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 inot 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.