Past exam of the rings and modules course of the University of Oxford / 2021 / 3 / c

Let $N=|G|$ be a square-free integer. If $N=1$, then $G$ is the trivial cyclic group. Otherwise, finiteness means that a direct sum of modules decomposition into cyclic groups has only finitely many nontrivial factors, each of finite group order. Omit trivial factors and writeThe integers $n_i,n_j$ are coprime integers whenever $i\ne j$. If a prime number $p$ divided both, then $p^2\mid N$, contradicting the square-free integer hypothesis.

Moreover, $n_1\cdots n_{j-1}$ and $n_j$ are coprime integers: any common prime divisor would divide one of the earlier factors as well as $n_j$. Thus part (b)(ii) can be applied repeatedly to combine factors, givingIn particular $G$ is a cyclic group. Under the displayed identifications, the element whose coordinate in every factor is the class of $1$ is a generator of a group.

Write $q=|\mathbb{F}|$. The unit group $U(\mathbb{F})=\mathbb{F}\setminus \{0\}$ is a finite abelian group of group order $q-1$, since multiplication in a field satisfies commutativity. The structure theorem for finitely generated abelian groups, applied to this finite group, gives invariant factors of a finitely generated moduleafter omitting trivial factors. If there are no factors, the unit group is trivial and hence a cyclic group.

Otherwise put $D=d_r$. Every element $u\in U(\mathbb{F})$ satisfies $u^D=1$, because the order of a group element in each coordinate divides $D$. Hence all $q-1$ distinct nonzero elements of $\mathbb{F}$ are roots of a polynomial $X^D-1$. The root bound for a polynomial over a field givesFor completeness, that bound follows by induction from the factor theorem: if a nonzero polynomial $f$ vanishes at $a$, write $f=(X-a)g$; every other root of a polynomial is a root of $g$, because a field has no zero divisors. Each new distinct root therefore uses one unit of polynomial degree.

On the other hand, the group order in the decomposition isEquality is forced, so $d_1\cdots d_{r-1}=1$. As every retained factor is greater than $1$, there can be only one. ConsequentlyThe same proof establishes the more general result finite multiplicative subgroup of a field is cyclic. The square-free integer result in part (c)(i) alone would not suffice, since $q-1$ need not be a square-free integer; the root bound for a polynomial rules out multiple invariant factors of a finitely generated module even then.