Proof: (ar)n/d=e,

• Let p be the order of ar.

• p|n/d, n/d|p

• p=n/d

• 2. Cyclic groups

• Definition 14: The group G is called a cyclic group if there exists gG such that h=gk for any hG , where kZ.We say that g is a generator of G. We denoted by G=(g).

• Example:group[{1,-1,i.-i};],1=i0,-1=i2,-i=i3,

• i and –i are generators of G.

• [Z;+]

• Example：Let the order of group G be n. If there exists gG such that g is of order n，then G is a cyclic group, and G is generated by g.

• Proof:

• Theorem 6.15: Let [G; *] be a cyclic group, and let g be a generator of G. Then the following results hold.

• (1)If the order of g is infinite, then [G;*] [Z;+]

• (2)If the order of g is n, then [G;*][ Zn;]

• Proof:(1)G={gk|kZ},

• :GZ, (gk)=k

• (2)G={e,g,g2,gn-1},

• :GZn, (gk)=[k]

### 6.4 Subgroups, Normal subgroups and Quotient groups

• 6.4.1 Subgroups

• Definition 15: A subgroup of a group (G;*) is a nonempty subset H of G such that *is a group operation on H.

• Example : [Z;+] is a subgroup of the group [R; +].

• G and {e} are called trivial subgroups of G, other subgroups are called proper subgroups of G.

• Theorem 6.16: Let [G;·] be a group, and H be a nonempty subset of G. Then H is a subgroup of G, iff

• (1) for any x, yH, x·yH; and

• (2) for any xH, x-1H.

• Proof: If H is a subgroup of G, then (1) and (2) hold.

• (1) and (2) hold

• eH

• Associative Law

• inverse

• Theorem 6.17: Let [G;·] be a group, and H be a nonempty subset of G. Then H is a subgroup of G, iff a·b-1H for a,b H.

• Example: Let [H1;·] and [H2;·] be subgroups of the group [G;·]，Then [H1∩H2;·] is also a subgroup of [G;·]

• [H1∪H2;·] ?

• Example:Let G ={ (x; y)| x,yR with x 0} , and consider the binary operation ● introduced by (x, y) ● (z,w) = (xz, xw + y) for (x, y), (z, w) G.

Let H ={(1, y)| yR}. Is H a subgroup of G? Why?

### 6.4.2 Coset

• Let [H;] is a subgroup of the group [G;]. We define a relation R on G, so that aRb iff for ab-1H for a,bG. The relation is called congruence relation on the subgroup [H;]. We denoted by ab(mod H)。

• Theorem 6.18 ：Congruence relation on the subgroup [H;] of the group G is an equivalence relation

• [a]={x|xG, and xa(mod H)}={x|xG, and xa-1H}

• Let h=xa-1. Then x=ha，Thus

• [a]={ha|hH}

• Ha={ha|hH} is called right coset of the subgroup H

• aH={ah|hH} is called left coset of the subgroup H

• Let [H;] be a subgroup of the group [G;], and aG. Then

• (1)bHa iff ba-1H

• (2)baH iff a-1bH

• Definition 16: Let H be a subgroup of a group G, and let aG. We define the left coset of H in G containing g,written gH, by gH ={g*h| h H}. Similarity we define the right coset of H in G containing g,written Hg, by Hg ={h*g| h H}.

• [E;+]

• Example:S3={e,1, 2, 3, 4, 5}

• H1={e, 1}; H2={e, 2}; H3={e, 3};

• H4={e, 4, 5}。

• H1

• Lemma 2：Let H be a subgroup of the group G. Then |gH|=|H| and |Hg|=|H| for gG.

• Proof: :HHg, (h)=hg

• NEXT : Lagrange's Theorem, Normal subgroups and Quotient groups

• Exercise:P371 (Sixth) ORP357(Fifth)22—26

• P376 10,12,21

• 1. Let G be a group. Suppose that a, and bG, ab=ba. If the order of a is n, and the order of b is m. Prove:

• (1)The order of ab is mn if (n,m)=1

• (2)The order of ab is LCM(m,n) if (n,m)1 and (a)∩(b)=. LCM(m,n) is lease common multiple of m and n