 Math and science::Algebra::Aluffi

# Groups. Order.

There are two related concepts of order: the order of an element, and the order of a group.

### Element order

Let $$g$$ be an element of group $$G$$. $$g$$ is said to have finite order iff [what condition?]. We say that the order of $$g$$ is $$k$$ iff $$k$$ is the [something]. We write $$|g| = k$$.

If there is no such natural number, $$g$$ does not have finite order and we write $$|g| = \infty$$.

Notation: $$e$$ is shorthand for $$e_G$$, the identity element of group $$G$$. $$g^n$$ is notation for $$g \bullet g \bullet ... g$$, $$g$$ composed with itself $$n$$ times.

### Group order

Let $$G$$ be a group. If $$G$$ has finite elements, then we say that the order of $$G$$ is [what?]. We write $$|G|$$ to denote the order of $$G$$. Otherwise, [when the opposite is true], we write $$|G| = \infty$$.

### 4 Lemmas

For a group $$G$$ and element $$g$$, [$$\quad ? \, \le \; ? \quad$$].

Proof. This is vacuously true if $$|G| = \infty$$. If $$G$$ has finite order, then consider $$|G| + 1$$ powers of $$g$$: $$g^0, g^1, g^2 ... g^{|G|}$$. All of these powers can't be unique, otherwise $$G$$ would have more than $$|G|$$ elements. Let $$g^i = g^j$$ be two repeated elements in the sequence (with $$i < j$$). Then $$g^{j-i}$$ must be the identity, and we have $$|g| \le j - i \le |G|$$.

Let $$g$$ be an element of a group. If $$g^n = e$$, then $$|g|$$ is a [what?] of $$n$$.

Aluffi cheats a little in his proof. I think the proof requires an inductive proof. I think it would be worth checking out how the inductive proof is set up.

An immediate consequence of the above lemma is the following corollary:

Let $$g$$ be an element with finite order, and let $$N \in \mathbb{R}$$. Then:

[$g^N = e \iff N \text{ is a (what?) of } |g|$]

Let $$G$$ be a group and $$g \in G$$ be an element of finite order. Then for any $$m > 0$$, $$g^m$$ has finite order. Specifically, the order of $$g^m$$ is related to the order of $$g$$ as follows:

[$|g^m| = \frac{\operatorname{lcm}(m, |g|)}{?} = \frac{?}{\operatorname{gcd}(m, |g|)}.$]