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For a field $K$, the nonzero elements form an abelian group under multiplication, called its multiplicative group. Products of nonzero elements are nonzero because a field has no nonzero zero divisors; multiplication is associative and commutative, the identity is $1$, and every nonzero element has an inverse. Thus every finite subgroup of $K^{\times }$ is a finite abelian group. The finite subgroup of a field multiplicative group is cyclic theorem uses the root bound for a polynomial to prove a stronger restriction on such subgroups.

The multiplicative group of a field is abelian, so its finite subgroup $G$ is a finite abelian group. Let $d\mid |G|$, and suppose $H$ is a subgroup of order $d$. By Lagrange's theorem, the order of each $h\in H$ divides $d$, hence $h^d=1$. Thus $H$ is contained in the set of roots in $k$ of $t^d-1$.

This nonzero polynomial has degree $d$, so it has at most $d$ roots. Since the $d$ distinct elements of $H$ are already roots, they comprise the whole root set. Any second subgroup $K$ of order $d$ would also consist of $d$ roots of the same polynomial, forcing $K=H$. Therefore there is at most one subgroup of each order dividing $|G|$.

Part (c) now applies to giveThe root bound remains valid in positive characteristic, even if the formal derivative of $t^d-1$ vanishes: the argument counts distinct roots, so no separability assumption is needed.

The multiplicative group $K^{\times }$ has order $p^n-1$, so Lagrange's theorem for finite groups gives $x^{p^n-1}=1$ for $x\ne 0$. Including zero, $x^{p^n}=x$ for every $x\in K$. Therefore ${\vartheta }^n$ is the identity. No positive smaller power is the identity by part (iii), soFor $n=1$ this says that Frobenius itself is the identity, as it should be.

The degree of a field extension $[K:F]$ is the dimension of $K$ as a vector space over $F$. The tower law states that for finite extensions $F\subseteq E\subseteq K$,If the left side is prime, one factor is one. Degree one forces equality of the corresponding fields, so there is no proper intermediate field.

Let $K$ have 32 elements. Its prime field is ${\mathbb{F}}_2$, and its dimension over that field is five, since a dimension-$d$ vector space over ${\mathbb{F}}_2$ has $2^d$ elements. Every element outside ${\mathbb{F}}_2$ has minimal polynomial of degree five: its generated intermediate field has degree greater than one and dividing five.

Every element of $K$ satisfies $x^{32}=x$, by the multiplicative group of a field of order 31 for nonzero elements. Hence $X^{32}-X$ splits completely over $K$ and has no repeated roots, since its derivative is minus one. Conversely any irreducible polynomial $q$ over ${\mathbb{F}}_2$ gives a quotient ring ${\mathbb{F}}_2[X]/(q)$ with 32 elements. Its defining element satisfies $X^{32}-X$, so $q$ divides that polynomial and therefore has all five distinct roots in $K$.

The 30 elements of $K\setminus {\mathbb{F}}_2$ are thus partitioned into disjoint five-element root sets of the irreducible polynomials. Their number isNo classification of the six individual polynomials is needed; the tower degree and separability already count them all.