at 30:30, Is the "intermediate group" Z3 just chosen randomly? I realize there are only two real options for subgroups of S3 (Z3 or a transposition subgroup), but it seems like we are missing a step where we would look at our polynomial and decide to use Z3 here.
I don't understand why the fact that p(x,y,z) is invariant under Z3, implies that it takes only 2 different values (1 value under the 3-cycles, ok this is by definition, but why do we know that it takes only one other value under all transpositions?) Why does the order of the quotient group Z2 (2 elements) tells us the number of different values p(x,y,z) can take?
I just finished your "Exploring Abstract Algebra II". It is the most excellent lectures in math I have taken.
I've seen the three videos and I have to congratulate you and to encourage every one to get until the end, EXCELENT JOB
at 30:30, Is the "intermediate group" Z3 just chosen randomly? I realize there are only two real options for subgroups of S3 (Z3 or a transposition subgroup), but it seems like we are missing a step where we would look at our polynomial and decide to use Z3 here.
This is a fantastic presentation. It's clear that a lot of thought and preparation went into it.
Hello, Can you help me to solve some problems in math
I don't understand why the fact that p(x,y,z) is invariant under Z3, implies that it takes only 2 different values (1 value under the 3-cycles, ok this is by definition, but why do we know that it takes only one other value under all transpositions?)
Why does the order of the quotient group Z2 (2 elements) tells us the number of different values p(x,y,z) can take?