On the computational relationship between the order of the cyclic group and the (generating element and subgroup) (2009 PKU Last question, no official definitive answer)

1. Here first Preheat:

Raw Narimoto: Less than 14, and 14 coprime of the number 1, 3, 5, 7, 11, 13, then its generating element is:;

number of subgroups: 14 of the factors are: 1, 2, 7, 14, then There are 4 subgroups, respectively:

2. The analysis of the PKU courseware together:

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the last of the 2009-year real title: My calculation results are | G|=4:

Set | G|=n,n have m factors, there are m subgroups (according to Theorem 17.13), because G has and has only one nontrivial subgroup H.

The subgroups of G are: G, <e>, H, and

the corresponding order of subgroups is: N, 1, | h|,

The factor of n is: N, 1, | h|,

also have first ask proof when I get | G| must be even, then, less than N and only one even, the even number is 2, then | h|=2,| G|=4.