TITLE: Does every finite cyclic group appear as a subgroup of the multiplicative group of a finite field?
QUESTION [6 upvotes]: Does every finite cyclic group appear as a subgroup of the multiplicative group of a finite field?
In other words, given any $d \in \mathbb{N}$, can we find a prime $p$ and $k \in \mathbb{N}$ such that the multiplicative group of the finite field of order $p^k$ has a subgroup of order $d$? Such a subgroup would be cyclic because every subgroup of the multiplicative group of a field is cyclic.
A finite cyclic group $G$ has a subgroup of order $d$ whenever $d$ divides $|G|$. Thus to prove the answer is yes it suffices to show that for every $d \in \mathbb{N}$ we can find a prime $p$ and $k \in \mathbb{N}$ such that $d$ divides $p^k - 1$.

REPLY [8 votes]: Given a natural number $d$, choose a prime $p$ which is relatively prime to $d$. By Euler's theorem, $p^{\varphi(d)}\equiv 1\bmod d$. Letting $k=\varphi(d)$, we see that $d$ divides $p^k-1$.
I guess that means the answer is yes.