# Degree of irreducible representation need not divide exponent

## Statement

We can have a finite group and an irreducible linear representation of the group over an algebraically closed field of characteristic zero (or more generally over a splitting field), such that the degree of the irreducible representation does not divide the exponent of the group.

This is a non-constraint on the Degrees of irreducible representations (?) of a finite group.

## Proof

### Example of big extraspecial groups

Consider an extraspecial group of order $p^7$ for any prime $p$. The exponent of this group is either $p$ or $p^2$. On the other hand, it admits a faithful irreducible representation of degree $p^3$.

For odd primes $p$, we can take an example of an extraspecial group of order $p^5$ and exponent $p$, which admits a faithful irreducible representation of degree $p^2$.

### Example of symmetric group of degree six

Further information: symmetric group:S6, linear representation theory of symmetric group:S6

The symmetric group of degree six has an irreducible representation of degree $16$ over the rational numbers, arising from the self-conjugate partition $6 = 3 + 2 + 1$. On the other hand, the exponent of the group is the lcm of the numbers from $1$ to $6$, which is $60$, and is not a multiple of $16$.