Equivalence of Definitions of Order of Group Element

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Theorem

The following definitions of the concept of Order of Group Element are equivalent:

Let $G$ be a group whose identity is $e$.

Let $x \in G$.

Definition 1

The order of $x$ (in $G$), denoted $\order x$, is the smallest $k \in \Z_{> 0}$ such that $x^k = e_G$.

Definition 2

The order of $x$ (in $G$), denoted $\order x$, is the order of the group generated by $x$:

$\order x := \order {\gen x}$

Definition 3

The order of $x$ (in $G$), denoted $\left\vert{x}\right\vert$, is the largest $k \in \Z_{\gt 0}$ such that:

$\forall i, j \in \Z: 0 \le i < j < k \implies x^i \ne x^j$


Proof

Let $k$ be the order of $x$ in $G$ according to Definition 1.

Let $l$ be the order of $x$ in $G$ according to Definition 3.


Definition $1$ is equivalent to Definition $3$

Aiming for a contradiction, suppose $k \ne l$.

Then from Ordering on 1-Based Natural Numbers is Trichotomy either $k < l$ or $k > l$.


According to Definition 3:

$\forall i, j \in \Z: 0 \le i < j < l \implies x^i \ne x^j$


If $k < l$, then letting $i = 0$ and $j = k$ yields a contradiction by Element to Power of Zero is Identity.


If $k > l$, then:

$\exists i,j \in \Z: 0 \le i < j < k: x^i = x^j$


But then $x^{j - i} = e$ and $j - i < k$, contradicting the fact that $k \in \Z_{> 0}$ is the smallest such that $x^k = e$.


Therefore $k = l$.

$\Box$


Definition $1$ is equivalent to Definition $2$

It follows straight away from List of Elements in Finite Cyclic Group that $\order {\gen a} = k$:

$\gen a = \set {a^0, a^1, a^2, \ldots, a^{k - 1} }$

$\blacksquare$


Sources

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