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 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

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


6 3 2 cyclic groups

  • 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


6 3 2 cyclic groups

  • 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;+]


6 3 2 cyclic groups

  • 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:


6 3 2 cyclic groups

  • 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 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.


6 3 2 cyclic groups

  • 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


6 3 2 cyclic groups

  • 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

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


6 3 2 cyclic groups

  • [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


6 3 2 cyclic groups

  • 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}.


6 3 2 cyclic groups

  • [E;+]

  • Example:S3={e,1, 2, 3, 4, 5}

  • H1={e, 1}; H2={e, 2}; H3={e, 3};

  • H4={e, 4, 5}。

  • H1


6 3 2 cyclic groups

  • 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


6 3 2 cyclic groups

  • 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


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