Let $Q_8$ be the quaternion group of order $8$. I would like to determine the algebra structure for $\mathbb{R}[Q_8]$.

I think $\mathbb{R}[Q_8] \cong \mathbb{R}^4 \oplus \mathbb{H}$. Maybe a simpler question to all of this is: why is $\mathbb{R}[Q_8] \not\cong \mathbb{R}^4 \oplus M_2(\mathbb{R})$?

My work so far:

By Maschke's Theorem, $\mathbb{R}[Q_8]$ is semisimple. By Artin-Wedderburn Theorem, since $\mathbb{R}[Q_8]$ is a finite-dimensional semisimple $\mathbb{R}$-algebra, it follows that $$\mathbb{R}[Q_8] \cong M_{n_1}(D_1) \oplus \dots \oplus M_{n_k}(D_k)$$ where each $n_i$ is a positive integer and $D_i$ is a division ring over $\mathbb{R}$.

By Frobenius theorem (of real division algebras), it follows that each $D_i$ is isomorphic to either $\mathbb{R}$ (1-dimensional), $\mathbb{C}$ (2-dimensional), or the quaternions $\mathbb{H}$ (4-dimensional).

Thus, we begin a combinatorial argument: $$\mathbb{R}[Q_8] \cong \mathbb{R}^a \oplus M_2(\mathbb{R})^b \oplus \mathbb{C}^c \oplus M_2(\mathbb{C})^d \oplus \mathbb{H}^e$$

(Note that $\mathbb{R}[Q_8]$ is an $8$-dimensional group algebra which is why $M_n(\mathbb{R})$ terms don't exist for $n>2$ and similar reasoning for $M_n(\mathbb{C})$ and $M_n(\mathbb{H})$.)

My argument will rely on the following facts:

(i) $Q_8$ has five conjugacy classes so $k=5=a+b+c+d+e$.

(ii) $\operatorname{dim}(\mathbb{R}[Q_8]) = 8 = a + 4b + 2c + 8d + 4e$

(iii) $\mathbb{R}[Q_8]$ is non-commutative because $Q_8$ is non-abelian so we must have at least one of $b$, $d$, or $e$ to be nonzero.

[Claim 1: d = 0] First of all, to satisfy (ii), $d$ must be either $0$ or $1$. If $d=1$, then we immediately get $a=b=c=e=0$ which contradicts (i) so $d=0$.

[Claim 2: c = 0] Similarly, if $c>1$ then we cannot simultaneously satisfy (i) and (ii) so $c$ is either $0$ or $1$. To satisfy both conditions with $c=1$, we are forced to have $a=4$ and $b=d=e=0$ which contradicts (iii). Thus, $c=0$.

[Claim 3: a = 4] Condition (ii) is now simplified to $8=a+4b+4e$. This equation can only be satisfied when $a\in \{ 0 ,4,8 \}$. However, $a=8$ contradicts (iii) and $a=0$ contradicts (i) so we must have that $a=4$.

At this point, I now have that $\mathbb{R}[Q_8] \cong \mathbb{R}^4 \oplus M_2(\mathbb{R})$ OR $\mathbb{R}[Q_8] \cong \mathbb{R}^4 \oplus \mathbb{H}$. Both possibilities satisfy (i)-(iii). Intuition tells me that I should get a copy of $\mathbb{H}$ and Wikipedia suggests this too. However, their reasoning seems to rely on more advanced machinery on irreducible characters/representations that I do not have. I thought about possibly arguing by nilpotent elements or the center of the algebra but I don't know how to rigorously argue that it should be isomorphic to the latter. Ideally, I am hoping there would be a fundamental property about these group algebras that I am overlooking to provide a simple argument rather than going into heavier machinary.

  • $\begingroup$ Studying this question for a qualifying exam. Do you have a good algebra textbook reference for statement (i) saying number of simple components in the decomposition is equal to the conjugacy classes of the group? My main resource is Hungerford, but it doesn't have this result. $\endgroup$ Jan 2, 2023 at 17:01
  • 1
    $\begingroup$ @JonathanMichala My school uses Lang. There is a chapter on Representations of Finite Groups that might be useful for you. Section XVIII Chapter 4 is where I got my information on (i) and (ii). $\endgroup$ Jan 2, 2023 at 17:54

1 Answer 1


Here is an argument that requires no explicit representation theory. As described on Wikipedia, the quaternion group $Q_8$ can be described by the following presentation using four (redundant) generators:

$$Q_8 \cong \langle i, j, k, c \mid c^2 = 1, i^2 = j^2 = k^2 = ijk = c \rangle.$$

I've named the fourth generator $c$ because it is central. It follows that $c$ spans a central subalgebra of $\mathbb{R}[Q_8]$ given by $\mathbb{R}[c]/(c^2 - 1) \cong \mathbb{R}^2$ where the two copies of $\mathbb{R}$ correspond to $c = 1$ and $c = -1$ respectively. It's a general fact that whenever a central subalgebra of an algebra splits as a direct product it induces a direct product splitting of the entire algebra: here this gives

$$\mathbb{R}[Q_8] \cong \mathbb{R}[Q_8]/(c = 1) \times \mathbb{R}[Q_8]/(c = -1).$$

The first factor is the group algebra of the quotient of $Q_8$ by the relation $c = 1$, which is the Klein four group $C_2 \times C_2$, and it's not too hard to show that this is isomorphic to $\mathbb{R}^4$. The second factor is what is known as a twisted group algebra; it is the algebra given by the presentation

$$\mathbb{R}[i, j, k]/(i^2 = j^2 = k^2 = ijk = -1)$$

which is exactly the usual presentation of the quaternions.

Representation theory provides more context for these sorts of calculations. Generally speaking, for $G$ a finite group, $\mathbb{R}[G]$ decomposes as a product of copies of the algebras $M_n(D)$ for every irreducible representation of $G$ over $\mathbb{R}$ with endomorphism algebra $D$ and dimension $n$ (over $D$). So this decomposition for $\mathbb{R}[Q_8]$ is equivalent to the claim that $Q_8$ has five irreducible real representations:

  • four of which have endomorphism algebra $\mathbb{R}$ and are $1$-dimensional (these are exactly the $1$-dimensional representations of the abelianization, which is the quotient $C_2 \times C_2$ that appears above), and
  • the fifth of which has endomorphism algebra $\mathbb{H}$ (a quaternionic representation) and which is $1$-dimensional over $\mathbb{H}$ (so $4$-dimensional over $\mathbb{R}$).

The quaternionic representation is given exactly by the quotient map $\mathbb{R}[Q_8] \to \mathbb{H}$ from $\mathbb{R}[Q_8]$ to the twisted group algebra above. These representations are distinguished by whether the central element $c$ acts by $1$ or $-1$; this is called the central character of a representation.

  • $\begingroup$ This makes a lot of sense. Thank you! $\endgroup$ Dec 22, 2022 at 23:39

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