# 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 $|x| = |H|$

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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