Linear representation theory of quaternion group

This article gives specific information, namely, linear representation theory, about a particular group, namely: quaternion group.
View linear representation theory of particular groups | View other specific information about quaternion group

The quaternion group is one of the few examples of a rational group that is not a rational-representation group. In other words, all its characters over the complex numbers are rational-valued, but not every representation of it can be realized over the rationals.

The character table of the quaternion group is the same as that of the dihedral group of order eight. Note, however, that the fields of realization for the representations differ, because one of the representations of the quaternion group has Schur index two.

Item	Value
Degrees of irreducible representations over a splitting field (such as $\mathbb{C}$ or $\overline{\mathbb{Q}}$)	1,1,1,1,2 maximum: 2, lcm: 2, number: 5, sum of squares: 8
Schur index values of irreducible representations	1,1,1,1,2 maximum: 2, lcm: 2
Smallest ring of realization for all irreducible representations (characteristic zero)	There are multiple candidates. $\mathbb{Z}[i]$ where $i$ is a square root of $-1$, equivalently $\mathbb{Z}[t]/(t^2+1)$, the ring of Gaussian integers is one candidate. Another is $\mathbb{Z}[\sqrt{-2}]$ or $\mathbb{Z}[t]/(t^2+2)$.
Smallest splitting field (i.e., field of realization) for all irreducible representations (characteristic zero)	There are multiple candidates. $\mathbb{Q}(i)$, equivalently $\mathbb{Q}[t]/(t^2+1)$ is one candidate. $\mathbb{Q}(\sqrt{-2})$, equivalently, $\mathbb{Q}[t]/(t^2+2)$ is another.
Smallest ring containing all character values (characteristic zero)	$\mathbb{Z}$
Smallest field containing all character vales (characteristic zero)	$\mathbb{Q}$ (hence it is a rational group)
Orbit structure of irreducible representations over splitting field under automorphism group	orbit sizes: 1 (degree 1 representation), 3 (degree 1 representations), 1 (degree 2 representation) number: 3
Degrees of irreducible representations over a non-splitting field, e.g., the field of rational numbers or field:F3	1,1,1,1,4
Condition for being a splitting field for this group	Necessary condition is unclear. Sufficient condition: The characteristic is not 2 and at least one of the polynomials $t^2+1$ and $t^2+2$ splits.
Smallest size splitting field	field:F3, i.e., the field of three elements.

Representations

Trivial representation

The trivial or principal representation is a one-dimensional representation that sends every element of the group to the 1-by-1 matrix 1. This representation makes sense over any field and in fact over any unital ring.

Element	Matrix	Characteristic polynomial	Minimal polynomial	Trace, character value
$1$	$(1)$	$t-1$	$t-1$	1
$-1$	$(1)$	$t-1$	$t-1$	1
$i$	$(1)$	$t-1$	$t-1$	1
$-i$	$(1)$	$t-1$	$t-1$	1
$j$	$(1)$	$t-1$	$t-1$	1
$-j$	$(1)$	$t-1$	$t-1$	1
$k$	$(1)$	$t-1$	$t-1$	1
$-k$	$(1)$	$t-1$	$t-1$	1

Sign representations with $i,j,k$-kernels

The quaternion group has three maximal normal subgroups: the cyclic subgroups generated by $i,j,k$ respectively. For each maximal normal subgroup, we obtain a one-dimensional representation with that subgroup as kernel. The representation sends elements inside the subgroup to 1, and elements outside the subgroup to -1.

The representations are detailed below:

Representation with $⟨i⟩$-kernel:

Element	Matrix	Characteristic polynomial	Minimal polynomial	Trace, character value
$1$	$(1)$	$t-1$	$t-1$	1
$-1$	$(1)$	$t-1$	$t-1$	1
$i$	$(1)$	$t-1$	$t-1$	1
$-i$	$(1)$	$t-1$	$t-1$	1
$j$	$(-1)$	$t+1$	$t+1$	-1
$-j$	$(-1)$	$t+1$	$t+1$	-1
$k$	$(-1)$	$t+1$	$t+1$	-1
$-k$	$(-1)$	$t+1$	$t+1$	-1

Representation with $⟨j⟩$-kernel:

Element	Matrix	Characteristic polynomial	Minimal polynomial	Trace, character value
$1$	$(1)$	$t-1$	$t-1$	1
$-1$	$(1)$	$t-1$	$t-1$	1
$i$	$(-1)$	$t+1$	$t+1$	-1
$-i$	$(-1)$	$t+1$	$t+1$	-1
$j$	$(1)$	$t-1$	$t-1$	1
$-j$	$(1)$	$t-1$	$t-1$	1
$k$	$(-1)$	$t+1$	$t+1$	-1
$-k$	$(-1)$	$t+1$	$t+1$	-1

Representation with $⟨k⟩$-kernel:

Element	Matrix	Characteristic polynomial	Minimal polynomial	Trace, character value
$1$	$(1)$	$t-1$	$t-1$	1
$-1$	$(1)$	$t-1$	$t-1$	1
$i$	$(-1)$	$t+1$	$t+1$	-1
$-i$	$(-1)$	$t+1$	$t+1$	-1
$j$	$(-1)$	$t+1$	$t+1$	-1
$-j$	$(-1)$	$t+1$	$t+1$	-1
$k$	$(1)$	$t-1$	$t-1$	1
$-k$	$(1)$	$t-1$	$t-1$	1

Two-dimensional representation over the complex numbers

The two-dimensional representation of the quaternion group can be described in a number of explicit ways. One such way is by viewing the Hamiltonian quaternions as a two-dimensional right vector space over the complex numbers, and viewing the actions of the elements of the quaternion group on this vector space by left multiplication.

The specific matrices for the representation depend on how we think of the Hamiltonians as a right vector space over the complex numbers. The typical way is to identify $\mathbb{C}$ as the subspace spanned by 1 and $i$, and take a basis as $1$ and $j$ for the vector space. Then, we have:

$1=(1,0);i=(i,0);j=(0,1);k=(0,-i)$

We can now compute the action of the elements $\pm 1,\pm i,\pm j,\pm k$ by left multiplication on this vector space, and write the matrices.

Below is such a description:

Element	Matrix	Characteristic polynomial	Minimal polynomial	Trace, character value
$1$	$\left(\begin{array}{cc}1& 0\\ 0& 1\\ \end{array}\right)$	$(t-1{)}^2=t^2-2t+1$	$t-1$	2
$-1$	$\left(\begin{array}{cc}-1& 0\\ 0& -1\\ \end{array}\right)$	$(t+1{)}^2=t^2+2t+1$	$t+1$	-2
$i$	$\left(\begin{array}{cc}i& 0\\ 0& -i\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0
$-i$	$\left(\begin{array}{cc}-i& 0\\ 0& i\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0
$j$	$\left(\begin{array}{cc}0& -1\\ 1& 0\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0
$-j$	$\left(\begin{array}{cc}0& 1\\ -1& 0\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0
$k$	$\left(\begin{array}{cc}0& -i\\ -i& 0\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0
$-k$	$\left(\begin{array}{cc}0& i\\ i& 0\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0

This two-dimensional representation can be interpreted over any field of characteristic not equal to 2 and containing a square root of -1 if we replace $i$ by one of the square roots of -1 and $-i$ by the other square root of -1.

There is an alternative equivalent representation that uses a square root of -2 instead of a square root of -1, and it makes sense over any field of characteristic not 2 and having a square root of -2. For concreteness, we consider the representation in characteristic zero and hence write the two square roots of -1 as $\sqrt{2}i$ and $-\sqrt{2}i$:

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.

Four-dimensional representation over the real numbers

The quaternion group has no irreducible two-dimensional representation over the reals. However, it has a four-dimensional representation over the reals, which splits over the complex numbers as a direct sum of two copies of the two-dimensional irreducible representation over the complex numbers. This representation is obtained by viewing the Hamiltonian quaternions as a four-dimensional vector space over the real numbers, and writing the matrices for left multiplication by the elements of the quaternion group. The typical choice of basis is $\{1,i,j,k\}$. Note that multiplication by anything other than $\pm 1$ gives a matrix with zeros on the diagonal, hence the character is zero on all elements outside the center.

Note that this representation is actually a representation over the rational numbers, and all its entries are signed permutation matrices, i.e., matrices with exactly one nonzero entry in every row and every column and that entry is $\pm 1$.

Element	Matrix	Characteristic polynomial	Minimal polynomial	Trace, character value
$1$	$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ \end{array}\right)$	$(t-1{)}^4=t^4-4t^3+6t^2-4t+1$	$t-1$	4
$-1$	$\left(\begin{array}{cccc}-1& 0& 0& 0\\ 0& -1& 0& 0\\ 0& 0& -1& 0\\ 0& 0& 0& -1\\ \end{array}\right)$	$(t+1{)}^4=t^4+4t^3+6t^2+4t+1$	$t+1$	-4
$i$	$\left(\begin{array}{cccc}0& -1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& -1\\ 0& 0& 1& 0\\ \end{array}\right)$	$(t^2+1{)}^2=t^4+2t^2+1$	$t^2+1$	0
$-i$	$\left(\begin{array}{cccc}0& 1& 0& 0\\ -1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& -1& 0\\ \end{array}\right)$	$(t^2+1{)}^2=t^4+2t^2+1$	$t^2+1$	0
$j$	$\left(\begin{array}{cccc}0& 0& -1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0\\ 0& -1& 0& 0\\ \end{array}\right)$	$(t^2+1{)}^2=t^4+2t^2+1$	$t^2+1$	0
$-j$	$\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& -1\\ -1& 0& 0& 0\\ 0& 1& 0& 0\\ \end{array}\right)$	$(t^2+1{)}^2=t^4+2t^2+1$	$t^2+1$	0
$k$	$\left(\begin{array}{cccc}0& 0& 0& -1\\ 0& 0& -1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\\ \end{array}\right)$	$(t^2+1{)}^2=t^4+2t^2+1$	$t^2+1$	0
$-k$	$\left(\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& -1& 0& 0\\ -1& 0& 0& 0\\ \end{array}\right)$	$(t^2+1{)}^2=t^4+2t^2+1$	$t^2+1$	0

This representation is irreducible over any formally real field (is it? verify).

Character table

This character table works over characteristic zero:

Rep/Conj class	1 (identity)	-1	$\pm i$	$\pm j$	$\pm k$
Trivial representation	1	1	1	1	1
$i$-kernel	1	1	1	-1	-1
$j$-kernel	1	1	-1	1	-1
$k$-kernel	1	1	-1	-1	1
2-dimensional	2	-2	0	0	0

The size-degree-weighted characters are given as follows (size-degree-weighted characters are algebraic integers):

Rep/Conj class	1 (identity)	-1	$\pm i$	$\pm j$	$\pm k$
trivial representation	1	1	2	2	2
$i$-kernel	1	1	2	-2	-2
$j$-kernel	1	1	-2	2	-2
$k$-kernel	1	1	-2	-2	2
2-dimensional	1	-1	0	0	0

Degrees of irreducible representations

Below is a description of the degrees of irreducible representations over fields of characteristic not equal to $2$.

Type of field	Condition on polynomials	Condition on $q$ for finite field of size $q$	Degrees of irreducible representations
splitting field	?	all finite fields of characteristic not equal to 2	1,1,1,1,2
not a splitting field	?	no finite field	1,1,1,1,4

Realizability information

Smallest ring of realization

Representation	Smallest ring of realization	Smallest set of elements occurring as entries of matrices
trivial representation	$\mathbb{Z}$ -- ring of integers	$\{1\}$
$i$-kernel	$\mathbb{Z}$ -- ring of integers	$\{1,-1\}$
$j$-kernel	$\mathbb{Z}$ -- ring of integers	$\{1,-1\}$
$k$-kernel	$\mathbb{Z}$ -- ring of integers	$\{1,-1\}$
two-dimensional irreducible (over splitting field)	$\mathbb{Z}[i]$ -- Gaussian integers is one candidate; $\mathbb{Z}[\sqrt{2}]$ is another	$\{0,1,-1,i,-i\}$ if over $\mathbb{Z}[i]$; $\{0,1,-1,\sqrt{2}i,-\sqrt{2}i\}$ if over $\mathbb{Z}[\sqrt{2}i]$
four-dimensional irreducible (over non-splitting field)	$\mathbb{Z}$ -- ring of integers	$\{0,1,-1\}$

Orthogonality relations and numerical checks

General statement	Verification in this case
number of irreducible representations equals number of conjugacy classes	Both numbers are equal to 5.
sufficiently large implies splitting: if a field has primitive $n^{th}$ roots where $n$ is the exponent of the group, it is a splitting field.	In this case, $n=4$, and having a primitive fourth root is equivalent to $-1$ having a square root, and a field has this property iff it is a splitting field.
number of one-dimensional representations equals order of abelianization	Both numbers are equal to 4. The derived subgroup is $\{1,-1\}$ (see center of quaternion group) and the quotient group (i.e., the abelianization) is a Klein four-group, having order four.
sum of squares of degrees of irreducible representations equals order of group	$1^2+1^2+1^2+1^2+2^2=8$.
degree of irreducible representation divides index of abelian normal subgroup	All degrees $1,1,1,1,2$ divide 2, which is the index of the abelian normal subgroup $⟨i⟩$ (and also of $⟨j⟩$ and $⟨k⟩$.
order of inner automorphism group bounds square of degree of irreducible representation	The center is $\{1,-1\}$ (see center of quaternion group) and the quotient group (i.e., the inner automorphism group) (a Klein four-group) has order four. The degrees of irreducible representations, 1,1,1,1,2, all have the property that the square is at most 4.
row orthogonality theorem and the column orthogonality theorem	can be verified from the character table.

Action of automorphisms

The automorphism group of the quaternion group permutes the three sign representations. In fact, this automorphism group permutes the sign representations in precisely the same way as it permutes the three maximal normal subgroups.

The trivial representation and the two-dimensional representations remain invariant under all automorphisms.

Relation with quotients

The quaternion group has six normal subgroups: the whole group, the trivial subgroup, center of quaternion group, and three cyclic maximal subgroups of quaternion group. The irreducible representations with kernel a particular normal subgroup correspond precisely to the faithful irreducible representations of the quotient group; the irreducible representations with kernel containing a particular normal subgroup correspond precisely to the irreducible representations of the quotient group. Information in this regard is presented below:

Normal subgroup in whole group	Normal subgroup isomorphism type	Quotient group	Number of such normal subgroups	Linear representation theory of quotient group	Degrees of irreducible representations of quotient group	List of irreducible representations of quotient group (corresponding representation of whole group)	Degrees of faithful irreducible representations of quotient group	List of faithful irreducible representations of quotient group (corresponding representation of whole group)
whole group	quaternion group	trivial group	1		1	trivial (trivial)	1	trivial (trivial)
cyclic maximal subgroups of quaternion group	cyclic group:Z4	cyclic group:Z2	3	linear representation theory of cyclic group:Z2	1,1	trivial (trivial), sign ($⟨i⟩$-kernel, $⟨j⟩$-kernel, and $⟨k⟩$-kernel sign depending on which subgroup is chosen)	1	sign ($⟨i⟩$-kernel, $⟨j⟩$-kernel, and $⟨k⟩$-kernel sign depending on which subgroup is chosen)
center of quaternion group	cyclic group:Z2	Klein four-group	1	linear representation theory of Klein four-group	1,1,1,1	trivial (trivial), three sign representations for the various copies of Z2 in V4 (correspond to the representations with $⟨i⟩$, $⟨j⟩$, and $⟨k⟩$ kernels)	--	--
trivial subgroup	triival group	quaternion group	1	current page	1,1,1,1,2	that's what this page is about	2	two-dimensional irreducible (two-dimensional irreducible)

Relation with subgroups

Induced representations from subgroups

Since the quaternion group is a finite nilpotent group, it is in particular a finite supersolvable group, and hence, it is a monomial-representation group: every irreducible representation can be realized as a monomial representation, i.e., every irreducible representation is induced from a degree one representation of a subgroup. (Point (5) below explains how the two-dimensional irreducible representation is induced).

1. The trivial representation on the center induces a representation obtained as a sum of the four one-dimensional representations.
2. The sign representation on the center (which comprises $\pm 1$) induces the double of the two-dimensional irreducible representation of the quaternion group.
3. The trivial representation on the cyclic subgroup generated by $i$ induces a representation on the whole group that is the sum of a trivial representation and the representation with the $i$-kernel. Analogous statements hold for $j,k$.
4. A representation on $⟨i⟩$ that sends $i$ to $-1$ induces a representation of the whole group that is the sum of the representations with $j$-kernel and $k$-kernel. Analogous statements hold for $j,k$.
5. A representation on $⟨i⟩$ that sends $i$ to $i$ (now viewed as a complex number) induces the two-dimensional irreducible representation.

Verification of Artin's induction theorem

Artin's induction theorem states that the characters induced from characters of cyclic subgroups span the space of class functions. This is easy to check for the quaternion group from the points made above. By point (2) or point (5), the two-dimensional irreducible character is in the span. Points (3) and (4) show that all pairwise sums of the four one-dimensional representations are in the span. Taking suitable linear combinations of these yields that all the four one-dimensional representations are in the span.