order

The order of a group is its cardinality – the number of elements in the group. It is denoted as $|G|$.

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Lemma: $(G,\circ )$ group, $a\in G,m<n:a^n=a^m⟹\exists k\in {\mathbb{N}}^{+}:a^k=e$

Proof:

$$\begin{array}{rl}{a}^n=a^m& ⟹a^n(a^{-m})=a^m(a^{-m})\\ & ⟹a^{n-m}=e\\ & ⟹k=n-m\end{array}$$

The rotation group of a square has 4 elements: rotations by 0°, 90°, 180°, and 270°.

Let $a$ be a 90° rotation.
Then: $a^5=a^1$
$n=5,m=1⟹a^4=e$ → four 90° rotations bring us back to the starting position $✓$.

Order of a group element

$(G,\circ )$ group, $a\in G:{\text{ord}}_G(a):=\{\begin{array}{l}\infty \text{if }n\ne m⟹a^n\ne a^m\\ \min \{k>0\mid a^k=e\}\end{array}$
In words: The order of an element $a$ in a group $G$ is infinite if all powers are different (i.e. different exponent = different result). Otherwise, it is the smallest positive number $k$ for which $a^k=e$.

The order captures how many times we need to apply an element to itself before returning to the identity element. For instance, in a rotation group, it tells us how many rotations it takes to get back to the starting position. If we never return to the identity element no matter how many times we apply the operation, the order is infinite.

EXAMPLE

Given $({\mathbb{Z}}_5\setminus \{\bar{0}\},\cdot )$ (congruence class modulo 5 without 0), let’s find the order of $\bar{2}$ by computing successive powers:
$\bar{2},\bar{4}={\bar{2}}^2$
${\bar{2}}^3=\bar{3}$
${\bar{2}}^4=\bar{1}$
→ ${\text{ord}}_{{\mathbb{Z}}_5\setminus \{\bar{0}\}\}}(\bar{2})=4$
This shows that after multiplying $\bar{2}$ by itself 4 times, we get back to $\bar{1}$ (the identity element in this multiplicative group). Each power of $\bar{2}$ gives us a different element until we complete the cycle.