in S6:
(1 3)(2 4 5)*(2 3 5) = (1 3 2)(5 4)
can anyone thoroughly explain how you get this answer? because i don't see it whatsoever!
(1 3)(2 4 5)*(2 3 5) = (1 3 2)(5 4)
can anyone thoroughly explain how you get this answer? because i don't see it whatsoever!
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First forget about the *, they're all multiplications. (EDIT: actually composition not multiplication, a minor detail)
Nowt, start by writing a 1 on the right hand side:
(1 3)(2 4 5)(2 3 5) = (1
Now let's see what happens to the 1. Work RIGHT TO LEFT. rightmost bracket (2 3 5) doesn't contain a 1, so ignore. Same for (2 4 5). Now the leftmost permutation (1 3) means 1 goes to 3 (i.e. an item in position 1 gets moved to position 3). You've reached the end, so 1 goes to 3, so write in a 3, like this:
(1 3)(2 4 5)(2 3 5) = (1 3
Next, see what happens to 3, working through the brackers right to left. (2 3 5) has 3 going to 5. So now you trace the 5. Next bracket, (2 4 5), has 5 going to 2. (2 4 5) is a cycle, i.e. identical to (4 5 2) or (5 2 4). So now you're tracing the 2. The leftmost bracket does nothing with the 2, so 2 is the final result (i.e. what 3 permutes to). So write a 2:
(1 3)(2 4 5)(2 3 5) = (1 3 2
Now trace 2 in the same way. Rightmost bracket: 2 goes to 3. Middle bracket: nothing done to 3. Leftmost bracket: 3 to 1. But we don't write 1 again, because the brackets indicate a cycle, so we close the cycle like this:
(1 3)(2 4 5)(2 3 5) = (1 3 2)
Now choose a new number for a new cycle. It doesn't matter which. By convention you usually choose the lowest remaining number, i.e. 4, but for some reason your text book chose 5. It doesn't matter, we;ll use 5. It's a new cycle, so write:
(1 3)(2 4 5)(2 3 5) = (1 3 2)(5
Now trace 5. Rightmost: 5 goes to 2. Middle: 2 goes to 4. Leftmost: nothing done with 4, so the result is 4, write 4:
(1 3)(2 4 5)(2 3 5) = (1 3 2)(5 4
Now trace 4. (This should be redundant because we've used all the numbers now, but it's good to check we get 5). Rightmost: nothing. Middle: 4 goes to 5. Leftmost: nothing. So result is 5. We've already got a 5 at the start of the cycle so we close the bracket like this:
(1 3)(2 4 5)(2 3 5) = (1 3 2)(5 4)
Nowt, start by writing a 1 on the right hand side:
(1 3)(2 4 5)(2 3 5) = (1
Now let's see what happens to the 1. Work RIGHT TO LEFT. rightmost bracket (2 3 5) doesn't contain a 1, so ignore. Same for (2 4 5). Now the leftmost permutation (1 3) means 1 goes to 3 (i.e. an item in position 1 gets moved to position 3). You've reached the end, so 1 goes to 3, so write in a 3, like this:
(1 3)(2 4 5)(2 3 5) = (1 3
Next, see what happens to 3, working through the brackers right to left. (2 3 5) has 3 going to 5. So now you trace the 5. Next bracket, (2 4 5), has 5 going to 2. (2 4 5) is a cycle, i.e. identical to (4 5 2) or (5 2 4). So now you're tracing the 2. The leftmost bracket does nothing with the 2, so 2 is the final result (i.e. what 3 permutes to). So write a 2:
(1 3)(2 4 5)(2 3 5) = (1 3 2
Now trace 2 in the same way. Rightmost bracket: 2 goes to 3. Middle bracket: nothing done to 3. Leftmost bracket: 3 to 1. But we don't write 1 again, because the brackets indicate a cycle, so we close the cycle like this:
(1 3)(2 4 5)(2 3 5) = (1 3 2)
Now choose a new number for a new cycle. It doesn't matter which. By convention you usually choose the lowest remaining number, i.e. 4, but for some reason your text book chose 5. It doesn't matter, we;ll use 5. It's a new cycle, so write:
(1 3)(2 4 5)(2 3 5) = (1 3 2)(5
Now trace 5. Rightmost: 5 goes to 2. Middle: 2 goes to 4. Leftmost: nothing done with 4, so the result is 4, write 4:
(1 3)(2 4 5)(2 3 5) = (1 3 2)(5 4
Now trace 4. (This should be redundant because we've used all the numbers now, but it's good to check we get 5). Rightmost: nothing. Middle: 4 goes to 5. Leftmost: nothing. So result is 5. We've already got a 5 at the start of the cycle so we close the bracket like this:
(1 3)(2 4 5)(2 3 5) = (1 3 2)(5 4)