Past exam of the algebra 2 course of the University of Oxford / 2014 / 3

The structure theorem for finitely generated modules over a principal ideal domain applies to a Euclidean domain because every Euclidean domain is a principal ideal domain. In invariant-factor form, every finitely generated module $M$ has a decompositionwhere $r\ge 0$, and each $d_i$ is a nonzero nonunit. The rank of a module $r$ is unique and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units. The zero module is allowed by taking $r=s=0$.

Every abelian group is a module over $\mathbb{Z}$, and a finite abelian group is finitely generated. Since $\mathbb{Z}$ is a Euclidean domain, the theorem applies. A nonzero free module summand would be infinite, so a finite group has $r=0$. Choosing positive integer representatives gives the finite abelian classificationThe ordered integers $d_i$ are uniquely determined. Splitting each $d_i$ into prime powers using the Chinese remainder theorem gives the equivalent elementary-divisor form as a direct sum of cyclic prime-power groups.

Use the Chinese remainder theorem to separate relatively prime factors:These five cyclic prime-power summands are the elementary-divisor form. Leave one $\mathbb{Z}/3\mathbb{Z}$ summand aside and combine the factors of orders $4,3,5,7$, which are pairwise coprime. Their direct sum of modules is cyclic of order $420$. The invariant-factor canonical form is thereforeIts group order is $3\cdot 420=1260=60\cdot 21$, and its exponent of a finite group is $420=\mathrm{lcm}(60,21)$, providing independent checks on the decomposition. Since its exponent is smaller than its order, $A$ itself is not cyclic.

We prove the unique-subgroup criterion for a finite abelian group to be cyclic using the invariant-factor classification from part (a). If $A$ is nontrivial, writewhere $C_m$ denotes a cyclic group of order $m$. If $s\ge 2$, choose a prime number $p$ dividing $d_1$; then $p$ also divides $d_2$. In the first cyclic factor, the element represented by $d_1/p$ has order of a group element $p$, and the corresponding element in the second factor has order $p$ as well.

The two elements supported in their separate coordinates generate distinct subgroups of order $p$: one has zero second coordinate, and the other has zero first coordinate. Both are subgroups of $A$, and $p\mid |A|$. This contradicts the assumed uniqueness for that divisor. Hence $s\le 1$, andIf $A$ is the trivial group, it is already cyclic, generated by its identity. Thus the argument covers every finite order.

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.