Problem 4.6. Do the two Latin squares in Ezample 4.3 have the same reduced form?

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
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Chapter10: Sequences, Series, And Probability
Section10.3: Geometric Sequences
Problem 52E
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Please just solve problem 4.6. The example it refers to is attached with another picture. Thank you!

Problems
Problem 4.1. Make a circulant Latin square for the symbol set {a, b, c, d, e, f, 9, h}.
Problem 4.2. Show that there are at least n! circulant Latin squares.
Problem 4.3. Prove that normalizing to the same Latin square is an equivalence relation.
Problem 4.4. Prove that being isotopic is an equivalence relation on Latin squares.
Problem 4.5. Show that, up to being isotopic, there is only one 3 x 3 Latin square.
Problem 4.6. Do the two Latin squares in Example 4.3 have the same reduced form?
Problem 4.7. Give the Latin square that is the Cayley table of S3, the symmetric group on three
letters.
Problem 4.8. Prove or disprove: the direct products of circulant Latin squares are circulant Latin
squares.
Transcribed Image Text:Problems Problem 4.1. Make a circulant Latin square for the symbol set {a, b, c, d, e, f, 9, h}. Problem 4.2. Show that there are at least n! circulant Latin squares. Problem 4.3. Prove that normalizing to the same Latin square is an equivalence relation. Problem 4.4. Prove that being isotopic is an equivalence relation on Latin squares. Problem 4.5. Show that, up to being isotopic, there is only one 3 x 3 Latin square. Problem 4.6. Do the two Latin squares in Example 4.3 have the same reduced form? Problem 4.7. Give the Latin square that is the Cayley table of S3, the symmetric group on three letters. Problem 4.8. Prove or disprove: the direct products of circulant Latin squares are circulant Latin squares.
Example 4.3. Applying the permutation (123) in cycle notation transforms an exemplary Latin
square as follows:
2
3 4 1
3
1.
4 2
3
1 2 4
2 3 4
4 3 2
4
1
3
4 2 1 3
4 3 2
1
2.
Transcribed Image Text:Example 4.3. Applying the permutation (123) in cycle notation transforms an exemplary Latin square as follows: 2 3 4 1 3 1. 4 2 3 1 2 4 2 3 4 4 3 2 4 1 3 4 2 1 3 4 3 2 1 2.
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