1. Let f = (x1, x2, ...., xr) in Sn. Show that o(f) = r.
(I have to do this using cycles or transpositions or something, but I don't know how it would work.)
2. Prove that Sn is non-abelian if n is greater than or equal to 3.
To show that $\displaystyle |f|=r$ you need to show that $\displaystyle f^r = \text{id}$ (identity permuation) and that $\displaystyle f^k \not = \text{id}$ if $\displaystyle k<r$. Also it would be helpful to write $\displaystyle f = (x_0,x_1,...,x_{r-1})$ (you will see why) instead.
First we show that $\displaystyle f^r$ is the identity permuation. This means we need to show $\displaystyle f^r(x) = x$ for all $\displaystyle x\in S_n$. We can assume that $\displaystyle x\in \{x_0,...,x_{r-1}\}$ because otherwise $\displaystyle f(x) = x$ and so there is no need to check that case. Remember that $\displaystyle f(x_j) = x_{j+1\bmod n}$ (this is why we rewrote the indicies differently). Therefore, $\displaystyle f^k (x_j) = x_{j+k\bmod n}$. If we want $\displaystyle f^k(x_j) = x_j$ it must mean that $\displaystyle j\equiv j+k(\bmod n) \implies n|k$. Therefore, the smallest such $\displaystyle k$ is $\displaystyle n$ and so $\displaystyle |f|=n$.
Compare $\displaystyle (12)(23)\text{ and }(23)(12)$.2. Prove that Sn is non-abelian if n is greater than or equal to 3.
You need to be careful, the problem said $\displaystyle S_n$ for $\displaystyle n\geq 3$ so we can find $\displaystyle (12),(23)$. However, we do not necessarily have $\displaystyle (24)$ because the group is not necessary $\displaystyle S_4$ it can be $\displaystyle S_3$.