I didn't find any good explanation how to perform multiplication on permutation group written in cyclic notation. For example, if $$ a=(1\,3\,5\,2),\quad b=(2\,5\,6),\quad c=(1\,6\,3\,4), $$ then why does $ab=(1\,3\,5\,6)$ and $ac=(1\,6\,5\,2)(3\,4)$?

  • 1
    $\begingroup$ In what order do you compose permutations? $\endgroup$
    – lhf
    Apr 8 '11 at 16:31
  • $\begingroup$ This question may be of help, if it is not a duplicate in some sense already =) $\endgroup$ 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 $1$, so we write $1$ down: $$(1,$$ 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 $1$, 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 $$(1,3,5,6)$$ 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 $2$, and since we started out with $2$, it again closes the cycle. So now we have $$(1,3,5,6)(2)$$ 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: $$ab = (1,3,5,2)(2,5,6)=(1,3,5,6)(2)(4) = (1,3,5,6)$$ $$\therefore (1,3,5,2)(2,5,6)=(1,3,5,6)$$

It's similar 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.

  • $\begingroup$ Good explanation. $\endgroup$
    – Ken Ingram
    Dec 30 '18 at 8:12

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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.