Powers, order of an element কী?

Example 7.15

In $(\C^*, \times)$: $\iu$ has order $4$, with $\langle \iu \rangle = \{1, \iu, -1, -\iu\} = \mathbb{U}_4$; more generally $\omega = \eu^{2\iu\pi/n}$ has order $n$ and $\langle\omega\rangle = \mathbb{U}_n$. In $(\Z, +)$, every $x \neq 0$ has infinite order. Why the claims in the definition hold: if $x$ has order $m$, divide any $k$ by $m$ ($k = mq + r$, $0 \leq r < m$, Theorem 6.2): $x^k = (x^m)^q x^r = x^r$, so the powers cycle with period $m$, the listed elements are pairwise distinct by minimality of $m$, and $x^k = e$ forces $r = 0$. Orders of permutations are computed in the weekend problem below.

Example 7.16 (Orders inside $\mathbb U_{12}$)

What is the order of $\omega^k$ in $\mathbb U_n$, for $\omega = \eu^{2\iu\pi/n}$? One has $(\omega^k)^m = 1$ iff $n \mid km$, and writing $d = \gcd(n, k)$, $n = dn'$, $k = dk'$ with $\gcd(n', k') = 1$: $n \mid km \iff n' \mid k'm \iff n' \mid m$ (Gauss’s lemma, Theorem 6.8). The least such $m \geq 1$ is $n' = \frac{n}{\gcd(n,k)}$. In $\mathbb U_{12}$ for instance, $\omega^8$ has order $\frac{12}{\gcd(12,8)} = 3$ (indeed $\omega^8 = \eu^{4\iu\pi/3} \in \mathbb U_3$), while $\omega^5$ has order $12$: it generates the whole group, though it is not the “standard” generator. Counting the generators — the $k$ with $\gcd(k, n) = 1$ — recovers the coprime counts of Example 2.25: group theory and counting meet.

Example 7.6 (The symmetries of a rectangle)

A (non-square) rectangle admits exactly four isometries onto itself: the identity $e$, the horizontal-axis reflection $h$, the vertical-axis reflection $v$, and the half-turn $r$ about the center. Composition makes this four-element set a group: each element is its own inverse ($h^2 = v^2 = r^2 = e$), and the product of any two distinct non-identity elements is the third ($hv = vh = r$: reflecting in both axes is the half-turn). The full table is symmetric, so the group is abelian — yet it is not the same group as the rotations $\mathbb U_4$ of Example 7.15: there, $\iu$ has order $4$, while here every element has order $\leq 2$. Two groups of the same size can thus have genuinely different multiplication structures — the figure below displays both tables side by side. This four-element group returns as $\{\pm1\} \times \{\pm1\}$, and Exercise 7.7 explains why any group with all squares trivial must, like this one, be abelian.