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

Summary

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 (characteristic zero) maximum: 2, lcm: 2 1,1,1,1,1 (characteristic other than 0,2)
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 $\mathbb{Z}[t]/(t^2+2)$. More generally, any ring of the form $\mathbb{Z}[\alpha ,\beta ]$ where ${\alpha }^2+{\beta }^2=-1$ is a ring of realization for all irreducible representations. In particular, $\mathbb{Z}[\sqrt{-m^2-1}]$ works.
Smallest splitting field (i.e., field of realization) for all irreducible representations (characteristic zero)	There are multiple candidates. $\mathbb{Q}(i)$ or $\mathbb{Q}[t]/(t^2+1)$ works, so does $\mathbb{Q}(\sqrt{2}i)$ or $\mathbb{Q}[t]/(t^2+2)$. More generally, $\mathbb{Q}(\alpha ,\beta )$ where ${\alpha }^2+{\beta }^2=-1$ is a splitting field. In particular, $\mathbb{Q}(\sqrt{-1-m^2})$ works for any rational $m$.
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 the field of real numbers	1,1,1,1,4
Condition for being a splitting field for this group	Sufficient condition: the characteristic is not two and there exist $\alpha ,\beta$ in the field such that ${\alpha }^2+{\beta }^2+1=0$. In particular, every finite field of characteristic not two is a splitting field, because every element of a finite field is expressible as a sum of two squares and in particular, $-1$ is a sum of two squares in any finite field.
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, determinant (all same because it is degree 1)
$1$	$(1)$	$t-1$	$t-1$	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, determinant (all same because it is degree 1)
$1$	$(1)$	$t-1$	$t-1$	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, determinant (all same because it is degree 1)
$1$	$(1)$	$t-1$	$t-1$	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 irreducible representation over a splitting field

Further information: Faithful irreducible representation of quaternion group

Suppose $\alpha ,\beta$ are elements of a field (or more generally, a commutative unital ring) such that ${\alpha }^2+{\beta }^2+1=0$, then the representation can be realized in terms of the entries $\{0,1,-1,\alpha ,-\alpha ,\beta ,-\beta \}$. The explicit representation involving the Hamiltonian quaternions is the special case of this where $\alpha$ is a square root of $-1$ and $\beta =0$. Another case is ${\alpha }^2=-2$, $\beta =1$. A third case of interest is $\mathbb{Q}(\sqrt{-3})$, which contains primitive cube roots of unity, where we set $\alpha$ and $\beta$ as distinct primitive cube roots of unity.

Note that the representation makes sense in all characteristics, but there are some problems interpreting it in characteristic two.

Element	Matrix using $\alpha ,\beta$ with ${\alpha }^2+{\beta }^2=-1$	Matrix using $i=\sqrt{-1}$. Obtained by setting $\alpha =i,\beta =0$	Matrix using $\sqrt{2}i=\sqrt{-2}$. Obtained by setting $\alpha =\sqrt{2}i,\beta =1$	Matrix using primitive cube roots of unity, which we denote by $\omega$ and ${\omega }^2$, so $\alpha =\omega$, $\beta ={\omega }^2$.	Characteristic polynomial	Minimal polynomial	Trace, character value	Determinant
$1$	$\left(\begin{array}{cc}1& 0\\ 0& 1\\ \end{array}\right)$	$\left(\begin{array}{cc}1& 0\\ 0& 1\\ \end{array}\right)$	$\left(\begin{array}{cc}1& 0\\ 0& 1\\ \end{array}\right)$	$\left(\begin{array}{cc}1& 0\\ 0& 1\\ \end{array}\right)$	$(t-1{)}^2=t^2-2t+1$	$t-1$	2	1
$-1$	$\left(\begin{array}{cc}-1& 0\\ 0& -1\\ \end{array}\right)$	$\left(\begin{array}{cc}-1& 0\\ 0& -1\\ \end{array}\right)$	$\left(\begin{array}{cc}-1& 0\\ 0& -1\\ \end{array}\right)$	$\left(\begin{array}{cc}-1& 0\\ 0& -1\\ \end{array}\right)$	$(t+1{)}^2=t^2+2t+1$	$t+1$	-2	1
$i$	$\left(\begin{array}{cc}\alpha & \beta \\ \beta & -\alpha \\ \end{array}\right)$	$\left(\begin{array}{cc}i& 0\\ 0& -i\\ \end{array}\right)$	$\left(\begin{array}{cc}i& 1\\ 1& -\sqrt{2}i\\ \end{array}\right)$	$\left(\begin{array}{cc}\omega & {\omega }^2\\ {\omega }^2& -\omega \\ \end{array}\right)$	$t^2+1$	$t^2+1$	0	1
$-i$	$\left(\begin{array}{cc}-\alpha & -\beta \\ -\beta & \alpha \\ \end{array}\right)$	$\left(\begin{array}{cc}-i& 0\\ 0& i\\ \end{array}\right)$	$\left(\begin{array}{cc}-\sqrt{2}i& -1\\ -1& -\sqrt{2}i\\ \end{array}\right)$	$\left(\begin{array}{cc}-\omega & -{\omega }^2\\ -{\omega }^2& \omega \\ \end{array}\right)$	$t^2+1$	$t^2+1$	0	1
$j$	$\left(\begin{array}{cc}0& -1\\ 1& 0\\ \end{array}\right)$	$\left(\begin{array}{cc}0& -1\\ 1& 0\\ \end{array}\right)$	$\left(\begin{array}{cc}0& -1\\ 1& 0\\ \end{array}\right)$	$\left(\begin{array}{cc}0& -1\\ 1& 0\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0	1
$-j$	$\left(\begin{array}{cc}0& 1\\ -1& 0\\ \end{array}\right)$	$\left(\begin{array}{cc}0& 1\\ -1& 0\\ \end{array}\right)$	$\left(\begin{array}{cc}0& 1\\ -1& 0\\ \end{array}\right)$	$\left(\begin{array}{cc}0& 1\\ -1& 0\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0	1
$k$	$\left(\begin{array}{cc}\beta & -\alpha \\ -\alpha & -\beta \\ \end{array}\right)$	$\left(\begin{array}{cc}0& -i\\ -i& 0\\ \end{array}\right)$	$\left(\begin{array}{cc}1& -\sqrt{2}i\\ -\sqrt{2}i& -1\\ \end{array}\right)$	$\left(\begin{array}{cc}{\omega }^2& -\omega \\ -\omega & -{\omega }^2\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0	1
$-k$	$\left(\begin{array}{cc}-\beta & \alpha \\ \alpha & \beta \\ \end{array}\right)$	$\left(\begin{array}{cc}0& i\\ i& 0\\ \end{array}\right)$	$\left(\begin{array}{cc}-1& i\\ i& 1\\ \end{array}\right)$	$\left(\begin{array}{cc}-{\omega }^2& \omega \\ \omega & {\omega }^2\\ \end{array}\right)$	$t^2+1$	$t^2+1$	0	1
Set of values used	$\{0,1,-1,\alpha ,-\alpha ,\beta ,-\beta \}$	$\{0,1,-1,i,-i\}$	$\{0,1,-1,\sqrt{2}i,-\sqrt{2}i\}$	$\{0,1,-1,\omega ,-\omega ,{\omega }^2,-{\omega }^2\}$	--	--	$\{2,-2,0\}$	$\{1\}$
Ring generated by values used (characteristic zero)	$\mathbb{Z}[\alpha ,\beta ]$	$\mathbb{Z}[i]$ or $\mathbb{Z}[t]/(t^2+1)$	$\mathbb{Z}[\sqrt{2}i]$ or $\mathbb{Z}[t]/(t^2+2)$	$\mathbb{Z}[t]/(t^2+t+1)$	--	--	$\mathbb{Z}$	$\mathbb{Z}$
Field generated by values used (characteristic zero)	$\mathbb{Q}(\alpha ,\beta )$	$\mathbb{Q}(i)$ or $\mathbb{Q}[t]/(t^2+1)$	$\mathbb{Q}(\sqrt{2}i)$ or $\mathbb{Q}[t]/(t^2+2)$	$\mathbb{Q}[t]/(t^2+t+1)$, same as $\mathbb{Q}[t]/(t^2+3)$ or $\mathbb{Q}(\sqrt{-3})$	--	--	$\mathbb{Q}$	$\mathbb{Q}$

Four-dimensional irreducible representation over a non-splitting field

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

FACTS TO CHECK AGAINST (for characters of irreducible linear representations over a splitting field):
Orthogonality relations: Character orthogonality theorem | Column orthogonality theorem
Separation results (basically says rows independent, columns independent): Splitting implies characters form a basis for space of class functions|Character determines representation in characteristic zero
Numerical facts: Characters are cyclotomic integers | Size-degree-weighted characters are algebraic integers
Character value facts: Irreducible character of degree greater than one takes value zero on some conjugacy class| Conjugacy class of more than average size has character value zero for some irreducible character | Zero-or-scalar lemma

This character table works over characteristic zero and over any other characteristic not equal to two once we reduce the entries mod the characteristic:

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)	$\mathbb{Z}[i]$; $\mathbb{Z}[\sqrt{2}i]$; $\mathbb{Z}[\alpha ,\beta ]$ where ${\alpha }^2+{\beta }^2=-1$	$\{0,1,-1,i,-i\}$ if over $\mathbb{Z}[i]$; $\{0,1,-1,\sqrt{2}i,-\sqrt{2}i\}$ if over $\mathbb{Z}[\sqrt{2}i]$; $\{0,1,-1,\alpha ,-\alpha ,\beta ,-\beta \}$ if over $\mathbb{Z}[\alpha ,\beta ]$
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.