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Section 3.2 Discovery guide

Discovery 3.1. Table of results for addition modulo \(12\).

Fill in the addition table (technically, the Cayley table) for addition modulo \(12\text{.}\) Each entry in the table should be the result of adding the value in the reference column to the left of the entry with the value in the reference row above the entry. For example, we have already entered the result

\begin{equation*} 7 + 9 \equiv 4 \end{equation*}

from Example 3.1.2 in Section 3.1.

\(+\) 12 1 2 3 4 5 6 7 8 9 10 11
12
1
2
3
4
5
6
7 4
8
9
10
11
Figure 3.2.1.

Now we would like to consider the collection

\begin{equation*} G_{12} = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 \} \end{equation*}

with the binary operation of addition modulo \(12\text{.}\) Clearly the closure axiom is satisfied, since we will always use the fact that the clock “resets” at \(12\) to reduce any addition result to a number between \(1\) and \(12\) (inclusive).

Discovery 3.2. Addition modulo \(12\) as a group operation.

(a) Associativity.

How about the associativity axiom? Choose a few examples of triples of values \(a,b,c\) from \(G_{12}\text{.}\) For each example, use your Cayley table from Discovery 3.1 to compute each of

\begin{equation*} (a + b) + c, \qquad a + (b + c), \end{equation*}

making sure to follow the order of operations implied by the brackets, and verify that both orders of computation result in the same result.

(b) Identity.

What number acts as the additive identity?

(c) Negatives.

Clearly \(G_{12}\) contains no negative values. Yet every element in an additive group must have a corresponding negative defined by the property that a number plus its negative results in the identity element. Using your Cayley table from Discovery 3.1 along with the identity element you identified in Task b, fill in the following table of negatives, so that \(x + (-x) = 0\) in all cases. Keep in mind that

  • \(0\) doesn't mean the number zero, it means the element of \(G_{12}\) that acts in the role of the additive identity;

  • all entries in the table must be values from \(G_{12}\).

\(x\) 12 1 2 3 4 5 6 7 8 9 10 11
\(-x\)
Figure 3.2.2.

The conclusions of Discovery 3.2 hold in general: for every positive value of \(n\text{,}\) the collection

\begin{equation*} G_n = \{ 1, 2, \dotsc, n \} \end{equation*}

forms a group under the operation of addition modulo \(n\) (i.e. clock arithmetic with an \(n\)-hour clock instead of a \(12\)-hour clock).

The pattern of Task 3.2.b also holds in general: the number \(n\) will act as the additive identity. However, mathematically it makes more sense to have an \(n\)-hour clock start at \(0\) instead of starting at \(n\text{.}\) So rather than having the number \(n\) act in the role of the number zero, we should just include the number zero in our group in its place.

With this in mind, we will replace \(G_n\) with the collection

\begin{equation*} \Z_n = \{ 0, 1, 2, \dotsc, n-1 \} \end{equation*}

of integers modulo \(n\). And the patterns of Discovery 3.2 allow us to conclude that \(\Z_n\) is always an additive group.

What about multiplication modulo \(n\text{?}\)

Discovery 3.3. Tables of results for multiplication modulo \(5\) and modulo \(6\).

Fill in the multiplication tables for multiplication modulo \(5\) and modulo \(6\text{.}\) Each entry in the tables should be the result of multiplying the value in the reference column to the left of the entry with the value in the reference row above the entry. In the table on the left these multiplication results should be “reduced” modulo \(5\) before being entered into the table, and in the table on the right the results should be “reduced” modulo \(6\text{.}\)

\(\times\) 0 1 2 3 4
0
1
2
3
4
Figure 3.2.3.
\(\times\) 0 1 2 3 4 5
0
1
2
3
4
5
Figure 3.2.4.

Now we consider whether the collection \(\Z_n\) is a group under the binary operation of multiplication modulo \(n\text{,}\) first in the example cases of \(n = 5\) and \(n = 6\text{.}\) Again, clearly the closure axiom is satisfied, since the modulo part of multiplication modulo \(n\) ensures that our calculation results will always be a value from \(\Z_n\) (keeping in mind that we are now always placing the value \(0\) “at the top of the clock”).

Discovery 3.4. Multiplication modulo \(5\) and modulo \(6\) as group operations.

(a) Associativity.

How about the associativity axiom? Choose a few examples of triples of values \(a,b,c\) from each of \(\Z_5\) and \(\Z_6\text{.}\) For each example, use the appropriate multiplication table Discovery 3.3 to compute each of

\begin{equation*} (a b) c, \qquad a (b c), \end{equation*}

making sure to follow the order of operations implied by the brackets, and verify that both orders of computation result in the same result.

(b) Identity.

What number acts as the multiplicative identity in \(\Z_5\text{?}\) What number acts as the multiplicative identity in \(\Z_6\text{?}\)

(c) Inverses.

Clearly \(\Z_n\) contains no fractions. Yet every element in an multiplicative group must have a corresponding inverse defined by the property that a number times its inverse results in the identity element. Using your multiplication tables from Discovery 3.3 along with the identity elements you identified in Task b, fill in the following tables of inverses, so that \(x \inv{x} = 1\) in all cases. Keep in mind that

  • \(1\) doesn't (necessarily) mean the number one, it means the element of \(\Z_n\) that acts in the role of the multiplicative identity;

  • all entries in the tables must be values from \(\Z_5\) or \(\Z_6\), as appropriate.

You will find that some elements of \(\Z_5\) and \(\Z_6\) do not have inverses. For those elements, enter the letter “s” as the \(\inv{x}\) entry, where “s” stands for singular.

\(x\) 0 1 2 3 4
\(\inv{x}\)
Figure 3.2.5.
\(x\) 0 1 2 3 4 5
\(\inv{x}\)
Figure 3.2.6.

In Discovery 2.3, you (hopefully) discovered that a collection with a binary operation that satisfies the Closure, Associativity, and Identity axioms can be made into a group by discarding those elements that do not have a corresponding “opposite”. Since Discovery 2.3 was about multiplicative collections, in that activity you would have kept only invertible elements (discarding all singular elements) to “fix” the collections as instructed.

In general, when \(M\) is a collection endowed with a multiplication operation that satisfies the Closure, Associativity, and Identity axioms but not the Inverses axiom, we will write \(\multunits{M}\) for the multiplicative group of invertible elements from \(M\text{.}\)

Discovery 3.5.

(a)

Based on Task 3.4.c, write out the elements of the multiplicative groups \(\multunits{\Z}_5\) and \(\multunits{\Z}_6\text{.}\)

(b)

Do you see a pattern to which values were included in and which were excluded from each of \(\Z_5\) and \(\Z_6\) to form \(\multunits{\Z}_5\) and \(\multunits{\Z}_6\text{,}\) respectively?

Hint.

The pattern has something to do with a particular relationship between each included/excluded value and the number \(n\) that defines the collection \(\Z_n\text{.}\)

(c)

Using your pattern from Task b, write out what you think the elements of \(\multunits{\Z}_{12}\) should be. Then write out the multiplication table for \(\Z_{12}\) and see if you were correct.