Exercise 1.1.35 (If $|x| = n$, then $\langle x \rangle = \{1,x,x^2,\ldots,x^{n-1}\}$)

Question

If $x$ is an element of finite order $n$ in $G$, use the Division Algorithm to show that {\bf any} integral power of $x$ equals one of the elements in the set $\{1,x,x^2,\ldots,x^{n-1}\}$ (so there are all the distinct elements of the cyclic subgroup (cf. Exercise 27 above) of $G$ generated by $x$).

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $x^a$ be any integeral power of $x$, with $a \in \mathbb{Z}$.

The Division Algorithm gives $q,r$ such that
$$a =n q + r,\qquad 0 \leq r < n.$$
Therefore
$$x^a = (x^n)^q\cdot x^r = x^r.$$
Hence
$$x^a \in \{1,x,x^2,\ldots,x^{n-1}\}.$$
(So $\langle x \rangle =  \{1,x,x^2,\ldots,x^{n-1}\}.$)

\end{proof}

Exercise 1.1.35 (If $|x| = n$, then $\langle x \rangle = \{1,x,x^2,\ldots,x^{n-1}\}$)

Answers

Comments

Exercise 1.1.35 (If $|x| = n$, then $\langle x \rangle = \{1,x,x^2,\ldots,x^{n-1}\}$)

Answers

Comments

Add answer