How do you prove cycles of length 3 don't generate the whole symmetric group? Let $S$ be some finite set. Let $Sym(S)$ be the symmetric group on $S$. Let $A$ be the subset of permutations $\sigma$ that are such that there exist $a,b,c \in S$ such that $\sigma(a) = b, \sigma (b) = c$ and $\sigma (c) =a$ and for every other element of $S$, the action of $\sigma$ is the identity. In short, let $A$ be the set of cycles of length 3.
How do you show that the subgroup generated by $A$ is not the whole of $Sym(S)$?
 A: Look at the sign of a permutation. If a permutation can be decomposed into a product of even number of transpositions (aka 2-cycles), we give it a sign of 1. Otherwise give it a sign of -1. It turns out that this is a well-defined notion. The group generated by the even permutations cannot cover the odd ones, thus cannot be the whole symmetric group.
P.S. The group generated by the even permutations is usually called the alternating group.
A: Consider the action of $S_n$ on the ring of polynomials $R=\mathbb R[x_1,\dots,x_n]$ given by permutation of the variables, and consider the element $$\Delta=\prod_{1\leq i<j\leq n}(x_i-x_j)\in R.$$ Then every $3$-cycle fixes $\Delta$, so the whole subgroup of $S_n$ generated by $3$-cycles acts trivially on $\Delta$. Yet there are elements of $S_n$, like all transpositions, which do not act trivially on $\Delta$ but map it to its opposite. Therefore $3$-cycles do not generate $S_n$.
One can alternatively use the action of $S_n$ on $\mathbb R^n$, again by permutation but now of the coordinates. Then the matrix corresponding to each $3$-cycle has determinant $1$, so every element of the subgroup generated by $3$-cycles is acts on $\mathbb R^n$ through a matrix with determinant $1$. Yet there are elements in $S_n$ whose corresponding matrix is of determinant $-1$: the same conclusion as before follows.
BN: The point of doing this in this way is to avoid having to check first that the parity of elements of $S_n$ is well-defined; all answers so far depend on doing this (I wonder if it is at all side-steppable...). In fact, this can be massaged easily into becoming a proof of that very same fact.
This is a more or less canonical example of why we want to study representations of groups. 
A: Consider the parity.
