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Section 8.2 Discovery guide

In this set of activities, we will explore how a group's operation can be used to permute the group itself.

Discovery 8.1.

Consider \(\Z_{12}\text{,}\) i.e. clock arithmetic on a \(12\)-hour clock. Normally we take \(\Z_{12}\) to be the collection

\begin{equation*} \{0,1,2,\dotsc,11\} \text{,} \end{equation*}

with the clock wrapping back around to \(0\) after \(11\text{,}\) but for this activity we will take

\begin{equation*} \Z_{12} = \{1,2,\dotsc,11,12\} \text{,} \end{equation*}

with the clock wrapping back around to \(1\) after \(12\text{,}\) and so that \(12\) takes on the role of identity element instead of \(0\text{.}\)

(a)

Adding \(8\) to every number in \(\Z_{12}\) (adhering to clock arithmetic in the process) permutes the elements of the group. (The “add \(8\)” process must be a bijective correspondence because we can reverse it by subtracting \(8\text{.}\))

Since \(\Z_{12}\) has twelve elements, this “add \(8\)” permutation must be an element of the group \(S_{12}\text{.}\) Express this permutation as a product of disjoint cycles.

Hint.

Use the usual procedure: start with

\begin{equation*} (\, 1 \cyclesep\phantom{9}\cyclesep\phantom{5} \,)\text{,} \end{equation*}

figure out where the permutation sends object \(1\) (in this case, by adding \(8\)) and place that result as the next number in the cycle, then figure out where the permutation sends that result and place that new result as the next number in the cycle, and so on. When the pattern returns to a result of \(1\text{,}\) the cycle is complete — start a new cycle with the lowest number in \(\Z_{12}\) that hasn't appeared yet. And so on.

(b)

Repeat Task a for each of

  • adding \(4\) to every number in \(\Z_{12}\text{,}\)

  • adding \(6\) to every number in \(\Z_{12}\text{,}\)

  • adding \(2\) to every number in \(\Z_{12}\text{,}\)

  • adding \(1\) to every number in \(\Z_{12}\text{,}\) and

  • adding \(12\) to every number in \(\Z_{12}\text{.}\)

(c)

Is it possible that every permutation in \(S_{12}\) is somehow the result of this “add a fixed number to every number in \(\Z_{12}\)” process?

Hint.

How many elements are there in \(S_{12}\text{?}\) How many different ways are there to add a fixed number to every number in \(\Z_{12}\)?

(d)

Is it possible that two different ways of carrying out the “add a fixed number to every number in \(\Z_{12}\)” process (that is, by choosing different numbers to use as the “fixed number” at the start) could end up leading to the same permutation in \(S_{12}\text{?}\)

Discovery 8.2.

Let's try the same sort of thing as in Discovery 8.1, but for the group \(D_3\text{.}\) In this instance, we will be multiplying by a fixed choice of element from \(D_3\text{,}\) not adding. Once again, this will create a permutation of the elements of \(D_3\text{.}\) Since there are \(6\) elements in \(D_3\text{,}\) we will be able to interpret this permutation as a group element of \(S_6\text{,}\) if we first choose a relabelling of the elements of \(D_3\) (as in Example 8.1.4). So let's set up such a relabelling:

\begin{align*} x_1 \amp = e, \amp x_2 \amp = r, \amp x_3 \amp = r^2, \amp x_4 \amp = s, \amp x_5 \amp = r s, \amp x_6 \amp = r^2 s \text{.} \end{align*}

(a)

Using the relabelling above, write the permutation of \(D_3\) that results from multiplying every element by \(r\) (on the left) as an element of \(S_6\) (as a product of disjoint cycles).

Hint.

You may wish to dig up your multiplication table for \(D_3\) from previous discovery activity sets. Or, remember that \(D_3\) is the group of rotational symmetries of a thin, triangular plate with equilateral face. Or, remember the algebra patterns of \(D_3\text{:}\)

\begin{align*} r^3 \amp = e, \amp s^2 \amp = e, \amp s r \amp = r^2 s. \end{align*}

(b)

Repeat Task a for multiplication by \(s\) (on the left).

(c)

Repeat Task a for multiplication by \(r s\) (on the left).

(d)

Multiply your \(S_6\) element from Task a against your \(S_6\) element from Task b, and simplify to a product of disjoint cycles.

Compare this result with your \(S_6\) element from Task c. Surprised?

(e)

What is the point the tasks in this activity? What previous concept from this course is being demonstrated?

Similar to Example 8.1.7, If \(G\) is an infinite group then multiplication (or addition, as appropriate) of all elements in the group by some fixed choice of group element still permutes the group elements, but we won't be able to use a relabelling of the group elements to interpret this permutation as an element of \(S_n\text{,}\) no matter how large \(n\) is taken to be.

Discovery 8.3.

Consider a number line with vertices at integer intervals.

The set of integers as vertices on a number line.
Figure 8.2.1. The set of integers as vertices on a number line.

Suppose \(g\) is a fixed choice of group element in \(\Z\text{.}\) Use Figure 8.2.1 to make a geometric interpretation of the permutation of \(\Z\) that results from adding \(g\) to every element of \(\Z\text{.}\)

Discovery 8.4.

Is it possible for the permutation of a group \(G\) that results from multiplying (or adding, as appropriate) every element of \(G\) by a fixed choice of group element \(g\) to have fixed points?

(A fixed point would be a group element that is not actually permuted by the process of multiplying by \(g\text{.}\) In the case that \(G\) is finite and the “multiplication by \(g\)” permutation is interpreted as an element of \(S_n\text{,}\) a fixed point would correspond to a number that does not appear in any of the cycles when the permutation is expressed as a product of disjoint cycles.)

We have seen through several examples now that, given an element \(g\) in a group \(G\text{,}\) we obtain an element

\begin{equation*} \funcdef{\inducedperm{g}}{G}{G} \end{equation*}

of the permutation group \(S_G\) by defining

\begin{equation*} \inducedperm{g}(x) = g x \text{.} \end{equation*}

If \(G\) is finite with \(n\) group elements, then we may use a relabelling of those elements to interpret \(\inducedperm{g}\) as an element of \(S_n\text{,}\) but this is merely a convenience.

But we can vary the choice of element \(g\) in \(G\) to create different permutations of \(G\) (just as we did in Task 8.1.b). This creates a mapping \(G \to S_G\) by

\begin{equation*} g \mapsto \inducedperm{g} \text{.} \end{equation*}

This mapping won't be an isomorphism from \(G\) onto \(S_G\text{,}\) because we have seen in several activities so far (Task 8.1.c, Discovery 8.4) that not every permutation in \(S_G\) will correspond to a specific element of \(G\) in this way. But what more can we say about those elements of \(S_G\) that do?

Discovery 8.5.

Is it possible for two different elements of \(G\) to create the same permutation of \(G\) through multiplication? That is, if \(g,h\) are elements in \(G\text{,}\) is

\begin{equation*} \inducedperm{g} = \inducedperm{h} \end{equation*}

possible?

Hint.

Remember that \(\inducedperm{g}\) and \(\inducedperm{h}\) are functions. How can we tell when two different function descriptions (in this case, multiplication by \(g\) versus multiplication by \(h\)) actually create the same function? We can check by testing whether they always produce the same output result when both are fed the same input. What particular input into these two different functions might give you relevant information about \(g\) and \(h\text{?}\) (Remember that inputs to a permutation \(\inducedperm{g}\) should be elements of \(G\text{,}\) not numbers.)

Discovery 8.6.

Given element \(g\) in group \(G\text{,}\) permutation \(\inducedperm{g}\) is an element in group \(S_G\text{.}\) Write \(\inducedpermset{G}\) for the collection of all such elements of \(S_G\text{.}\)

Is the collection \(\inducedperm{G}\) a subgroup of \(S_G\text{?}\) This activity will lead you through the steps of The Subgroup Test (Version 1). (Clearly this collection is nonempty, since it contains an \(\inducedperm{g}\) for each element \(g\) of \(G\text{.}\))

(a)

Check closure under multiplication: Suppose we have two elements \(\inducedperm{g}\) and \(\inducedperm{h}\) of \(\inducedpermset{G}\text{.}\) Is \(\inducedperm{g} \circ \inducedperm{h}\) also an element of \(\inducedpermset{G}\text{?}\) Be sure that you can properly justify your answer.

Hint.

Every element of \(\inducedpermset{G}\) “looks like” \(\inducedperm{x}\) for some single element \(x\) of \(G\text{.}\) So to demonstrate that \(\inducedperm{g} \circ \inducedperm{h}\) is an element of \(\inducedpermset{G}\) you must come up with a proposed element \(x\) so that

\begin{equation*} \inducedperm{g} \circ \inducedperm{h} = \inducedperm{x} \text{,} \end{equation*}

and you must properly justify that these two functions are actually equal by considering how they permute the elements of \(G\text{.}\)

To get an idea of what to use for \(x\text{,}\) look back to the patterns uncovered for the example \(G = D_3\) in Discovery 8.2.

(b)

Check closure under taking inverses: Suppose we have an element \(\inducedperm{g}\) of \(\inducedpermset{G}\text{.}\) Is the inverse permutation \(\invinducedperm{g}\) also an element of \(\inducedpermset{G}\text{?}\) Be sure that you can properly justify your answer.

Hint.

Again, to demonstrate that \(\invinducedperm{g}\) is an element of \(\inducedpermset{G}\) you must come up with a proposed element \(x\) (probably based on \(g\)) so that

\begin{equation*} \invinducedperm{g} = \inducedperm{x} \text{,} \end{equation*}

and you must properly justify that these two functions are actually equal by considering how they permute the elements of \(G\text{.}\)

To get an idea of what to use for \(x\text{,}\) consider that since \(\inducedperm{g}\) is defined by permuting each \(y\) in \(G\) to \(g y\text{,}\) the inverse permutation \(\invinducedperm{g}\) has to permute \(g y\) back to \(y\text{.}\)

Discovery 8.7.

If \(\inducedpermset{G}\) is a subgroup of \(S_G\text{,}\) it must contain the identity permutation of \(G\) (i.e. the permutation of \(G\) that doesn't actually “mix up” any of the elements of \(G\)). So there must be a particular element \(x\) of \(G\) so that \(\inducedperm{x}\) is the identity permutation of \(G\text{.}\)

What is this particular element \(x\text{?}\) Be sure that you can properly justify your answer.