2 deleted 3 characters in body

The analogy with finite groups is not quite as strong as Greg suggests. If you're looking at a braided fusion category then the dimensions of objects all divide the the global dimension (the sum of the squares of the dimensions).

The actual theorem (thanks Noah!) is: given $X$ an object of $Z(\mathcal{C})$, the double of a fusion category $\mathcal{C}$, $\operatorname{dim}(X)$ divides $|\mathcal{C}|$. When $\mathcal{C}$ is already braided, $Z(\mathcal{C})$ is equivalent to includes into $\mathcal{C}$, and this gives the statement above.

However, for general fusion categories this is only a conjecture, and indeed a conjecture with a proposed counterexample (we're still working on this one...)

1

The analogy with finite groups is not quite as strong as Greg suggests. If you're looking at a braided fusion category then the dimensions of objects all divide the the global dimension (the sum of the squares of the dimensions).

The actual theorem (thanks Noah!) is: given $X$ an object of $Z(\mathcal{C})$, the double of a fusion category $\mathcal{C}$, $\operatorname{dim}(X)$ divides $|\mathcal{C}|$. When $\mathcal{C}$ is already braided, $Z(\mathcal{C})$ is equivalent to $\mathcal{C}$, and this gives the statement above.

However, for general fusion categories this is only a conjecture, and indeed a conjecture with a proposed counterexample (we're still working on this one...)