0
$\begingroup$

I've been given the problem to write f = (1,2,3,4,5,6) as a product of permutation g of order 2 and a permutation h of order 3.

It doesn't make sense to me how a permutation of order 6 can be written as one of order 2 and one of order 3 because in doing this aren't we eliminating an element of f?

I understand what has to be done but I just don't get how I can break it down into two smaller order permutations without losing information?

$\endgroup$
  • $\begingroup$ Being a permutation of order $2$ or a permutation of order $3$ doesn't imply that only two or only three elements appear. The permutation $(1,3,5)(2,4,6)$ for example is of order $3$. Written in a different format, it is $\begin{pmatrix} 1&2&3&4&5&6\\3&4&5&6&1&2\end{pmatrix}$. I agree that $f$ couldn't be written as a product of a disjoint $2$-cycle and a disjoint $3$-cycle, but that's not what the problem asks for. $\endgroup$ – JMoravitz Oct 24 '15 at 16:10
  • $\begingroup$ @JMoravitz that definitely helps, thanks. So I something in the form of (x,y) and (a,b,c)(d,e,f) works since the first is order 2 and the second is order 3? $\endgroup$ – CKCMathCS613 Oct 24 '15 at 16:25
  • $\begingroup$ Or even something of the form (x,y)(u,v)(s,t) and (a,b,c)(d,e,f). The order of a permutation is the least common multiple of the lengths of the cycles present when representing it as a product of disjoint cycles. Since $2$ and $3$ are prime, that implies that $g$ will only have $1$-cycles (fixed points, which usually aren't written) and disjoint $2$-cycles. Similarly for $h$. The problem now is to find which ones work. $\endgroup$ – JMoravitz Oct 24 '15 at 16:27
  • $\begingroup$ @JMoravitz ooooooh of course! so (1,2)(2,3)(3,4) and (4,5,6) would work? $\endgroup$ – CKCMathCS613 Oct 24 '15 at 16:35
  • $\begingroup$ actually no because we are doing the product of two... forget I said that! $\endgroup$ – CKCMathCS613 Oct 24 '15 at 16:36
1
$\begingroup$

Hint: (expanded from those given in comments above)

$\sigma = \begin{pmatrix} 1&2&3&4&5&6\\3&4&5&6&1&2\end{pmatrix} = (1,3,5)(2,4,6)$ is of order three.

$\tau = \begin{pmatrix} 1&2&3&4&5&6\\4&5&6&1&2&3\end{pmatrix} = (1,4)(2,5)(3,6)$ is of order two.

See what their product is. What is $\sigma\circ \tau$? What about $\tau\circ\sigma$? It won't be quite what you are looking for, but maybe it can be fixed. Do you see how?

As $\tau$ sends $1\mapsto 4$ and $\sigma$ sends $4\mapsto 6$, you have $\sigma\circ\tau$ sends $1\mapsto 6$. Similarly we see $\sigma\circ\tau$ sends $2\mapsto 1$, $3\mapsto 2$,... so we get $\sigma\circ\tau = \begin{pmatrix}1&2&3&4&5&6\\6&1&2&3&4&5\end{pmatrix} = (1,6,5,4,3,2)$ This should look very close to what you wanted. What is wrong about it?

Being a product of two permutations doesn't require them to be written as the product of two cycles. In the above, $\sigma\circ\tau = \left((1,3,5)(2,4,6)\right)\left((1,4)(2,5)(3,6)\right)$.

$\endgroup$
0
$\begingroup$

Hint: $$(1\,2)f=(2\,3\,4\,5\,6).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.