Degree of irreducible representation divides index of abelian normal subgroup

For an algebraically closed field of characteristic zero
Let $$G$$ be a finite group and $$\rho$$ an irreducible representation of $$G$$ over an algebraically closed field of characteristic zero. Let $$N$$ be an fact about::abelian normal subgroup of $$G$$. Then the degree of $$\rho$$ divides the index $$[G:N]$$.

For a splitting field of characteristic zero
Let $$G$$ be a finite group and $$\rho$$ an irreducible representation of $$G$$ over a splitting field for $$G$$ of characteristic zero. Let $$N$$ be an fact about::abelian normal subgroup of $$G$$. Then the degree of $$\rho$$ divides the index $$[G:N]$$.

Note that the two statements are equivalent because any representation of $$G$$ that is irreducible over a splitting field is also irreducible over the algebraic closure.

For a splitting field, any characteristic not dividing the group order
Let $$G$$ be a finite group and $$\rho$$ an irreducible representation of $$G$$ over a splitting field for $$G$$ of any characteristic (which must not divide the order of $$G$$). Let $$N$$ be an fact about::abelian normal subgroup of $$G$$. Then the degree of $$\rho$$ divides the index $$[G:N]$$.

This is equivalent to the other two formulations because degrees of irreducible representations are the same for all splitting fields.

Similar divisibility facts

 * Degree of irreducible representation divides group order
 * Degree of irreducible representation divides index of center (or equivalently, it divides the order of the inner automorphism group)
 * Degree of irreducible representation divides index of abelian subgroup in finite nilpotent group
 * Square of degree of irreducible representation divides order of inner automorphism group in finite nilpotent group

Combinatorial/inequality facts

 * Degree of irreducible representation is bounded by index of abelian subgroup
 * Order of inner automorphism group bounds square of degree of irreducible representation
 * Sum of squares of degrees of irreducible representations equals order of group

Converse

 * Jordan-Schur theorem on abelian normal subgroups of small index

Opposite facts

 * Degree of irreducible representation need not divide index of abelian subgroup

Applications

 * Ito-Michler theorem: The easy direction of the theorem follows immediately from this statement. It basically states that if a finite group has an abelian normal $$p$$-Sylow subgroup, then $$p$$ does not divide the degrees of any of its irreducible representations. The converse, which is the hard part of the Ito-Michler theorem, relies on much heavier machinery including the classification of finite simple groups.

Breakdown for a field that is not algebraically closed
Let $$G$$ be the cyclic group of order three and $$\R$$ (real numbers) be the field. Then, there are two irreducible representations of $$G$$ over $$\R$$: the trivial representation, and a two-dimensional representation given by the action by rotation by multiples of $$2\pi/3$$. The two-dimensional representation has degree $$2$$, and this does not divide the order of the group, which is $$3$$.

This representation does not remain irreducible over $$\mathbb{C}$$: in fact, it decomposes as a direct sum of two irreducible representations, both of which have degree $$1$$, which satisfies the condition of dividing the index of any abelian normal subgroup.

Examples
Note that the result is true for trivial reasons for a finite abelian group. We thus concentrate on non-abelian groups. In the table below, we list, for each group, the indexes of all subgroups that are maximal among abelian normal subgroups, the gcd of these indexes, the degrees of irreducible representations, and the lcm of these degrees. By the result stated, the lcm of degree of irreducible representations must divide the gcd of indices of all subgroups that are maximal among abelian normal subgroups. Note that in fact we do not need to take gcd -- we can take the minimum, because every finite group has an abelian normal subgroup whose order is divisible by the orders of all abelian normal subgroups.

Note that among the examples below, special linear group:SL(2,3) is the only case where the lcm of degrees of irreducible representations is strictly smaller than the gcd of index values for abelian normal subgroups.

Facts used

 * 1) uses::Third isomorphism theorem
 * 2) uses::Index is multiplicative
 * 3) uses::Normality satisfies image condition
 * 4) uses::Abelianness is quotient-closed
 * 5) uses::Isotypical-or-induced lemma: Let $$G$$ be a finite group, $$N$$ a normal subgroup, and $$k$$ an algebraically closed field of characteristic zero (remove those assumptions on $$k$$?). Then if $$\rho$$ is an irreducible representation of $$G$$ one of the following must hold:
 * 6) * $$\rho$$ is induced from an irreducible representation of $$H$$, where $$H$$ is a proper subgroup of $$G$$ containing $$N$$
 * 7) * The restriction of $$\rho$$ to $$N$$ is an isotypical representation
 * 8) uses::Normality satisfies intermediate subgroup condition
 * 9) uses::degree of irreducible representation divides index of center: This result states that the degree of any irreducible representation must divide the index of the center, or equivalently, the order of the inner automorphism group.

Proof
We prove the statement by a strong form of induction on the order.

Statement to be proved by induction on $$n$$: For any finite group of order $$n$$, and any abelian normal subgroup of the group, and any algebraically closed field $$k$$ of characteristic zero, the degree of any irreducible linear representation of the finite group over $$k$$ must divide the index of the abelian normal subgroup.

Our strong form uses that the statement is true for all smaller numbers.

Inductive hypothesis: The statement to be proved by induction is true for all $$m < n$$. In particular, it is true for all proper divisors of $$n$$.

Inductive goal:

Given: A group $$G$$ of order $$n$$. An abelian normal subgroup $$N$$ of $$G$$. An irreducible linear representation $$\rho$$ of $$G$$ over an algebraically closed field $$k$$ of characteristic zero.

To prove: The degree of $$\rho$$ divides the index $$[G:N]$$.

Proof: Suppose $$\rho$$ has kernel $$K$$. Let $$L = G/K$$.