## Proof That Every Subgroup of a Cyclic Group is Cyclic

Every subgroup of a cyclic group is cyclic. If generates the group (so that ) and the order of is so that then the order of any subgroup of is a divisor of and for each divisor of the group has exactly one subgroup of order – the subgroup generated by Proof: Let and suppose that is a subgroup of If consists of the identity element then is cyclic, so assume and contains some element  then Let be the least positive integer such that The claim is that Let be an arbitrary member of Since we have for some Apply the division algorithm to and to obtain a quotient and remainder satisfying so that but and hence but is the least power of in and since we must have so that Conversely suppose that is any subgroup of We have already shown that for some Since we have that the order of is a divisor of hence the order of  is a divisor of Now let be any divisor of and let  so no smaller power of equals so has order  is the only subgroup of order since if is any other subgroup of order then where is the least positive integer such that Write with We have so that Then and Also so and  