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.
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.