# [SOLVED] Expressing as a product of transpositions

#### 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 $$\displaystyle b \in \{ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 \}$$

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

Thanks

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#### tonio

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

craig

#### craig

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

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

#### tonio

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

#### craig

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.

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.

#### tonio

Yeh must of missed that one out.

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

#### craig

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 )