1. Symmetric Groups Questions

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.

2. Originally Posted by Janu42
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.)
To show that $|f|=r$ you need to show that $f^r = \text{id}$ (identity permuation) and that $f^k \not = \text{id}$ if $k. Also it would be helpful to write $f = (x_0,x_1,...,x_{r-1})$ (you will see why) instead.

First we show that $f^r$ is the identity permuation. This means we need to show $f^r(x) = x$ for all $x\in S_n$. We can assume that $x\in \{x_0,...,x_{r-1}\}$ because otherwise $f(x) = x$ and so there is no need to check that case. Remember that $f(x_j) = x_{j+1\bmod n}$ (this is why we rewrote the indicies differently). Therefore, $f^k (x_j) = x_{j+k\bmod n}$. If we want $f^k(x_j) = x_j$ it must mean that $j\equiv j+k(\bmod n) \implies n|k$. Therefore, the smallest such $k$ is $n$ and so $|f|=n$.

2. Prove that Sn is non-abelian if n is greater than or equal to 3.
Compare $(12)(23)\text{ and }(23)(12)$.

3. And for 2,
(12)(24) and (24)(12) is different as well, and so on... correct?

Like to prove it, I could take an x greater than or equal to 3, and do
(12)(2x) does not equal (2x)(12)?

4. Originally Posted by Janu42
And for 2,
(12)(24) and (24)(12) is different as well, and so on... correct?
You need to be careful, the problem said $S_n$ for $n\geq 3$ so we can find $(12),(23)$. However, we do not necessarily have $(24)$ because the group is not necessary $S_4$ it can be $S_3$.