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I have two permutations $(2,7,9,4,3)$ and $(2,3,6)$ of $S_9$.

According to this post I want to multiplay these two. Unfortunately multiplication is described there only for disjoint permutations.

The algorithm presented there does not work for me.

$(2,7,9,4,3)(2,3,6)$

Following that instruction I get $(2,2 ...$

I have started with $2$:

$$(2,$$

The first cycle, $(2,3,6)$ sends $2$ to $3$. The second cycle, $(2,7,9,4,3)$ sends $3$ again to $2$ so I get:

$$(2,2 ...$$

And I am stuck in an infinite loop here.

Question: How do I multiply permutations that are not disjoint? Or is it the same way as I should deal with disjoint permutations and pay attention to some additional rules?

Community
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jublikon
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3 Answers3

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Writing both the permutations as product of transpositions, we get

$(2,7,9,4,3)=(2,3)(2,4)(2,9)(2,7)$

and

$(2,3,6)=(2,6)(2,3)$. So their multiplication is

$(2,3)(2,4)(2,9)(2,7)\cdot (2,6)(2,3)$

Now we do it like this :- Start with $2$. So $2\mapsto 3$ due to the rightmost transposition in the product and $3$ is fixed in all the other elements of the product and finally $3$ is mapped to $2$ because of the transposition $(2,3)$. So the overall effect is that $2$ is fixed in this multiplication.

Now let's start with $3$. $3\mapsto 2$ (from the rightmost permutation) which in turn is mapped to $6$ and then $6$ is fixed throughout, so the overall effect is that $3\mapsto 6$.

Doing the same trick for all the elements we see that

$(2,3)(2,4)(2,9)(2,7)\cdot (2,6)(2,3)=(3,6,7,9,4)$.

seeker
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  • why does wolfram alpha give for permutation (2,7,9,4,3)(2,3,6) = (2,7,9,4,6)? – jublikon Jan 08 '17 at 18:29
  • @jublikon because you have wrongly calculated $(2,3,6)(2,7,9,4,3)$ and $not$ $(2,7,9,4,3)(2,3,6)$. Since the group is non-abelian, so order matters. – seeker Jan 08 '17 at 18:31
  • @seeker jublikon is correct, Wolfram Alpha indeed gives the stated multiplication. Whether multiplication of permutations is right-to-left or left-to-right is not consistent in the literature. – Joshua Mundinger Jan 08 '17 at 18:33
  • @Alqatrkapa : yes you're right. I didn't know that as I have never calculated such a product on Wolfram. – seeker Jan 08 '17 at 18:33
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Let $\pi_1=(2\,7\,9\,4\,3)$, $\pi_2=(2\,3\,6)$, $\pi=\pi_1\pi_2$.

You have worked out that $\pi_2(2)=3$ and $\pi_1(3)=2$; hence $\pi(2)=2$, i.e. $\pi$ fixes $2$.

Now do the same for $3$. $\pi_2(3)=6$ and $\pi_1$ fixes $6$, so $\pi(3)=6$. Now look at $6$. $\pi_2(6)=2$ and $\pi_1(2)=7$; therefore $\pi(6)=7$. Continue with $7$. $\pi_2(7)=7$, $\pi_1(7)=9$; therefore $\pi(7)=9$. Carry on, you get $\pi(9)=4$, $\pi(4)=3$ which is what we started with. Hence: $$(2\,7\,9\,4\,3)\,(2\,3\,6)\ =\ (3\,6\,7\,9\,4)$$ The numbers which don't appear ($1$, $5$, $8$) are fixed.

George Law
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0

The algorithm works. Start with $2$:

$ (2,7,9,4,3)(2,3,6)=(2)(3,6.7,9,4) $

Since the any cycle $(k)$ is the same as the identity you can just drop it; in the present case drop $(2)$.

Hagen Knaf
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