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Section 13.1 Pre-read

Subsection Reversing Lagrange's Theorem

Recall.

The converse of this statement is the following.

Unfortunately, Conjecture 13.1.2 is false, as we have already seen in Discovery 11.7.

However, consider the following partial converse.

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.

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

\begin{equation*} D_3 = \{ e, r, r^2, s, r s, r^2 s \} \text{,} \end{equation*}

as usual. Let's relabel these elements as

\begin{equation*} D_3 = \{ e, r_1, r_2, s_0, s_1, s_2 \} \text{,} \end{equation*}

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

Then as words we will consider

\begin{equation*} r_2 s_0, \qquad s_2 \end{equation*}

to be different, though of course they compute to the same element of \(D_3\text{:}\)

\begin{equation*} r_2 s_0 = r^2 s = s_2 \text{.} \end{equation*}

Similarly, as words the expressions

\begin{equation*} s_0 r_1, \qquad e s_2 s_0 e s_0 \text{,} \end{equation*}

are different, but again they compute to the same element of \(D_3\) (check!).