?

Abstract Algebra

Free Version

Upgrade subject to access all content

Difficult

Number of Roots of Unity in an Abelian Group

ABSALG-XZWEJ4

In answering this question, you may use the following true facts:

Fact 1 - If $A$ is a finite abelian group and $H$ is the largest cyclic subgroup of $A$, then:

the order of any other cyclic subgroup of $A$ divides the order of $H$.

Fact 2 - In the group $A=({\mathbb Z}/p{\mathbb Z})^\ast$, $p\ge 2$ prime, there are:

at most $m$ solutions to $x^m=1$, for every $m\ge 1$.

Which of the following is FALSE?

(Notation: For $N\ge 2$ an integer, we denote by $({\mathbb Z}/N{\mathbb Z})^\ast$ the group of integers mod $N$ coprime to $N$.)

A

If $A$ is an abelian group, with neutral element $1$, such that, for every $m\ge 1$, there are at most $m$ solutions to $x^m=1$, then:

every finite subgroup of $A$ is cyclic

B

The group $({\mathbb Z}/p^2{\mathbb Z})^\ast$ is cyclic, for $p\ge 2$ prime

C

The group $({\mathbb Z}/pq{\mathbb Z})^\ast$ is cyclic, for $p$, $q$ positive odd primes and $p\not=q$.

D

The group $({\mathbb Z}/18{\mathbb Z})^\ast$ contains at most $m$ solutions to $x^m=1$ for every integer $m\ge 1$

E

In a finite cyclic group with neutral element $1$ there are at most $m$ solutions to $x^m=1$, for every integer $m\ge 1$