UA-AUG-AUMAT229 Discovery guide

Symmetric. Verify that if group element $x$ is conjugate to group element $y$ (i.e. there exists a group element $g$ so that $y = g^{- 1} x g$ ), then $y$ is also conjugate to $x$ (i.e. there exists a group element $h$ so that $x = h^{- 1} x h$ ).

🔗

(c)

Transitive. Verify that if group element $x$ is conjugate to group element $y,$ and $y$ is conjugate to group element $z,$ then also $x$ is conjugate to $z .$

🔗

Since conjugacy is an equivalence relation, the equivalence classes (called conjugacy classes in this case) partition the group.

🔗

Computing conjugacy classes is tedious but not difficult: to compute the class

[x],

compute

g^{- 1} x g

for all group elements

g,

and collect the results in the class. Then move on to calculating another class by choosing an element

y

that doesn't appear in

[x]

(since otherwise you would just be calculating

[x]

all over again), and computing

g^{- 1} y g

for all group elements

g .

And so on.

🔗

Discovery 14.4.

Compute the conjugacy classes of the following groups.

🔗

(a)

$Z_{6} .$

🔗

(b)

$Z .$

🔗

(c)

$D_{3} .$

🔗

(d)

$D_{4} .$

🔗

Discovery 14.5.

What happened for the Abelian examples in Discovery 14.4? Would the same happen in every Abelian group? Justify your answer.

🔗

Discovery 14.6.

Use the patterns of the conjugacy classes in $D_{3}$ and in $D_{4}$ from Discovery 14.4 to guess what the conjugacy classes in $D_{5}$ and in $D_{6}$ should be.

🔗

Subsection Conjugacy in $S_{n}$

🔗

For permutations

p

and

q

S_{n},

there is a trick to computing a permutation

g

so that

q = g^{- 1} p g .

🔗

Discovery 14.7.

Consider the following two elements of $S_{11} .$

\begin{matrix} p = (1 5 9 11) (2 4 6) (7 10) (3) (8) \\ ↑ g ↓ g^{- 1} \\ q = (1 2 3 4) (5 6 7) (8 9) (10) (11) \end{matrix}

Notice that $p$ and $q$ have the same cycle structure (and this time we have included the $1$ -cycles) and that we have written $q$ above $p$ so that cycles of the same size are above/below one another.

As indicated by the arrows, take $g$ to be the permutation that sends each number to the one that appears above it. For example, $g$ fixes $1,$ sends $2$ to $5,$ and so on.

Write out the permutations $g$ and $g^{- 1}$ as products of disjoint cycles. Then compute $g^{- 1} p g$ to confirm that the result is $q .$

🔗

The existence of the trick in Discovery 14.7 suggests that permutations with the same cycle structure are always conjugate. As we will read in the textbook, the converse is also true.

🔗

Theorem 14.2.1. Conjugacy in $S_{n}$ .

Two permutations in $S_{n}$ are conjugate precisely when they have the same cycle structure. That is, two permutations are conjugate when they have the same cycle structure, and not conjugate when they do not have the same cycle structure.

🔗

Discovery 14.8.

Instead of lots of tedious calculations, use Theorem 14.2.1 to write down all of the conjugacy classes of $S_{4} .$ (Remember that $S_{4}$ has $24$ elements, so you'll know when you're done by counting elements. Also, note that a single $1$ -cycle by itself is just the identity permutation.)

🔗

Warning 14.2.2.

The pattern of Theorem 14.2.1 does not work in $A_{n},$ because in order for two permutations $x$ and $y$ to be conjugate in $A_{n},$ there must exist an even permutation $g$ so that $y = g^{- 1} x g .$