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Sorry for stupid question but I didn't find any good explanation how to perform multiplication on permutation group written in cyclic notation.

For example, $a=(1352)$, $b=(256)$, $c=(1634)$, $ab=(1356)$, $ac=(1652)(34)$.


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In what order do you compose permutations? – lhf Apr 8 '11 at 16:31
This question may be of help, if it is not a duplicate in some sense already =) – Adrián Barquero Apr 8 '11 at 16:41

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: $$(1,3,5,2)(2,5,6).$$ Now, moving from right to left, see what happens to each number in each cycle.

For instance, start with $1$: 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 $$(1,3,$$ 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 $$(1,3,5,$$ 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 $$(1,3,5,6,$$ 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 $1$, which closes the cycle we have; so the product so far is $$(1,3,5,6)$$ 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 $2$. That is, we have $$ab = (1,3,5,2)(2,5,6) = (1,3,5,6)(2).$$ And finally we check what happens $4$: $(2,5,6)$ fixes $4$, as does $(1,3,5,2)$, so $4$ is fixed. So we have: $$ab = (1,3,5,2)(2,5,6)=(1,3,5,6)(2)(4) = (1,3,5,6).$$

Similarly with $ac$. Here we have: $$(1,3,5,2)(1,6,3,4).$$ 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.

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great explanation!!! Thanks! – com Apr 8 '11 at 17:00

The method I use for multiplying permutations like this is to think of each cycle as a set of mappings. a (in your example) maps 1 to 3, 3 to 5, 5 to 2, and 2 to 1. Also, remember that ab means "apply b, then apply a." So, here, we want to see where ab maps each number 1-6.

Start with 1: b fixes 1 (maps it to itself) and a maps 1 to 3. So we can begin writing ab = (13...

Now do 3: b fixes 3, and a maps 3 to 5. Put a 5 in: ab = (135...

Now 5: b maps 5 to 6 and a fixes 6, so ab = (1356...

Now 6: b maps 6 to 2 and a maps 2 to 1, so ab = (13561... = (1356).

Notice that ab fixes 4 since both a and b fix 4, but ab actually also fixes 2. This is because b maps 2 to 5, and a maps 5 right back to 2.

Hopefully you can use this method to check the other products.

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very helpful, thank you! – com Apr 8 '11 at 17:01

There is a small example on this page . Basically multiplication of permutation groups is applying permutations from right to left on an unaltered sequence.

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