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?

  • $\begingroup$ 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

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, $\sigma\circ\tau = \left((1,3,5)(2,4,6)\right)\left((1,4)(2,5)(3,6)\right)$.


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


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