2
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Not sure what you mean by "multiplicative group of integers modulo $n$". You mean "group of invertible elements of $\mathbb{Z}_n$"? $\endgroup$
    – Minkowski
    Feb 12 '21 at 19:51
  • $\begingroup$ 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. $\endgroup$ Feb 13 '21 at 3:10
  • $\begingroup$ Yes the term refers to the units in the ring $\mathbb{Z}_n$ $\endgroup$
    – jaykopp
    Feb 13 '21 at 13:28
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks, now that I see a counter example it seems obvious $\endgroup$
    – jaykopp
    Feb 13 '21 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.