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.
Best Answer
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:
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.