We denote $\mathbb{Z}_n^*$ be the multiplicative group of integers modulo $n$ (i.e. $\{a∈\mathbb{Z}/n\mathbb{Z}:\gcd (a,n)=1\}$). We know that if $n=p_1^{k_1}\dots p_r^{k_r}$, then
$$\mathbb{Z}_n^*\simeq \mathbb{Z}_{p_1^{k_1}}^*\times\dots\times \mathbb{Z}_{p_r^{k_r}}^*.$$
What I am interested in is the structure of the quotient groups of $\mathbb{Z}_n^*$, in other words $\mathbb{Z}_n^*/\langle q\rangle$ for some $q \in \mathbb{Z}_n^*$. I want to the determine the structure when $\langle q\rangle$ is a subgroup of two or more of the factors $\mathbb{Z}_{p_i^{k_i}}^*$.
Say we have $n=2^{15}-1=7\cdot31\cdot151$, so we know that the order of 2 is 15 by $2^{15}\equiv 1\mod n$. We also know that $\mathbb{Z}_n^*\simeq \mathbb{Z}_6\times \mathbb{Z}_{30} \times \mathbb{Z}_{150}$. Then there are two options, either $\mathbb{Z}_n^*/\langle 2\rangle$ is isomorphic to $\mathbb{Z}_6\times \mathbb{Z}_{2} \times \mathbb{Z}_{150}$ or to $\mathbb{Z}_6\times \mathbb{Z}_{30} \times \mathbb{Z}_{10}$. These two groups are not isomorphic. How can I know which one is the case? Is there a general method we can employ by looking at the group structure?
Best Answer
"How can I know which one is the case?"
You can search for an element of order $25$. One group has it, the other not.