Degree of irreducible representation divides index of abelian normal subgroup

This article gives the statement, and possibly proof, of a constraint on numerical invariants that can be associated with a finite group

This article states a result of the form that one natural number divides another. Specifically, the (degree of a linear representation) of a/an/the (irreducible linear representation) divides the (index of a subgroup) of a/an/the (abelian normal subgroup).
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This fact is related to: linear representation theory
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Statement

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 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 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 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.

Related facts

Similar divisibility facts

Further information: degrees of irreducible representations

• 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

Further information: cyclic group:Z3, linear representation theory of cyclic group:Z3

Let $G$ be the cyclic group of order three and $\mathbb {R}$ (real numbers) be the field. Then, there are two irreducible representations of $G$ over $\mathbb {R}$: 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.

Group	Linear representation theory	Indexes of maximal among abelian normal subgroups	gcd of index values	Degrees of irreducible representations	lcm of degrees
symmetric group:S3	linear representation theory of symmetric group:S3	2	2	1,1,2	2
dihedral group:D8	linear representation theory of dihedral group:D8	2,2,2	2	1,1,1,1,2	2
quaternion group	linear representation theory of quaternion group	2,2,2	2	1,1,1,1,2	2
dihedral group:D10	linear representation theory of dihedral group:D10	2	2	1,1,2,2	2
alternating group:A4	linear representation theory of alternating group:A4	3	3	1,1,1,3	3
symmetric group:S4	linear representation theory of symmetric group:S4	6	6	1,1,2,3,3	6
special linear group:SL(2,3)	linear representation theory of special linear group:SL(2,3)	12	12	1,1,1,2,2,2,3	6
alternating group:A5	linear representation theory of alternating group:A5	60	60	1,3,3,4,5	60

Facts used

1. Third isomorphism theorem
2. Index is multiplicative
3. Normality satisfies image condition
4. Abelianness is quotient-closed
5. 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:
   • $\rho$ is induced from an irreducible representation of $H$, where $H$ is a proper subgroup of $G$ containing $N$
   • The restriction of $\rho$ to $N$ is an isotypical representation
6. Normality satisfies intermediate subgroup condition
7. 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

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).

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 $L=G/K$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$L$ has an abelian normal subgroup $NK/K$ whose index divides the index $[G:N]$.	Facts (1), (2)	$N$ is abelian normal in $G$.		[SHOW MORE] $[G/K:NK/K]$is the same as $[G:NK]$ by Fact (1). Since $N\leq NK\leq G$, Fact (2) yields that this divides $[G:N]$. Thus, $[L:NK/K]$ divides $[G:N]$. By Facts (3) and (4), and the observation that $NK/K$ is the image of $N$ under the quotient map by $K$, we obtain that $NK/K$ is an abelian normal subgroup of $L$.
2	$\rho$ descends to an irreducible linear representation $\rho '$ of $L$, having the same degree as $\rho$.		$\rho$ is irreducible, has kernel $K$		[SHOW MORE] We simply define $\rho '$ at any element of $L$ as the image under $\rho$ of any of its pre-images. The notion of invariant subspace remains the same, hence irreducibility of $\rho$ forces the irreducibility of $\rho '$.
3	If $\rho$ is not faithful, i.e., $K$ is nontrivial, then the degree of $\rho$ divides $[G:N]$.			Steps (1), (2)	[SHOW MORE] If $\rho$ is not faithful, $K$ is nontrivial, so the order of $L$ is strictly less than $n$. In particular, by induction and Steps (1) and (2), we obtain that the degree of $\rho '$ divides the index $[L:NK/K]$ of the abelian normal subgroup $NK/K$ of $L$. By Step (2), $\deg \rho =\deg \rho '$ and by Step (1) again, $[L:NK/K]$ divides $[G:N]$. Combining, we obtain that $\deg \rho$ divides $[G:N]$.
4	Either $\rho$ is induced from an irreducible representation of $H$, where $H$ is a proper subgroup of $G$ containing $N$, or the restriction of $\rho$ to $N$ is isotypical.	Fact (5)	$\rho$ is irreducible over an algebraically closed field (or splitting field) of characteristic zero, $G$ is finite, and $N$ is normal in $G$.		Fact-direct
5	If $\rho$ is induced from an irreducible representation $\sigma$ of $H$, where $H$ is a proper subgroup of $G$ containing $N$, then $\deg(\rho )=[G:H]\deg(\sigma )$	Definition of induced representation
6	If $\rho$ is induced from an irreducible representation $\sigma$ of $H$, we must have that $\deg \sigma$ divides $[H:N]$.	Fact (6)		Step (5) (for notation)	[SHOW MORE] Since $H$ is a proper subgroup of $G$, its order is strictly smaller than that of $G$.we can apply induction to the situation of the irreducible representation $\sigma$ of $H$ and the abelian normal subgroup $N$ of $H$ (normal by fact (6)).
7	If $\rho$ is induced from an irreducible representation $\sigma$ of $H$, we obtain that the degree of $\rho$ divides $[G:N]$.	Fact (2)		Steps (5), (6)	[SHOW MORE] The degree of $\rho$ divides $[G:H]$ times the degree of $\sigma$, which in turn divides $[H:N]$. Combining, we obtain that the degree of $\rho$ divides $[G:H][H:N]$, which by Fact (2), is $[G:N]$.
8	If $\rho$ is faithful and the restriction of $\rho$ to $N$ is isotypical, then $N$ is contained in the center $Z(G)$ of $G$.		$N$ is abelian.		[SHOW MORE] Since $N$ is abelian, all irreducible representations are one-dimensional, and hence all isotypical representations give scalar matrices. In particular, this means that elements of $N$ get mapped to scalars, and hence commute with $\rho (g)$ for all $g\in G$. With the assumption of faithfulness, we obtain that elements of $N$ commute with elements of $G$, so $N$ is in the center.
9	If $\rho$ is faithful and the restriction of $\rho$ to $N$ is isotypical, then the degree of $\rho$ divides $[G:N]$.	Facts (2),(7)		Step (8)	[SHOW MORE] By Step (8), $N\leq Z(G)$, so by Fact (2) $[G:Z(G)]$ divides $[G:N]$. By Fact (7), $\deg \rho$ divides $[G:Z(G)]$. Combining, we obtain the required result.
10	In every case, the degree of $\rho$ divides $[G:N]$.			Steps (3), (4), (7), (9)	[SHOW MORE] Of the two cases in Step (4), the induced case is tackled in Step (7). For the isotypical case, note that if $\rho$ is not faithful, Step (3) completes the proof. If $\rho$ is faithful, Step (9) completes the proof.