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Let $n=2^7 \cdot 3^5 \cdot 11^3 \cdot 35$. In how many ways can the cyclic group $C_n$ can be written as a direct product of two or more nontrivial groups? List all these direct products.

Can someone guide me how to do this question please. I am not looking for a straight answer obviously.

Also, I know what cyclic groups are when its like for $\mathbb{Z}_4$ for example but what does it mean by $C_n$ in this case?

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You can think of $C_{n}$ to be $\mathbb{Z}_{n}$; they are both cyclic groups of the same order.

I think this question wants you to use the fact that $\mathbb{Z}_{km} \cong \mathbb{Z}_{k} \times \mathbb{Z}_{m}$ iff $\gcd(k,m) = 1$ for $k,m\in\mathbb{Z}_{\geq 2}$. Using this you should be able to decompose $\mathbb{Z}_{2^7\cdot 3^5 \cdot 11^3 \cdot 35}$ into several direct products of smaller cyclic groups.

For example, $\mathbb{Z}_{2^7\cdot 3^5} \cong \mathbb{Z}_{2^7} \times \mathbb{Z}_{3^5}$, and $\mathbb{Z}_{35} = \mathbb{Z}_{5\cdot 7} \cong \mathbb{Z}_{5}\times \mathbb{Z}_{7}$.

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  • $\begingroup$ I don't think that's quite right; you're not counting that $35 = 5\cdot 7$. For $n = 2^7\cdot 3^5\cdot 11^3\cdot 35$, it's actual prime factorization is $n = 2^7\cdot 3^5\cdot 5\cdot 7\cdot 11^3$. The fact that $\mathbb{Z}_{km} \cong \mathbb{Z}_{k}\times\mathbb{Z}_{m}$ iff $\gcd(k,m) = 1$ makes it so that for any factorization of $\mathbb{Z}_{n}$ the factors must have orders relatively prime to each other. So you cannot have more than 5 factors in a factorization for $\mathbb{Z}_{n}$ since there are only 5 distinct prime powers in the prime factorization of $n$. $\endgroup$ – catfish Oct 31 '15 at 0:45
  • $\begingroup$ So we need to find number of ways to have 2 products of your actual prime factorization of n, then no. of ways to have 3 products, then 4, then 5? and then add the number of ways together? $\endgroup$ – snowman Oct 31 '15 at 0:49
  • $\begingroup$ Yes, I believe that's correct. Though to count that way we assume that two products with the same factors are the same (so not counting permutations, i.e. we would treat $\mathbb{Z}_{k}\times\mathbb{Z}_{m}$ as the same factorization as $\mathbb{Z}_{m}\times\mathbb{Z}_{k}$). $\endgroup$ – catfish Oct 31 '15 at 0:52
  • $\begingroup$ Would it be $5 C 2 + 5 C 3 + 5 C 4 + 5 C 5$? C as in 'choose' $\endgroup$ – snowman Oct 31 '15 at 0:57
  • $\begingroup$ Oops, actually I think it's Stirling numbers of the second kind instead of combinations; you want to count the number of ways to partition the set $\{2^7,3^5,5,7,11^3\}$ into two or more nonempty subsets, each partition would correspond to a distinct factorization. $\endgroup$ – catfish Oct 31 '15 at 1:16

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