# Is the order of a quotient of the multiplicative group of integers a prime power if the group is cyclic?

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?

• Not sure what you mean by "multiplicative group of integers modulo $n$". You mean "group of invertible elements of $\mathbb{Z}_n$"? Feb 12 '21 at 19:51
• A $\it{unital}$ ring ($\it{algebra}$) is a ring with a multiplicative identity element called the $\it{unity}$ of that ring. A unital ring element that has a multiplicative inverse is called a $\it{unit}$ of that ring. The units group, archaically denoted $\mathcal U(n)$ for the ring $\Bbb Z_n$, of a unital ring is the multiplicative group made of that ring's units. Feb 13 '21 at 3:10
• Yes the term refers to the units in the ring $\mathbb{Z}_n$ Feb 13 '21 at 13:28

## 1 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.

• Thanks, now that I see a counter example it seems obvious Feb 13 '21 at 13:25