Problems with permutations and transpositions

[morrowcosom]

Homework Statement

1)Consider the permutation in S3 = ( 1 2 3 )
( 1 2 3 ) NOTE: the two pairs of parenthesis are
meant to be one pair that encases both rows

Write as a product of transpositions

Homework Equations

The Attempt at a Solution

(1 3) (1 2), which was wrong. I randomly plugged in numbers until I ended up with the correct solution of (1,3) (1,2) (1,3) which contradicts all the examples of transpositions I have seen, like (1 3 2 4) = (1 4) (1 2) (1 3).
How does one go about finding the solution to my problem?

 

[HallsofIvy]

Homework Statement

1)Consider the permutation in S3 = ( 1 2 3 )
( 1 2 3 ) NOTE: the two pairs of parenthesis are
meant to be one pair that encases both rows

So
\begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3\end{pmatrix}
This is the identity permutation. Normally it is written just as "i". If you really want to write it as a product of transpositions, you could use (1 2)(1 2). That is, transpose 1 and 2, then transpose them again, going back to what you had originally. (1 3)(1 3) will also work as will (2 3)(2 3). Or could do two "transpositions and back": (1 3)(1 3)(1 3)(2 3)(2 3). Since each "transposition and back" requires two transpositions, no matter how many times we do that, we will have an even number of transpositions.

Write as a product of transpositions

Homework Equations

The Attempt at a Solution

(1 3) (1 2), which was wrong. I randomly plugged in numbers until I ended up with the correct solution of (1,3) (1,2) (1,3) which contradicts all the examples of transpositions I have seen, like (1 3 2 4) = (1 4) (1 2) (1 3).
How does one go about finding the solution to my problem?

No, (1 3)(1 2)(1 3) will not work. (1 3) takes (1 2 3) to (3 2 1), then (1 2) take it to (3 1 2), and finally, (1 3) to (1 3 2), not (1 2 3). Since the identity permutation is an even permutation, it can only be written as a product of an even number of transpositons, not 3.

The identity transformation is pretty trivial. Are you sure you have copied the problem correctly?

 

[morrowcosom]

No, (1 3)(1 2)(1 3) will not work. (1 3) takes (1 2 3) to (3 2 1), then (1 2) take it to (3 1 2), and finally, (1 3) to (1 3 2), not (1 2 3). Since the identity permutation is an even permutation, it can only be written as a product of an even number of transpositons, not 3.

When you say (1 3) takes (1 2 3) to (3 2 1), then (1 2) takes it to (3 1 2) could you explain the process of how this occurs? I am fine with multiplying permutatations and disjointed cycles, transposition confuses me though. Yes, the identity transformation was copied right.

I got this problem through cow.temple.edu. and am doing independent study.

Thanks