Multiplication in Permutation Groups Written in Cyclic Notation

I didn’t find any good explanation how to perform multiplication on permutation group written in cyclic notation. For example, if
then why does ab=(1356) and ac=(1652)(34)?


You are thinking of the permutations as functions, so when you write “ab“, you mean that you perform the permutation b first, and the permutation a second.

Here’s one way to do it: write the disjoint cycle expressions for both a and b, in the given order:
Now, moving from right to left, see what happens to each number in each cycle.

For instance, start with 1, so we write 1 down:
The first cycle, (2,5,6), does nothing to 1, so it stays 1. Then the next cycle, (1,3,5,2), sends 1 to 3. So, in total, 1 is sent to 3. We write
Now consider 3. The first cycle, (2,5,6), does nothing to 3. The second maps 3 to 5. So the product maps 3 to 5. So now we have
Now 5. The first cycle, (2,5,6), sends 5 to 6; the second cycle does nothing to 6, so in total, 5 is sent to 6. So for the product we now have
Next, what happens to 6? The first cycle sends 6 to 2; and then the next cycle sends 2 to 1. So 6 is sent to 1, and because we started out with 1, this now closes the cycle we have; thus, we also close the bracket. So the product so far is
Now we consider the “next” number that hasn’t been described yet, 2. The first cycle, (2,5,6), sends 2 to 5; then we check what the next cycle does to 5, which is that it sends it back to 2. So 2 maps to 2, and since we started out with 2, it again closes the cycle. So now we have
Finally we check what happens 4, as it’s the remaining number: (2,5,6) fixes 4 (it doesn’t do anything to it – it remains as it is), as does (1,3,5,2), so 4 is overall fixed. So now finally we have:
\therefore (1,3,5,2)(2,5,6)=(1,3,5,6)

It’s similar with ac. Here we have:
First consider 1: the first cycle maps it to 6, the second cycle fixes 6. So 1\mapsto 6. Then 6 is sent to 3 by the first cycle, and 3 to 5 by the second cycle (reading right to left, remember), so 6\mapsto 5. Then 5 is fixed by the first cycle and sent to 2 by the second cycle, so 5\mapsto 2. Then 2 is fixed by the first cycle and sent to 1 by the second, which means 2\mapsto 1, closing the cycle: we have (1,6,5,2). The next number not already covered is 3; 3 is mapped to 4 by the first cycle (by b), and 4 is fixed by a, so 3\mapsto 4. Then 4 is sent to 1 by the first cycle, and 1 is sent to 3 by the second cycle, so this closes this second cycle as (3,4). Putting the two together we get
(1,3,5,2)(1,6,3,4) = (1,6,5,2)(3,4)
as given.

Source : Link , Question Author : com , Answer Author : Skeleton Bow

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