The problem states: Consider the following permutations:
$\alpha = (145)$, $\beta = (145)(3267)$
It then wants me to find the following permutations:
$(12)\alpha(12)$, $(12)\beta(12)$, $\alpha(12)\alpha^{-1}$, and $\beta(12)\beta^{-1}$
I am horrible at cycle notation, so I use the longer way to find disjoint cycles. An example of what I mean is this:
If I were to find the disjoint cycles of $(123)(23)$ I would do the following:
$(123)(23)$ = $\begin{bmatrix}1 & 2 &3\\2 & 3 &1\end{bmatrix}$ $\begin{bmatrix}1 & 2 &3\\1 & 3 & 2\end{bmatrix}$ = $\begin{bmatrix}1 & 2 &3\\2 & 1 &3\end{bmatrix}$ = $(12)(3)$
I was wondering if you could take a similiar approach to a problem like this. But I'm unsure how one would do so.
Like would: $(12)\alpha(12)$ = $\begin{bmatrix}1 & 2 &3&4&5\\2 & 1 &3&4&5\end{bmatrix}$ $\begin{bmatrix}1 & 2 &3&4&5\\4 & 2 &3&5&1\end{bmatrix}$ $\begin{bmatrix}1 & 2 &3&4&5\\2 & 1 &3&4&5\end{bmatrix}$ = $...$
Or am I just taking the wrong approach entirely?
Once I figure this out I can figure out a way to all $(ab)x(ab)$.
Any help appreciated!
I guess your approach will work. But if you just apply the permutations successively, starting from the right, you get, for instance: $(12)\alpha (12)=(12)(145)(12)=(1)(245)=(245)$. Be sure to account for where each of $1,\dots,5$ goes.