Here is an exposition of the ``alternative hack'' to which David Speyer alluded. For concreteness, I have fixed a group; the argument will go through in general, of course.
Let $g = (12) \in S_3$, a transposition in the symmetric group on three elements. The conjugacy class of $g$ consists of all three transpositions. Form $a = \frac{1}{3}\left[(12) + (13) + (23) \right] \in \mathbb{C}S_3$, the average of $g$'s conjugacy class. It is clear that for any $x \in G$,
$$xax^{-1} = a$$
since conjugating $a$ will simply rearrange the terms in the sum. Equivalently,
$$xa = ax,$$
and $a$ lies in the center of $\mathbb{C}S_3$. It follows that if $\rho: S_3 \longrightarrow \mbox{Aut}(V)$ is an irreducible representation, the matrix
$$\rho(a) = \frac{1}{3}\left[\rho(12) + \rho(13) + \rho(23) \right]$$
commutes with any $\rho(x)$. Schur's lemma tells us that $\rho(a)$ is a scalar matrix, that is, $\rho(a) = \lambda I$ for some $\lambda \in \mathbb{C}$.
Summarizing, for each irreducible representation $\rho$, we may define a class function $\psi_{\rho}$ which associates to any $g \in S_3$ the scalar $\lambda$ by which $a=\frac{1}{|S_3|}\sum_{x \in S_3} xgx^{-1}$ acts on $V$.
As it happens, $\psi_{\rho}(g)$ can be computed easily in terms of the character $\chi^{\rho}$: after all, every element conjugate to $g$ has the same trace; by linearity,
$\mbox{Tr}(\rho(a)) = \chi^{\rho}(a) = \chi^{\rho}(g)$.
It follows that
$$\psi_{\rho}(g) = \frac{\chi^{\rho}(g)}{\mbox{dim}V}.$$
A natural next step is to eliminate reference to a particular $g$ using an inner product:
$$\langle \psi_{\rho},\chi^{\rho} \rangle = \frac{1}{\mbox{dim}V}.$$
Expanding the definition of the inner product on class functions,
$$\mbox{Tr}\left[\frac{1}{|S_3|^2}\sum_{g \in S_3} \left(\sum_{x \in S_3} \rho(xgx^{-1}) \right) \rho(g^{-1}) \right] = \frac{1}{\mbox{dim}V},$$
and
$$\frac{1}{|S_3|^2}\sum_{g \in S_3} \sum_{x \in S_3} \chi^{\rho}(xgx^{-1}g^{-1}) = \frac{1}{\mbox{dim}V}.$$
We are led to consider the formal sum $d=\sum_{g,h \in S_3} ghg^{-1}h^{-1}$ mentioned in David Speyer's post. This sum is invariant under any automorphism of $S_3$ (in particular inner automorphisms) and so lies in the center of the group algebra $\mathbb{C}S_3$. Schur's lemma tells us that the image of $d$ under any irreducible representation $\rho$ is a scalar. By the above we get
$$\sum_{g \in S_3} \sum_{h \in S_3} \rho(ghg^{-1}h^{-1}) = \left[\frac{|S_3|}{\mbox{dim}V}\right]^2I.$$
Considering now the regular representation $\mathbb{C}S_3$, we see that the element $d$ acts with rational number eigenvalues. In other words, its characteristic polynomial splits completely over $\mathbb{Q}$. But we can also see by inspection that $d$ acts by an integer matrix. The characteristic polynomial is a monic integer polynomial and splits into linear factors, so its roots are integers. Since $\mathbb{C}S_3$ contains every irreducible representation as a summand at least once, we see that each
$$\frac{|S_3|}{\mbox{dim}V} \in \mathbb{Z}.$$
In particular, $1$, $2$, and $1$ all divide $6$.
