Subgroup of cyclic group

• The subgroup of the cyclic group is still the cyclic group
• The subgroups of the infinite cyclic group are of infinite order except those generated by {e}
• For each positive factor d of order n, there is exactly one subgroup of order d.
  $$\begin{gathered} Firstto{a}^{\frac{n}{d}}Healthintothegroup\; ofasetisGofd-ordersub-groups \\ The\; needtolicenseouttheirCDasex:fakeset\; up{a}^{m}alsobornintoad-ordersub-cluster,(A^m{)}^d=e,It\; isanintegerexceptm\ast d\Rightarrow m=k\ast \frac{n}{d} \\ {a}^m=a^{k\ast \frac{n}{d}}=(a^{\frac{n}{d}}{)}^k,Asto{a}^{m}bornintothegroupof\; thesetstogetheris{a}^{\frac{n}{d}}Healthintotheset\; ofengagementof\; thesub-set, \\ Butbecauseoftwogroupsthe\; ordernumberwiththe\; same,theordertwothsettogetherwithother \end{gathered}$$

$Example:,tryfinda8-ordercirculationloopgroupofthehavegreentoelementandaretheresub-groups.$
Solution: Let G be a cyclic group of order 8, and a is its generator. G = {e, a, a 2,……, a 7} The necessary and sufficient condition for ak to be a generator of G is that k and 8 are relatively prime, so a, a 3, a 5, a 7 are all generators of G Because the subgroup of the cyclic group is also a cyclic group, and the order of the subgroup is a factor of the order of G. The subgroup of G can only be of order 1, 2, 4, or 8, d = 1, 2, 4, 8, and are generated by groups, so all subgroups of G in addition to two trivial subgroups, there are {e, a 4}, {e, a 2, a 4, a 6} Solution: Let G be Cyclic group of order 8, a is its generator. G=\{ e,a,a^2,……，a^7 \}\\ a^k is the generator of G. The necessary and sufficient condition is that k and 8 are relatively prime, so a,a^3,a^ 5,a^7 is all generators of G\Because the subgroups of the cyclic group are also cyclic groups, and the order of the subgroup is a factor of the order of G\ The subgroup of G can only be of order 1, 2. Order, 4 or 8, d=1,2,4,8, a group generated by a^{\frac{n}{d}}, so all subgroups of G except for two trivial subgroups In addition, there are \{e,a^4 \}, \{e,a^2,a^4,a^6 \}solution:Provided G is an 8 -order cycle cycloalkyl group , A is that it is born into the element . G={ e,a,a2,……，a7}ak isGthegreeninto theelementof thechargetobarmemberiskand8mutuallyprime,so theA,a3,a5,a7 is aGinthatthere arestudentsintoelementBecause as circulation loop group of the sub- group is also a circulation ring group , and the sub- group of order number is G of the order number of result subG of the sub- group only can be an order of , 2 order of , four order of or 8 step of ,d=1,2,4,8,adn​Born into the groupTherefore, G is has a sub- group except the two th level where the sub- group of external , but also there { E ,a4}，{ e,a2,a4,a6 }
Theorem 3: The order of elements and the order of subgroups