[SOLVED] Expressing as a product of transpositions

**craig** · May 18th 2010, 08:37 AM

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

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

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

Thanks

**tonio** · May 18th 2010, 08:55 AM

Originally Posted by craig

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

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

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 $(i_1\,i_2\,\ldots,i_n)=(1\,i_1)(1\,i_n)(1\,i_{n-1})\ldots(1\,i_2)(1\,i_1)$

Tonio

**craig** · May 18th 2010, 08:58 AM

Originally Posted by tonio

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

Tonio

Cheers for the reply.

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

**tonio** · May 18th 2010, 09:01 AM

Originally Posted by craig

Cheers for the reply.

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

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

**craig** · May 18th 2010, 09:09 AM

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

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

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

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

**tonio** · May 18th 2010, 11:21 AM

Originally Posted by craig

Yeh must of missed that one out.

In the answers they have:

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

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

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

**craig** · May 19th 2010, 12:33 AM

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 $(a \; b) = (1 \; a)(1 \; b)(1 \; a)$ , we have:

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

Thanks again for the help, (and patience

)

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