Order of the unit group of a finite field F if for all two subgroups of F one is contained in the other.

Question

Let $F$ be a finite field. Prove that the following are equivalent:

i) $A \subset B$ or $B \subset A$ for each two subgroups $A,B$ of $F^*$.

ii) $\#F^*$ equals 2, 9, a Fermat-prime or $\#F^* -1$ equals a Mersenne prime.

Any ideas for i => ii ? I don't know where to start, except for remarking that $\#F^*$ is cyclic. Thanks.

Answer 1

Hints/suggestions:

Let $F=GF(q)$ with $q=p^m$. For $F^*$ to have property i) it is necessary and sufficient that $q-1=\ell^n$ for some prime $\ell$ and non-negative integer $n$.

1. If $p>2$, then $q-1$ is an even integer, so we must have $\ell=2$. Also all factors of $q-1$ must be powers of two. In particular we must have that $p-1$ and $p^{m-1}+p^{m-2}+\cdots+p+1$ must both be powers of two. The first implies that $p$ is a Fermat prime, and the second implies that $m$ must be an even integer or equal to one. The case $m=1$ is already covered. The case $m>1, 2\mid m$ implies that $(p^2-1)\mid (p^m-1)$. As $p^2-1=(p-1)(p+1)$ both $p-1$ and $p+1$ must be powers of two. The only Fermat prime for which that happens is $p=3$. A higher power of three can be eliminated by repeating this argument, and showing that we must have $4\mid m$ and consequently also $p^2+1$ should be a a factor of $p^m-1$ and hence also a power of two.
2. If $p=2$, then the Mersenne prime cases obviously work, and the goal is to show that nothing else works. Well, we have $$ 2^m-1=\ell^n, $$ so $2^m=\ell^n+1$. If $n$ is even, then $\ell^n+1\equiv 2\pmod4$, which cannot be a power of two. Therefore $n$ is odd, and $\ell+1$ is a factor of the r.h.s. and hence also a power of two. So $\ell$ is a Mersenne prime, say $\ell=2^r-1$ for some prime $r$. If $n=2k+1$ is odd, then $$ \ell^n=(2^r-1)^n=(2^r-1)((2^r-1)^2)^k\equiv(2^r-1)\pmod{2^{r+1}}, $$ so $\ell^n+1\equiv 2^r\pmod{2^{r+1}}$ cannot be a power of two unless $n=1$.

A lot of details left for you to check.

Answer 2

Jyrki's comment is a huge hint: if $\;\,p,q\,\mid\; |\Bbb F^*|\;$ for two different primes $\,p,q\,$, then there exist cyclic subgroups $\,P,Q\,$ of order $\,p,q\,$ resp., and clearly neither $\,P\subset Q\,$ nor $\,Q\subset P\,$, so $\,|F^*|\,$ is divisible only by one single prime (or it is the trivial group... trivial case), say $\;p\,$.

Since $\,|F|=r^n\,$, for some prime $\,r\,$, then $\,|F^*|=r^n-1=p^m\,$, yet

$$r^n-1=(r-1)(r^{n-1}+r^{n-2}+\ldots+r+1),$$

so in particular $\,p\,\mid\, r-1\,$, and the only numbers that fulfill this are...