TITLE: Kaplansky's 6th conjecture: dim(Irrep) | dim(algebra) - for semi-simple Hopf algebras 
QUESTION [16 upvotes]: Let $H$ be a semisimple Hopf algebra. One of the Kaplansky's conjectures states that the dimension of any irreducible $H$-module divides the dimension of $H$. 
In which cases the conjecture is known to be true?

REPLY [6 votes]: There is a new survey on Kaplansky's sixth conjecture by L. Dai and J. Dong, available on the arxiv. Among other results, it mentions the following (always assuming $\operatorname{char} k=0$):
Special primes:

If a semisimple Hopf algebras has a simple module of dimension $p$, where $p=2$ or $p=3$, then its dimension is divisible by $p$.

Low dimension:

Semisimple Hopf algebras of dimension less than $60$ satisfy Kaplansky's sixth conjecture.

Particular properties:

A semisimple Hopf algebra $H$ satisfies Kaplansky's sixth conjecture if it satisfies one of the following conditions:

It is quasitriangular.
Its characters are central in $H^*.$
It is semisolvable.


Dividing other invariants:

$H$ a semisimple Hopf algebra. $A$ a transitive $H$-module algebra (e.g. $A=\operatorname{End}_k(V)$ for a simple $H$-module $V$).  Then $\dim A$ divides $(\dim V)^2\dim H$.

For references, see the survey mentioned above.