Where N=10,
I'm really struggling even starting both questions.
Thanks to anyone who can help.
$\displaystyle (N 2) = \left(\begin{array}{cccc}1&2&...&N\\1&N&...&2\end{ array}\right)$
$\displaystyle \alpha=(N 2)(12 (N-2) N) = \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&N&3&...&6&...&N-2&N-1&2\\2&1&3&...&6&...&N&N-1&N-2\end{array}\right) $
same to find beta
$\displaystyle (23(N-1)) = \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&3&N-1&...&6&...&N-2&2&N\end{array}\right)$
$\displaystyle \beta = (23(N-1))(6(N-1)N) = \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&3&N-1&...&6&...&N-2&2&N\\1&3&N&...&N-1&...&N-2&2&N\end{array}\right)$
want to find n such that
$\displaystyle \beta = \alpha \cdot n $
let's write alpha and beta to see how we can find n
$\displaystyle \alpha \cdot n=\beta$
$\displaystyle \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\2&1&3&...&6&...&N&N-1&N-2\end{array}\right)n=$ $\displaystyle \left(\begin{array}{ccccccccc}1&2&3&...&6&...&N-2&N-1&N\\1&3&N&...&N-1&...&N-2&2&N\end{array}\right)$
RIGHT, so you can solve it algebraically then sub in N=10 at the end?
EDIT: To clarify, I was starting off with:
Alpha = (10,2) (1,2,8,10)
Beta = (2,3,9) (6,9,10)
but it was the whole cycle notation/product of disjoint cycles thing that really threw me off. I find it very hard to understand.
For question 2:
$\displaystyle \forall a, b \in G$:
$\displaystyle abba = ab^2a = a(1)a = aa = a^2 = 1$
$\displaystyle abab = (ab)^2 = 1$
Let $\displaystyle ab = c$.
Then $\displaystyle cba = cab$
$\displaystyle c^{-1}cba = c^{-1}cab$
$\displaystyle ba = ab$