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? 發問於 科學及數學數學 · 1 十年前

Math Group

1. If g is an element of order 12 in a group, howmany distinct elements are there in the cyclic group generated by g, i.e.<g>. What are the orders of each element in <g>?2. The additive group of integers, (Z,+) is acyclic group generated by 1. Let H be a subgroup of Z and m the smallest positiveinteger in H. Proves, using the Division Algorithm that H is a cyclic groupgenerated by m, i.e. H=<m>.3. Let G be a finite cyclic group of order n with agenerator x. Prove, using the ideas in problem 2 that every subgroup of G isalso cyclic. 4. (a) Find the number of generators of a cyclic groupof order 24.(b) Find the number of generators of a cyclicgroup of order 60.

1 個解答

  • ?
    Lv 5
    1 十年前


    If g is of order 12, then g^12 = e.

    Therefore, the order of is at most 12.

    If the order of is less than 12,then g^12 = g^k for some positive integer k<12. then g^(12-k) = e,contracting the fact that order of g is 12.

    Hence the order of is exactly 12.

    For each element g^k (1<12), let="" h="" be="" the="" greatest="" common="" factor="" of="" k="" and="" 12,="" l="12/h," j="k/h.</span">

    (g^k)^l = g^(kl) = g^(jlh) = g^(j 12) = e^j =e

    Therefore the order of g^k is at most l.

    Now suppose (g^k)^m = e

    g^(m k) = e

    therefore mk is a multiple of 12, <12),>

    Hence m j is a multiple of l.

    As j is coprime with l, m is a multiple of l.

    It follows that the order of g^k is exactly l.


    By Division Algorithm,

    any element n of H can be expressed as

    n = qm + k where 0 <=k < m

    If n is not a multiple of m, then k>0.

    As n - qm = k is also an element of H.

    This contracdicts with the fact that m is the smallestelement of H.

    Therefore n must be a multiple of m.


    Let x^k be the element of G with smallest index k andx^m be another element of G.Then by Division Algorithm,

    m = qk + h where 0<=h < kIf m isnot a multiple of k, then h>0.Therefore x^m/x^gk = x^h is in G,contracting the fact that k is the smallest index.Therefore G is a cyclic group withgenerator x^k. 4.From 1., the element g^k of a cyclic groupwith order equal to order of g if and only if k is coprime with the order of g.(a) Number ofgenerators = φ(24) =φ(8) xφ(3) = 4 x 2 = 8whereφ(n) is the Euler’s Phi Function which represents the number of integers lessthan n and coprime with n.(b) Number ofgenerators = φ(60) =φ(5) xφ(4) xφ(3)= 4 x 2 x 2 = 16