Yahoo 知識+ 將於 2021 年 5 月 4 日 (美國東岸時間) 停止服務。從 2021 年 4 月 20 日 (美國東岸時間) 起，Yahoo 知識+ 網站將轉為僅限瀏覽模式。其他 Yahoo 資產或服務，或你的 Yahoo 帳戶將不會有任何變更。你可以在此服務中心網頁進一步了解 Yahoo 知識+ 停止服務的事宜，以及了解如何下載你的資料。
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.
- ?Lv 51 十年前最愛解答
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