Write permutation as disjoint cycles $(5 1 2)(3 1 2)(4 1 3)$

Question

I'm really confused with permutations overall, I know the "algorithm" yet every time I get a mistake...

Write this permutation in $S_6$ as disjoint cycles and find its order: $(5 1 2)(3 1 2)(4 1 3)$

My try:
$$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 2 & 3 & 4 & 5 & 1 & 6\\ \end{pmatrix}$$ so $(5 1 2)(3 1 2)(4 1 3)=(1 2 3 4 5)$.

Yet the solution in WolframAlpha states: $(1 4)(2 5)$ (order $2$ yet in my case the order is $5$). Am I mistaken?

Is there any good website for checking this sort of calculation (or is WolframAlpha actually correct)?

Thank you in advance.

Answer 1

WolframAlpha evaluates permutations from left to right. Your calculation is correct when evaluating from right to left.

Answer 2

I've been there. It takes a while to get used to working with permutations, and cycle notation isn't the easiest to work with. I would write out each cycle in 2-line notation: $$ (512) = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 5 & 3 & 4 & 1 & 6\end{pmatrix}, (312) = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 4 & 5 & 6\end{pmatrix}, (413) = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5& 6\end{pmatrix} $$ To "multiply" these (what we are really doing is composing them), you can put them "on top of each other"

$$ \begin{matrix} \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 5 & 3 & 4 & 1 & 6\end{pmatrix} \\ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 4 & 5 & 6\end{pmatrix} \\ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix} \end{matrix} $$

and then align the top row of the second permutation with the second row of the first one.

$$ \begin{matrix} \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 2 & 5 & 3 & 4 & 1 & 6\end{pmatrix} \\ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 2 & 3 & 1 & 4 & 5 & 6\end{pmatrix} \\ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix} \end{matrix} = \begin{matrix} \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 2 & 5 & 3 & 4 & 1 & 6\end{pmatrix} \\ \begin{pmatrix}2 & 5 & 3 & 4 & 1 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix} \end{matrix} = \begin{matrix} \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 2 & 5 & 3 & 4 & 1 & 6\\ 2 & 5 & 3 & 4 & 1 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix} \end{matrix} = \begin{matrix} \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix} \end{matrix} $$

and now do the same thing again:

$$ \begin{matrix} \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 2 & 4 & 1 & 5 & 6\end{pmatrix} \end{matrix} = \begin{matrix} \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 1 & 4 & 2 & 6\end{pmatrix} \\ \begin{pmatrix}3 & 5 & 1 & 4 & 2 & 6 \\ 4 & 5 & 3 & 1 & 2 & 6\end{pmatrix} \end{matrix} = \begin{matrix} \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 3 & 5 & 1 & 4 & 2 & 6 \\ 3 & 5 & 1 & 4 & 2 & 6 \\ 4 & 5 & 3 & 1 & 2 & 6 \end{pmatrix} \end{matrix} $$ which equals: $$ \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6\\ 4 & 5 & 3 & 1 & 2 & 6\end{pmatrix} $$

Now that we have our permutation we can write it in cycle notation:[ 1->4, 4-> 1 ][ 2->5,5->2 ][ 3->3 ] gives (14)(25)(3). People normally drop the (3), so it can be written as (14)(25).

The order is 2 because if you compose the permutation (14)(25) with itself you get the identity.