non-disjoint permutations So, since apparently every calculator I encounter to check my answer does not help with joint cycles, I came humbly to ask if I did this correctly. $\alpha$ = (1, 3, 2, 5)(6, 4, 7, 3) y $\beta$ = (3, 6, 5, 2)(1, 4, 7, 3) So, they ask for these: $α$, $\beta$, $α^2$, $α^2$$β$, Here are my answers: > Writing it as disjoint cycles: > >> $\alpha$ = (1 3 6 4 7 2 5) >> >> $\beta$ = (1 4 7 6 5 2 3) > > Then operating: > >> $\alpha^2$ = (6 1 4 2 3 7 6) >> >> $α^2$$β$ = (2 4 6 5 1 3 7) Hopefully I did this right.

You're confusing two different notations. There is a long way to describe a permutation and a short way. The long way list which element maps to which:

$$\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 \\\ 1 & 4 & 2 & 3 \\\ \end{pmatrix}$$ Here we have $1$ as a fixed element, while $\sigma(2)=4, \sigma(3)=2$, and $\sigma(4)=3$.

The short way is to write out cycles; the same permutation can be written $$\sigma = (1)(2 4 3)$$ showing that $1$ is fixed, while we have $2 \to 4 \to 3 \to 2$.

In your answer for $\alpha^2$ and $\alpha^2\beta$, you've written out the bottom row of the long notation, which doesn't make sense when read as cycle notation.