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Abstract Algebra

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Easy

Primitive Roots mod m

ABSALG-YVLLPE

Let $G$ be a group.

For $g\in G$, let $\langle g\rangle=\{g^n\mid n\in{\mathbb Z}\}$ be the cyclic subgroup generated by $g$.

The order of $h\in G$ is the order of the cyclic subgroup $\langle h\rangle$ it generates.

For example, if the group $({\mathbb Z}/m{\mathbb Z})^\ast$ of residue classes mod $m$ of the integers coprime to $m$ is cyclic, a generator is called a primitive root mod $m$, and the existence of a primitive root mod $m$ of course implies that $({\mathbb Z}/m{\mathbb Z})^\ast$ is cyclic.

Which of the following is correct? You don't need a calculator.

A

$5$ mod $18$ and $(-5)$ mod $18$ are both primitive roots mod $18$

B

$5$ mod $18$ and $7$ mod $18$ are both primitive roots mod $18$

C

$({\mathbb Z}/18{\mathbb Z})^\ast$ has $2$ elements of order $6$, $2$ elements of order $3$, and $1$ element of order $2$

D

$({\mathbb Z}/18{\mathbb Z})^\ast$ has 2 elements of order $6$, $1$ element of order $3$, and $2$ elements of order $2$

E

$({\mathbb Z}/18{\mathbb Z})^\ast$ has 1 element of order $6$, $2$ elements of order $3$, and $2$ elements of order $2$