Every cyclic group is isomorphic to Zn. Now what is the total number of isomorphic direct product groups of the form Zn1*Zn2*Zn3*........Where n1*n2*n3*.....=n. It is known that (Zp*Zq) is cyclic if and only if p and q are relatively prime.
@Dinesh Valluri . No you can also have other decompositions. For example Z/30Z can be decomposed as Z/6Z x Z/5Z but 6 is not a prime power. More generally, let p1^e1 x ... x pk^ek be the factorisation of n, then each ni has to be a product of integers of the form pj^ej . So the number of ways this can be done will only depend on the number k. In order to be able to answer the question more concretely I'm going to slightly modify it, namely I call two such direct product decompositions n1x n2 x ... and m1 x m2 x ... the same if the mi and the ni are a permutation of each other. If you count them in this way then the number will be the bell number https://en.wikipedia.org/wiki/Partition_of_a_set
ps. Without my modification you can also answer it concretely, but then the answer will be the number of "ordered" partitions of {1,...,k} instead of the number of partitions. http://en.wikipedia.org/wiki/Ordered_partition_of_a_set
Let k be the number of pairwise distinct prime numbers dividing n. I assume here "different" is up to permutations, that is you do not consider Z2*Z3 different from Z3*Z2.Then the answer is the number of partitions of a set of k elements; this is the Bell number B_k (you can find some informations on it on wikipedia). If the order of the factors matter, you get the ordered Bell number.
The reason is the following: Zn1*Zn2*...*Z_nk is cyclic if and only if n1, n2... nk are pairwise relatively prime. If you decompose n as product of prime numbers p1^a1*...pk^ak, then, for the product to be cyclic, we need the following: if ni is divisible by pj, the others are not, so p_j must be divisible by the whole pj^aj. So you must only decide how to distribute your primes pj among the factors ni.
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