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 $g ∙ g ∙ . . . g$ , $g$ 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 $g$ , [ $? \leq ?$ ].

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 $| g | \leq j - i \leq | G |$ .

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 $N \in R$ . Then:

[

g^{N} = e ⟺ N 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{lcm (m, | g |)}{?} = \frac{?}{\gcd (m, | g |)} .

]

12.11.2020