Multiplicative group of a finite field

The multiplicative group of a finite field ${\mathbb{F}}_q$ is the group ${\mathbb{F}}_q^{\times }$ of its $q-1$ nonzero elements under multiplication.

A nonzero element $a$ of a finite field is a quartic residue when $a=b^4$ for some nonzero $b$. If $q\equiv 1(\mathrm{mod}4)$, the multiplicative group of a finite field is cyclic of order divisible by four, so the quartic residues form a subgroup of index four; equivalently $a^{(q-1)/4}=1$.

The nonzero elements of every finite field form a cyclic group. For a field with $q$ elements, combine elements realizing each prime-power factor of the group exponent to obtain an element of exponent order; the root bound for $X^m-1$ then forces that order to be $q-1$.

Every finite subgroup $H$ of the multiplicative group of a field is cyclic. If $m$ is the exponent of $H$, commutativity lets one combine elements whose orders realize the prime-power factors of $m$, producing an element of order $m$. Every element of $H$ is a root of $X^m-1$, so the Lagrange root bound over a field gives $|H|\le m$. Since $m\le |H|$, equality holds and that element generates $H$.

In a finite field with $q$ elements, the equation $x^k=1$ has exactly $\gcd (k,q-1)$ solutions. Indeed, after choosing a generator $g$ of the multiplicative group, $x=g^j$ is a solution exactly when $(q-1)\mid kj$.

For odd $p$, the nonzero squares form an index-two subgroup of ${\mathbb{F}}_p^{\times }$. Given nonzero $a,b$, either $a$ or $b$ is a square, or both occupy the nontrivial coset and $ab$ is a square. Taking $a=2$ and $b=3$ shows that one of $2,3,6$ is a square, apart from the immediate zero case $p=3$.