Cyclic groups of order 1 and n < a >; = {e, a, a 2, ..., a (n- 1)}, then a n = e, and e is the unit element. In addition to a, the generator can also be k (1 < k < n). As for higher powers, there is no point in discussing them, because there must be a (n+k) = a k, k < n), then k must be coprime with n. As long as you find all powers of B = A K that are not greater than n- 1, you will find that B 0 = E, b, B 2, ..., b (n-/.

For example, when n= 15, k can take the value of 2, so the results of various powers of b = a 2 are: b 0 = e, b 2 = a 4, b 3 = a 6, b 4 = a 8, b 5 =. b^9=a^ 18=a^3,b^ 10=a^20=a^5,b^ 1 1=a^22=a^7,b^ 12=a^24=a^9,b^ 13=a^26=a^ 1 1,b^ 14=a^28=a^ 13。 therefore

2. The order of the group refers to the number of elements. The order r of subgroup h of n-order group must be a factor of n. There is only one element in 12 >=<0 & gt={0}, so it is naturally a subgroup of order 1.

3. Group G has two special subgroups, one is subgroup {e} of order 1, and the other is its own G containing all elements. These two subgroups are called trivial subgroups.

G =<a> is a cyclic group and subgroup of order 15.