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 $p^k - 1$.


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.

  • $\begingroup$ I certainly like this elementary solution, but also wanted to point out that allowing the use of Dirichlet's theorem, we can in fact take $k=1$. $\endgroup$ – RghtHndSd Mar 17 '14 at 22:43
  • $\begingroup$ If you want to go down that road, we don't even need the full power of Dirichlet's Theorem: you can prove using cyclotomic polynomials that there are infinitely many primes congruent to $1\bmod d$ for any integer $d$. $\endgroup$ – user134824 Mar 17 '14 at 23:04

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