[SOLVED] Expressing as a product of transpositions

Apr 2008
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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
 
Last edited:
Oct 2009
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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
 
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Apr 2008
748
159
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 = \(\displaystyle (1 \; 7\; 8)(3 \;10)(6 \;11\; 12)\)?
 
Oct 2009
4,261
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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
 
Apr 2008
748
159
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.
 
Oct 2009
4,261
1,836
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
 
Apr 2008
748
159
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 ;) )