How do you multiply permutations?

[JessBrown]

Hi
I'm having a little bit of trouble understanding about permutations, and how you multiply them? Say I have the permutations a = (45)(67), b = (46)(57) how do I multiply them?
Thanks heaps!
Jess

 

[DrDu]

Starting from a, [4567], the first permutation yields b, [5476]. Appying the second permutation yields c, [7654]. Hence the permutation which permutes a to c is (47)(56).

 

[JessBrown]

Thanks, I'm having trouble with another question though, when it says that if I have three different permutations,
a = (45)(67)
b = (46)(57)
c = (47)(56)
I have to multiply them to show that if multiply any two, you'll get the third...
Taking what you've said above,

4567 x (45)(67) gives 5476, then x (46)(57) = 7654 which does not equal (47)(56)

where am I going wrong?

Thanks

 

[DrDu]

So, a, [4567] becomes c, [7654], hence 4->7 5->6 6->5 7->4. You see that there are two cycles, the first one is formed by 4->7 and 7->4, (47) for short. The other one is 5->6 and 6->5, (56) for short. So the total permutation is (47)(56). Maybe you are mixing up the set of numbers to be permuted (which I put in square brackets) and the cycle notation?

 

[JessBrown]

Thanks,
I think its just the wording and I'm interpreting it wrong, cos it says show how when
(45)(67) is multiplied by (46)(57) the result is (47)(56)

So to write it out I say (45)(67) goes to (54)(76) then when multiplied by (46)(57) this becomes (76)(54) which then when apply c (47)(56) becomes (45)(67) which is the original answer? I think I sort of get it just the wording is really difficult!

If you could help me with the wording to show how a x b = c, that would be amazing! Thanks

 

[HallsofIvy]

Perhaps it is the word "multiply" that is bothering you. What we are really doing is "combining" the permutations, not "multiplying" in the arithmetic sense.

4567 x (45)(67) means "first apply the (45)(67) permutation, then the 4567 permutation".

(45)(67) means that we swap 4 and 5 and swap 6 and 7 while leaving all other number the same- 1234567 is mapped to 1235476.

4567 means 4 is changed to 5, 5 to 6, 6 to 7, and 7 to 4- 1234567 is mapped to 1235674

Putting those together, (45)(67) maps 4 to 5 and then 4567 changes that 5 to 6: 4 is mapped to 6. (45)(67) maps 5 to 4 and then 4567 changes that 4 to 5: 5 is mapped to 5. (45)(67) maps 6 to 7 and then 4567 changes that 7 to 4: 6 is mapped to 4. (45)(67) maps 7 to 6 and then 4567 changes that 6 to 7: 7 is mapped to 7. In other words, all number except 4 and 6 remain the same and they are swapped:
(4567) (45)(67)= (46).

DrDu is using a different convention going from left to right rather than right to left.

If we apply (4567) first, then (45)(67)
1) 4 is changed to 5 then back to 4: 4->4
2) 5 is changed to 6 then to 7: 5-> 7
3) 6 is changed to 7, then back to 6: 6->6
4) 7 is changed to 4, then to 5: 5-> 7
(4567) (45)(67)= (47).

I guess the real question is, when you say "4567 x (45)(67) gives 5476", how are you getting that?

 

[JessBrown]

Thank you for replying and clearing that up, I still don't entirely understand (I'm so sorry, I'm really trying to!)

So if I'm combining (45)(67) and (46)(57) to get (47)(56)

I go (45)(67) (the original) becomes (54)(76)
then combined with (46)(57) becomes (76)(54) which then I am stuck as I don't see how this results in (47)(56)

Thank you so much for your patience with me!

 

[DrDu]

"I go (45)(67) (the original) becomes (54)(76)"
You are trying to permute the numbers in the cycles of the permutation symbols. That's wrong!
Only apply the permutations to an independent set of numbers (I wrote [4567]).

 

[JessBrown]

Ok so if I go a random set of numbres, [7654]
by applying the first permutation 'a = (45)(67)', this becomes [6745]
then apply the second permutation 'b=(46)(57)', it becomes [4567] which is equal to taking the first set of numbers [7654] and applying 'c=(47)(56)' to get [4567].

Is this right? Finally! Thank you so much for all your help everyone!

 

[DrDu]

Sounds good :-)
You should learn to read of the cycles directly from the unpermuted and the permuted set of numbers.

 

[JessBrown]

Thank you so much for your help! I kinda get it now! Thanks!