Conjugate of permutations

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Christian Lawson-Perfect

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Christian Lawson-Perfect

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10 years, 1 month ago

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Christian Lawson-Perfect 10 years, 1 month ago

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Christian Lawson-Perfect 10 years, 8 months ago

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Name	Status	Author	Last Modified
Conjugate of permutations	Ready to use	Christian Lawson-Perfect	01/06/2016 10:06
Lois's copy of Conjugate of permutations	draft	Lois Rollings	19/08/2020 13:06

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Question statement

Give any introductory information the student needs.

			Name	Type	Generated Value

conjugate1	permutation	perm("(1,9,4,8,3,5,2)")
conjugate2	permutation	perm("(1,4,6,3,5)")
tau2	permutation	perm("(1,2,5)(3,6,4)")
tau1	permutation	perm("(1,9,6,4,3,8,7,5,2)")
sigma1	permutation	perm("(1,6,3,4,7,5,2)")
sigma2	permutation	perm("(2,5,6,3,4)")

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
tau1*sigma1*inverse(tau1)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `permutation`

perm("(1,9,4,8,3,5,2)")

← Depends on:

sigma1
tau1

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Match text pattern
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Part b) Gap-fill
1. Gap Match text pattern
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Where $\sigma = \var{sigma1}$ and $\tau = \var{tau1}$ , write down the permutation $\tau \sigma \tau^{-1}$ .

$\tau \sigma \tau^{-1}$ = Unnamed gap

Advice

You could find $\tau \sigma \tau^{-1}$ by first finding $\tau^{-1}$ and then performing the composition by hand, but there's a quicker way: $\tau \sigma \tau^{-1}$ has the same cycle structure as $\sigma$, but each $n$ is replaced by $\tau(n)$.

Proof

Assume, without loss of generality, that $\sigma$ is a single cycle $(b_1, \dots, b_n)$, and let $\pi = \tau \sigma \tau^{-1}$. Look at what happens to $\tau(b_i)$ when acted on by $\sigma$:

\[ \tau(b_i) \overset{\tau^{-1}}\mapsto b_i \overset{\sigma}\mapsto b_{i+1} \overset{\tau}\mapsto \tau(b_{i+1}) \]

So, for any $1 \lt i \leq n$, $\tau(b_i) \overset{\pi}\mapsto \tau(b_{i+1})$.

For any $x \notin \{b_1,\dots,b_n\}$, $\sigma(x) = x$, so $\pi(x) = x$.

Hence, $\pi$ is the cycle $(\tau(b_1), \dots, \tau(b_n))$.

a)

$\var{conjugate1}$

b)

$\var{conjugate2}$

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This question is used in the following exam:

Permutations by Christian Lawson-Perfect in Christian's workspace.

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Taxonomy: mathcentre

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Taxonomy: Context

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Checkpoint description

Christian Lawson-Perfect 10 years, 1 month ago

Christian Lawson-Perfect 10 years, 8 months ago

Error in variable testing condition

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Generated value: `permutation`

Parts

Gap-fill

Proof

a)

b)

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Conjugate of permutations

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 10 years, 1 month ago

Christian Lawson-Perfect 10 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: permutation

Parts

Gap-fill

Proof

a)

b)

Generated value: `permutation`