Smallest $\mathbb R$-algebra which contains a subgroup isomorphic to $A_4$ $A_4$ (the alternating group on $4$ elements) can be thought of as the group of direct Euclidean isometries of a regular tetrahedron. This shows that there is a subgroup of the algebra of $3\times3$ matrices which contains it. This algebra is $9$ dimensional. There is no subgroup of the quaternion algebra $\mathbb H$ which is isomorphic to $A_4$, because there is only one element in $\mathbb H$ which has order exactly $2$, and that is $-1$, and there are three elements of $A_4$ of order $2$. It's possible that a subgroup of the $2 \times 2$ real matrices could work, but it looks like a lot of work to show that, if it is indeed the case.
What is the smallest associative and unital algebra over $\mathbb R$ which contains a subgroup isomorphic to $A_4$?
 A: Any $k$-algebra $A$ with a subgroup isomorphic to $G$ gives you an algebra homomorphism $kG \to A$. Since you are looking for the smallest algebra, you immediately get surjectivity, i.e. $A$ is a quotient of $kG$. Since you're in characteristic zero, $kG$ is a direct product of matrix algebras so that $A$ itself must be a product of matrix algebras. In particular, that would give you a representation of $G$ over $k$ that is faithful by assumption. And conversely, any faithful representation gives you an algebra homomorphism from $kG$ into a product of matrix rings that is injective on $G$.
A look at the character table of $A_4$ shows that there is a (unique) faithful 3-dimensional representation over $\mathbb{C}$, but no smaller one. In particular, there are also no faithful representations over $\mathbb{R}$ of dimension $<3$. The representation in question is the one coming from the permutation representation, so that it is indeed already defined over $\mathbb{R}$.
In other words: Dimension $9$ is the best you can do.
A: It is a consequence of Clifford's theorem that it is not possible to embed $A_{4}$ in such a real algebra of dimension less than $9$. There must be an involution of $A_{4}$ acting non-trivially in the associated representation. Then since all involutions of $A_{4}$ are conjugate, every involution in $A_{4}$ must act non-trivially. Let $V$ be the normal Sylow $2$-subgroup of $A_{4}$. Then the three non-trivial $1$-dimensional representations of $V$ are permuted transitively by $A_{4}.$
Thus $A_{4}$ acts (absolutely) irreducibly on a $3$-dimensional submodule of the underlying module, and its elements span the endomorphism algebra of this submodule, which is $9$-dimensional.
