# Order of element and order of cyclic group coincide

## Homework Statement

Let ##G## be a group and ##x \in G## any element. Prove that if ##|x| = n##, then ##|x| = |\{x^k : k \in \mathbb{Z} \}|##.

## The Attempt at a Solution

Let ##H = \{x^k : k \in \mathbb{Z} \}##. I claim that ##H = \{1,x,x^2, \dots , x^{n-1} \}##. First, we show that these sets are equal. Let ##m \in \mathbb{Z}##. By the Euclidean algorithm, ##m = nq + r##, where ##r \in [1, n)##. Then ##x^m = x^{nq+r} = (x^n)^qx^r = x^r \in \{1,x,x^2, \dots , x^{n-1} \}##. The reverse containment is obvious. Now, we show that ##H## has ##n## distinct elements. Let ##i,j \in [1,n)## and suppose that ##x^i = x^j##. Then ##x^{i-j}=1##. So ##n ~ | ~ i-j##, which implies that ##i \equiv j ~ (\bmod n)##. But since ##1 \le i,j < n##, ##i = j##. Hence, all of the elements of ##H## are distinct and ##|x| = |H|##

fresh_42
Mentor

## Homework Statement

Let ##G## be a group and ##x \in G## any element. Prove that if ##|x| = n##, then ##|x| = |\{x^k : k \in \mathbb{Z} \}|##.

## The Attempt at a Solution

Let ##H = \{x^k : k \in \mathbb{Z} \}##. I claim that ##H = \{1,x,x^2, \dots , x^{n-1} \}##. First, we show that these sets are equal. Let ##m \in \mathbb{Z}##. By the Euclidean algorithm, ##m = nq + r##, where ##r \in [1, n)##. Then ##x^m = x^{nq+r} = (x^n)^qx^r = x^r \in \{1,x,x^2, \dots , x^{n-1} \}##. The reverse containment is obvious. Now, we show that ##H## has ##n## distinct elements. Let ##i,j \in [1,n)## and suppose that ##x^i = x^j##. Then ##x^{i-j}=1##. So ##n ~ | ~ i-j##, which implies that ##i \equiv j ~ (\bmod n)##. But since ##1 \le i,j < n##, ##i = j##. Hence, all of the elements of ##H## are distinct and ##|x| = |H|##
Looks good. Only one minor mistake: The lower bound for the remainders is ##0##, not ##1##, so ##r,i,j \in [0,n)## and ##0\leq i,j <n ## which doesn't harm the proof, as ##x^0=1\,.##

Mr Davis 97