How many subgroups does (Z35,+) have? What are they Solution all subgroups of Z

Question

How many subgroups does (Z35,+) have? What are they

Explanation / Answer

all subgroups of Z are of the form kZ for some k>0 together with the correspondence theorem. You may proceed as follows: Observe Z35 is isomorphic to Z/35Z. Let J be a subgroup of Z/35Z, by the correspondence theorem this subgroup is of the form J = S/35Z where S is a subgroup of Z containing 35Z. So it all boils down to finding the subgroups of Z containing 35Z. Now let S be a subgroup of Z, then by the first remark S is of the form kZ for some k>0. But we need that kZ contains 35Z. It is easy to show this can only happen when k divides 35. Now the divisors of 18 are 1,5,7,35. Therefore by the correspondence theorem the subgroups of Z35 are: Z/35Z , 5Z/35Z, 7Z/35Z In class notation the subgroups are: , < [5]>, , Here < > means the cyclic subgroup generated by. In fact there's a standard result: If G is a cyclic group of order n there is a unique subgroup (hence cyclic) of order d for each divisor d of n. Now Z35 is cyclic and has order 35, so in this particular case n=35 and 35 has 4 divisors so we get 4 cyclic subgroups.

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