UA-AUG-AUMAT229 Pre-read

Recall that a group of prime order must be cyclic (Discovery 11.5). And a cyclic group must be generated by an element of the same order as the group. So Conjecture 13.1.3 is equivalent to the following.

🔗

Theorem 13.1.4 (Cauchy).

If prime $p$ divides evenly into $| G |,$ then $G$ has an element of order $p .$

🔗

Notice that we have labelled this a theorem instead of a conjecture, because it is true. We will explore the reason it is true in the Discovery guide.

🔗

Subsection Words

🔗

A word is a string of symbols. In a multiplicative group, if we temporarily ignore the inverse operation, we perform group algebra calculations by first forming a word in the group elements, and then using the properties of the group operation to “simplify” to a specific element as a the calculation result.

🔗

Example 13.1.5. Words in $D_{3}$ .

Take the elements of $D_{3}$ to be

D_{3} = {e, r, r^{2}, s, r s, r^{2} s},

as usual. Let's relabel these elements as

D_{3} = {e, r_{1}, r_{2}, s_{0}, s_{1}, s_{2}},

where the labels match up with the original element expressions by position in the two lists.

Then as words we will consider

r_{2} s_{0}, s_{2}

to be different, though of course they compute to the same element of $D_{3} :$

r_{2} s_{0} = r^{2} s = s_{2} .

Similarly, as words the expressions

s_{0} r_{1}, e s_{2} s_{0} e s_{0},

are different, but again they compute to the same element of $D_{3}$ (check!).

Section 13.1 Pre-read

Subsection Reversing Lagrange's Theorem

Theorem 13.1.1 (Lagrange).

Conjecture 13.1.2. Converse of Lagrange's Theorem.

Conjecture 13.1.3.

Theorem 13.1.4 (Cauchy).

Subsection Words

Example 13.1.5. Words in D3.

Example 13.1.5. Words in $D_{3}$ .