# [SOLVED] Expressing as a product of transpositions

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• May 18th 2010, 08:37 AM
craig
[SOLVED] Expressing as a product of transpositions
Let $\displaystyle \beta = (1 \; 7\; 8)(3 \;10)(6 \;11\; 12)$ in the symmetric group $\displaystyle S_{12}$

Express $\displaystyle \beta$ as a product of transpositions in the form $\displaystyle (1 \; b)$, with $\displaystyle b \in \{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 \}$

I can express it as normal transpositions, but not sure how to go about it with the first number only being 1.

Thanks
• May 18th 2010, 08:55 AM
tonio
Quote:

Originally Posted by craig
Let $\displaystyle \beta = (1 \; 7\; 8)(3 \;10)(6 \;11\; 12)$ in the symmetric group $\displaystyle S_{12}$

Express $\displaystyle \beta$ as a product of transpositions in the form $\displaystyle (1 \; b)$, with $\displaystyle b \in \{ 2, 3, 4, 5, 6, 7, 9, 10, 11, 12 \}$

I can express it as normal transpositions, but not sure how to go about it with the first number only being 1.

Thanks

Lemma (boring): we have $\displaystyle (i_1\,i_2\,\ldots,i_n)=(1\,i_1)(1\,i_n)(1\,i_{n-1})\ldots(1\,i_2)(1\,i_1)$

Tonio
• May 18th 2010, 08:58 AM
craig
Quote:

Originally Posted by tonio
Lemma (boring): we have $\displaystyle (i_1\,i_2\,\ldots,i_n)=(1\,i_1)(1\,i_n)(1\,i_{n-1})\ldots(1\,i_2)(1\,i_1)$

Tonio

Cheers for the reply.

How would I apply that to the problem where instead of $\displaystyle (i_1\,i_2\,\ldots,i_n)$, you have = $\displaystyle (1 \; 7\; 8)(3 \;10)(6 \;11\; 12)$?
• May 18th 2010, 09:01 AM
tonio
Quote:

Originally Posted by craig
Cheers for the reply.

How would I apply that to the problem where instead of $\displaystyle (i_1\,i_2\,\ldots,i_n)$, you have = $\displaystyle (1 \; 7\; 8)(3 \;10)(6 \;11\; 12)$?

Apply the lemma for each of the cycles...!! Of course, you need 8 (which you didn't include among your possible b's in your OP. This must be a mistake, of course)

Tonio
• May 18th 2010, 09:09 AM
craig
Quote:

Originally Posted by tonio
Of course, you need 8 (which you didn't include among your possible b's in your OP. This must be a mistake, of course)

Tonio

Yeh must of missed that one out.

In the answers they have:

So $\displaystyle (1 \; 7\; 8)(3 \;10)(6 \;11\; 12)$ =

$\displaystyle (1\;8)(1\;7)(1\;3)(1\;10)(1\;3)...$

How is it that they've got 2 $\displaystyle (1\;3)$s?

Sorry if this is something stupid, thanks again for the replies.
• May 18th 2010, 11:21 AM
tonio
Quote:

Originally Posted by craig
Yeh must of missed that one out.

In the answers they have:

So $\displaystyle (1 \; 7\; 8)(3 \;10)(6 \;11\; 12)$ =

$\displaystyle (1\;8)(1\;7)(1\;3)(1\;10)(1\;3)...$

How is it that they've got 2 $\displaystyle (1\;3)$s?

Sorry if this is something stupid, thanks again for the replies.

I wouldn't say it is stupid but it definitely seems to be careless...please do look carefully at the lemma in my first post!

Tonio
• May 19th 2010, 12:33 AM
craig
Quote:

Originally Posted by tonio
I wouldn't say it is stupid but it definitely seems to be careless...please do look carefully at the lemma in my first post!

Tonio

Ahh sorry, just had a look again this morning, not sure how I didn't get it last night.

Using the fact that $\displaystyle (a \; b) = (1 \; a)(1 \; b)(1 \; a)$, we have:

$\displaystyle (1 \; 7\; 8)(3 \;10)(6 \;11\; 12) =$
$\displaystyle (1 \; 8)(1 \; 7)(1 \; 3)(1 \; 10)(1 \; 3)(6 \; 11)(11 \; 12) =$
$\displaystyle (1 \; 8)(1 \; 7)(1 \; 3)(1 \; 10)(1 \; 3)(1 \; 6)(1 \; 11)(1 \; 6)(1 \; 11)(1 \; 12)(1 \; 11)$

Thanks again for the help, (and patience ;) )