I am looking at the multiplicative group of integers modulo $n$, denoted $\mathbb{Z}_n^*$, and its subgroups/quotient groups. For a number $p \in \mathbb{Z}$ we denote $\bar{p} \in \mathbb{Z}_n^*$. We are given that $\mathbb{Z}_n^*/H$ is cyclic, where $H=\langle \bar{p} \rangle$ for a prime number $p$ which does not divide $n$. Say $|\mathbb{Z}_n^*|=l\cdot d$ where $|H|=d$. What can we say about the number $l$?
I know if $n = p_1^{k_1}\dots p_r^{k_r}$ then $\mathbb{Z}_n^* \simeq \mathbb{Z}_{p_1^{k_1}}^*\times \dots \times \mathbb{Z}_{p_1^{k_r}}^*$. For odd primes each of these factors are cyclic (as well as for $p_i=2, k_i = 0,1,2$). We know that $l$ isn't necessarily prime, since we could get $n=p_1p_2^k, k>1$ and that $|H|=p_1$, but can $l$ be something else than a prime power? Maybe it is easier to prove that $d$ and $l$ are coprime?
Best Answer
No. The size of the quotient need not be a prime power. Consider the field $\Bbb F_{31}:=\frac{\Bbb Z}{31\Bbb Z}$ and denote by $\mathbf x$ the residue class $x+31\Bbb Z$. As presecribed in the OP the prime $2$ is a representative of the residue class $\mathbf 2$ but $$\mathbf 2^5=\mathbf{32}=\mathbf 1\;\;\;\;\;\therefore\;|\langle\mathbf 2\rangle_{*}|=5$$ where $\Bbb F_{31}^*:=\Bbb F_{31}\setminus\{\mathbf 0\}$ is the $\Bbb F_{31}$cyclic units group of size $30$ and $\langle\mathbf 2\rangle_*$ is the $\Bbb F_{31}^*$ subgroup generated by $\mathbf 2$. However, $|\frac{\Bbb F_{31}^*}{\langle\mathbf 2\rangle_{*}}|=\frac{30}{5}=6$ but $6$ is not a prime power.