prove disjoint cycles commute

[galactus]

Hey All:

I am going to ask a question this time. Regarding some abstract algebra I am self-studying a little as I get time.

Suppose that $\displaystyle \alpha, \;\ \beta \;\ \in \;\ S_{n}$ are disjoint cycles. Prove that disjoint cycles commute. Of course, that is, $\displaystyle {\alpha}{\beta}={\beta}{\alpha}$.

I was thinking maybe we can say that $\displaystyle \alpha=(a_{1}a_{2}\cdot\cdot\cdot a_{k}), \;\ and \;\ \beta=(b_{1}b_{2}\cdot\cdot\cdot b_{n})$

Every permutation is a cycle, but every permutation can be expressed as the product of disjoint cycles.

Say we have the permutation $\displaystyle \begin{pmatrix}1&2&3&4&5&6&7\\5&1&7&2&4&6&3\end{pmatrix}$ in $\displaystyle S_{7}$.

We can find an element that is not mapped to itself and trace where it is sent.

$\displaystyle \begin{pmatrix}1&2&3&4&5&6&7\\5&1&7&2&4&6&3\end{pmatrix}=(1542)(37)$

1 maps to 5, and then 5 maps to 4, then 4 maps to 2, then since 2 maps to 1 we start over with 3 maps to 7 and since 7 maps to 3, that's it.

We can also see that 3 maps to 7. This is commutative, but for this case.

Is there are more general way to go about it?. I always like to learn more about abstract algebra. Interesting stuff. I think so anyway. I don;t know what good it is in the real world, but that does not matter to me.

 

[daon]

I don't rmember eactly how we proved it, but I believe we used the assumption that you can write cycles as products of 2-cycles. I could be wrong about that.

Anyway, the easiest way might be cases: an element belongs to alpha, beta, or neither. Here's what the first case might look like:

If $\displaystyle \alpha(a_i) = a_j$, $\displaystyle \beta(a_j) = a_k$.... Just giving names to the images.

If $\displaystyle a_i \in \alpha$ then $\displaystyle \beta(a_j)=a_j$ (meaning k=j, but this would hold for any) otherwise beta contains one of alphas entries. Hence:

$\displaystyle \beta \alpha (a_i) = \beta(a_j) = a_j$

$\displaystyle \alpha \beta (a_i) = \alpha(a_i) = a_j$

 

[galactus]

Thanks, Daon. I got it.