0

How would I go about computing $$(1 2 3)\cdot(12)(34)$$

I know the definitions but I do not know how to apply them here. This is rather strange and odd-looking to me. I know I have to construct a natural group (1234), relate it to the product, but then what?

D. N.
  • 2,221
Don Larynx
  • 4,703

3 Answers3

2

If $(1 2)(3 4) \in S_n$, $(12)(34)$ is notation for the function $f : \{1, \dots, n\} \to \{1, \dots, n\}$, $f(1) = 2$, $f(2) = 1$, $f(3) = 4$, $f(4) = 3$, $f(i) = i$ for $5 \leq i \leq n$. Likewise, $(1 2 3)$ is notation for the function $g : \{1, \dots, n\} \to \{1, \dots, n\}$, $g(1) = 2$, $g(2) = 3$, $g(3) = 1$, $g(j) = j$ for $4 \leq j \leq n$. Then $(1 2 3)\cdot(1 2)(3 4)$ is the function $g\circ f : \{1, \dots, n\} \to \{1, \dots, n\}$.

2

(123)⋅(12)(34)

  • $1$ is sent to $2$, which is in turn sent to $3$. Thus $1$ goes to $3$.

  • $2$ is sent to $1$, which is sent to $2$. Thus $2$ goes to $2$.

  • $3$ is sent to $4$, which is no longer messed with. Thus $3$ goes to $4$.

What does $4$ go to? What is the resulting permutation?

  • Can you explain the reason for your bullet premises? – Don Larynx Sep 27 '13 at 21:44
  • @Jossie, sure. This is the standard way of computing products of cycles. (34) means send 3 to 4 and 4 to 3. (12) means sending 1 to 2 and 2 to 1. (123) means sending 1 to 2 and 2 to 3 and 3 to 1. So, what happens if you apply all three in turn (first applying 34, then 12, then 123)? Well, you have to see where each individual number will be sent. For example, 1 will first be sent to 2, and then it will be sent to 3 by the third permutation (since it is now a 2). Thus the final permutation should send 1 to 3. – Caleb Stanford Sep 28 '13 at 08:25
  • @Jossie please see here for some good more detailed explanations. – Caleb Stanford Sep 28 '13 at 09:25
0

The operators represent the translations relative to an initial starting position $1234$. If you want to apply it successively, you have to suppose that the numbers in the operators are not the numbers in what's currently displayed. You could use letters to represent the starting point, like $abcd$.

In any case, $(1,2)(3,4) \cdot 1,2,3,4$ gives $2,1,4,3$. Applying $(1,2,3) \cdot 2,1,4,3$ rotates the first three elements, to give $1,4,2,3$.

You can directly derive $1,4,2,3$ from $1,2,3,4$ by rotating the triplet $2,3,4$, ie $(2,3,4)$

So the operation $(1,2,3) \cdot (1,2)(3,4) = (2,3,4)$