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# Computing the Order of an Element in a Group.

ABSALG-SVGA68

Let $G$ be a group with neutral element $e$. An element $g\in G$ is said to have finite order if we can compose $g$ a finite number of times with itself and get $e$.

When $g$ has finite order, the order of $g$ is defined to be the order of the cyclic subgroup $\langle g\rangle$ of $G$, generated by $g$. It is the smallest positive integer $k$ such that $g$ composed with itself $k$ times equals $e$.

If the cyclic subgroup $\langle g\rangle$ of $G$ has infinite order, then we say that the order of $g$ is infinite.

Which of the following are TRUE?

You do not need a calculator.

Select ALL that apply.

A

$3$ mod $11$ has order $5$ in the group $({\mathbb Z}/11{\mathbb Z})^\ast=\{a\,{\rm mod}\,11\mid a\not\equiv 0\, {\rm mod}\,11\}$ under multiplication mod $11$.

B

$2$ mod $11$ has order $5$ in the group $({\mathbb Z}/11{\mathbb Z})^\ast=\{a\,{\rm mod}\,11\mid a\not\equiv 0\, {\rm mod}\,11\}$ under multiplication mod $11$.

C

The element $\sqrt{2}$ of the group $[0,1)$ under addition mod $1$ has finite order ($x+y$ mod $1$ is the unique element $z$ of $[0,1)$ such that $x+y-m=z$ for some integer $m$).

D

The matrix

$\begin{pmatrix}0&1&1\cr0&0&1\cr0&0&0\end{pmatrix}$

generates a cyclic group of order $3$ under matrix multiplication.

E

The matrix

$\begin{pmatrix}0&0&1&0\cr0&0&0&1\cr-1&0&0&0\cr0&-1&0&0\end{pmatrix}$

generates a cyclic group of order $4$ under matrix multiplication.