Draw Lattice Diagram for K: Solving Algebra I Mapping Homework

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SUMMARY

The discussion centers on solving the algebraic mapping problem K = {x C S7 | 2x=2, {1,4}x={1,4}, {1,5,7}x={1,5,7}} and drawing a corresponding Lattice Diagram. The participant initially calculated 140 unique solutions using the formula 3!3!2!2! - 4 but later clarified that the mappings imply specific relationships between elements. Ultimately, the simplified combinations of {5,7} and {3,6} lead to the final mappings: (57)(36), I(36), (57)I, and I sub 7.

PREREQUISITES
  • Understanding of algebraic mappings and functions
  • Familiarity with Lattice Diagrams in mathematics
  • Knowledge of combinatorial calculations, specifically factorials
  • Basic set theory concepts
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  • Study the construction of Lattice Diagrams in algebra
  • Explore combinatorial mathematics, focusing on factorial applications
  • Learn about algebraic mappings and their properties
  • Investigate advanced set theory and its implications in algebra
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Students of algebra, mathematicians interested in combinatorial problems, and educators teaching algebraic concepts and Lattice Diagrams.

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Homework Statement



K = {x C S7 | 2x=2, {1,4}x={1,4}, {1,5,7}x={1,5,7}}.
Draw Lattice Diagram for K.2. The attempt at a solution

I've looked at this for about 30 minutes and came to the conclusion that there are 140 unique solutions to this mapping, and I know for a fact that the professor does not want us to draw all 140 of these. Essentially, I found that 3!3!2!2! - 4 = 140. At this point I have no idea how to express this in a far easier form then what is given and I'm not sure where to go with the Lattice Diagram. Help would def be appreciated.
 
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Never mind. Mods can delete this thread because I've figured it out. The statement above states that {1,4}x = {1,4}, but similarly {1,5,7}x = {1,5,7}, which implies that for this statement to be true, 1 maps to 1, and 4 maps to 4, therefore all we are left with is combinations of {5,7} and {3,6}, which are simply (57)(36), I(36), (57)I, and I sub 7.
 

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