Think of an equilateral triangle where you label the corners 1, 2, 3 starting from one vertex and moving counterclockwise. The permutation (1 2 3) will rotate the triangle 120 degrees counterclockwise sending corner 1 to position 2, corner 2 to position 3 and corner 3 to position 1. If you perform this same move again (the equivalent of permuting by (1 2 3)² in total), then corner 1 will go to position 3, corner 2 will go to position 1, and corner 3 will go to position 2. This is the same as if you had rotated 120 clockwise from the beginning, which is given by the permutation (1 3 2). Alternatively, you can just write down (1 2 3) ⚬ (1 2 3) and compute its value on each number: [(1 2 3) ⚬ (1 2 3)](1) = (1 2 3)(2) = 3 [(1 2 3) ⚬ (1 2 3)](2) = (1 2 3)(3) = 1 [(1 2 3) ⚬ (1 2 3)](3) = (1 2 3)(1) = 2 and we see that (1 2 3) ⚬ (1 2 3) = (1 3 2).
The permutation on the left should be read in two pieces, from right to left. The first piece says to swap 1 and 3, the second to swap 1 and 2. Think of the numbers as representing positions of objects. So when you swap 1 and 3 it means to swap the objects in positions 1 and 3. So where does the object starting in position 1 go. The (1 3) moves it to position 3. The (1 2) then leaves it alone. In total, the object moved from position 1 to position 3. Now, where does the object in position 3 go. The (1 3) moves it to position 1 and then the (1 2) moves it to position 2. In total, in moved from position 3 to position 2. Finally, the object in position 2 is left alone by (1 3) and then moved to position 1 by (1 2). Thus, the object in position 2 is moved to position 1. We summarize all of this with (1 3 2). Also, consider watching my video on composing permutations: ruclips.net/video/ii5NXuoH148/видео.html
The set you listed is not closed. The square of (1 2 3) needs be contained in the subgroup. But the square equals (1 3 2), which is not in the set you gave.
so an S7 is going to be a pain in the ass
Thank you, it's crystal clear!
Why is (123)^2 = (132)?
Think of an equilateral triangle where you label the corners 1, 2, 3 starting from one vertex and moving counterclockwise. The permutation (1 2 3) will rotate the triangle 120 degrees counterclockwise sending corner 1 to position 2, corner 2 to position 3 and corner 3 to position 1. If you perform this same move again (the equivalent of permuting by (1 2 3)² in total), then corner 1 will go to position 3, corner 2 will go to position 1, and corner 3 will go to position 2. This is the same as if you had rotated 120 clockwise from the beginning, which is given by the permutation (1 3 2).
Alternatively, you can just write down (1 2 3) ⚬ (1 2 3) and compute its value on each number:
[(1 2 3) ⚬ (1 2 3)](1) = (1 2 3)(2) = 3
[(1 2 3) ⚬ (1 2 3)](2) = (1 2 3)(3) = 1
[(1 2 3) ⚬ (1 2 3)](3) = (1 2 3)(1) = 2
and we see that (1 2 3) ⚬ (1 2 3) = (1 3 2).
Sir
Write the elements of length of the group of S3
Please answer
I don't understand the question. Would you clarify what it is you want to know?
can you explain why (1 2)(1 3) =( 1 3 2) ?
The permutation on the left should be read in two pieces, from right to left. The first piece says to swap 1 and 3, the second to swap 1 and 2. Think of the numbers as representing positions of objects. So when you swap 1 and 3 it means to swap the objects in positions 1 and 3. So where does the object starting in position 1 go. The (1 3) moves it to position 3. The (1 2) then leaves it alone. In total, the object moved from position 1 to position 3. Now, where does the object in position 3 go. The (1 3) moves it to position 1 and then the (1 2) moves it to position 2. In total, in moved from position 3 to position 2. Finally, the object in position 2 is left alone by (1 3) and then moved to position 1 by (1 2). Thus, the object in position 2 is moved to position 1. We summarize all of this with (1 3 2).
Also, consider watching my video on composing permutations: ruclips.net/video/ii5NXuoH148/видео.html
Thanks Adam
Thank you
Write subgroup of s4 plz
Make video on s4
Thanks 😘
Why H6 not be just { e, (1 2 3) }
The set you listed is not closed. The square of (1 2 3) needs be contained in the subgroup. But the square equals (1 3 2), which is not in the set you gave.
@@AdamGlesser Tks u very much sir, now i see that you already mention it on the video. Great work btw!
@@mongky9903 Glad to help!
Thank you