The question is asking for the primary decomposition of $Z_{1001}$ as an abelian group under multiplication. So I did the following.
By Euler $\phi$ function, I count the number of integers relatively prime to 1001 and it resulted in 720 which means this group is of order 720. So 720=$2^4\times 3^2\times 5$. Now one can check the elements of this group has an exponent of 60. $3^2=3\times 3$ to get the factor of 3 in 60. But $2^4=4\times 4=4\times 2\times 2$ to get the factor of 4. The factor 5 comes directly from 5. So we get two kind of structure $Z_4\times Z_2\times Z_2\times Z_3\times Z_3\times Z_5\cong Z_2\times Z_6\cong Z_{60}$ or $Z_4\times Z_4\times Z_3\times Z_3\times Z_5\cong Z_{12}\times Z_{60}$. However I do not know how to proceed any more. Did I do anything really wrong here. Any hints without answering question directly will be helpful? And is there a good way to get exponent(i.e. 60) of this group without plug it into mathematica?
Hint By the Chinese Remainder Theorem $Z_{1001} \sim Z_{7} \times Z_{11} \times Z_{13}$. Therefore $U(Z_{1001})\sim U(Z_{7}) \times U(Z_{11}) \times U(Z_{13})$.
The rest is easy, especially if you know that the multiplicative group of a field is cyclic.