# 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:

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: