# Getting a one to one morphism $p : G \to GL_n(\mathbb{C})$

I am studying representation theory, and I would like to find an algorithm that finds, given a finite group $$G$$, a one to one morphism $$p : G \to GL_n(\mathbb{C})$$ (the integer $$n$$ is also found by the algorithm). I don't necessarily want this algorithm to be efficient. First is it possible to make such an algorithm ?

Now I am wondering how to begin with. I thought about finding the character table of the group first (I know this not an easy task) and find its irreductible representations, but then from here I don't see how I can find a solution to my problem. If I have all the irreductible representations of my group I don't see how I can get $$p$$. I mean the characters only give information about the sum on the digonal of $$p(g)$$, but how to get from this information the other entries in the matrices ?

Thank you !

• "given a group $G$" is important. i removed the parentheses – mathworker21 Jun 3 at 11:02
• Take a look at Theorem 2.12 of "Introduction to Representation Theory" by Etingof (for finite groups only). Since the field is $\mathbb{C}$, $A=\mathbb{C}[G]$ is semisimple, and therefore there is an isomorphism of algebras $A\cong \oplus_i End(V_i)$, where $V_i$ are the irreducible representations of $G$. – Javi Jun 3 at 11:04
• @mathworker21 when I say given a group $G$ it means any type of group theoric description of $G$ (for example Cayley table) – bonjour1 Jun 3 at 11:04
• If $G$ is finite you can just use $n=\#G$ with the regular representation. For infinite $G$ we may not have finite dimensional faithful representation. – user10354138 Jun 3 at 11:08
• Matrices of the regular representations are easily extracted from the Caley table. Try at hand with the symmetric group of order 6. – InfiniteLooper Jun 3 at 11:22

Every finite group embeds into a permutation group; this is Cayley's theorem. The proof of the theorem is essentially constructive, with $$G$$ embedding into $$S_{|G|}$$. That is, there is an algorithm with input a Cayley table* of a finite group $$G$$ and output a permutation group $$\operatorname{Perm}(G)$$ isomorphic to $$G$$. You can then easily embed $$\operatorname{Perm}(G)$$ into $$\operatorname{GL}_{|G|}(\mathbb{C})$$ using permutation matrices, and again this can be made algorithmic.

*The input form can vary, but the OP explicitly mentions Cayley tables in the comments.

• Thankfully, the OP also explicitly mentioned that the algorithm didn't need to be efficient ! (well this one is very quick to find the morphism, but the $n$ it produces is most often far from optimal) – Max Jun 3 at 14:55
• The constructive proof of Cayley's Theorem is essentially the content of my answer, except that instead of embedding in $S_{|G|}$, I've further mapped $S_{|G|}$ into $SO(|G|)$ so that you get an actual faithful representation of $G$ in the orthogonal group. – John Hughes Jun 6 at 10:45

Take a group, with elements $$g_1, \ldots, g_n$$.

Associate to $$g_1$$ the vector $$e_1 \in \Bbb R^n$$.

Now for element $$g_1$$, compute the $$n$$ items $$g_1 \cdot g_1, g_1 \cdot g_2, \ldots, g_1 \cdot g_n.$$ These will be the elements of $$G$$, in some new order, such as $$g_7, g_2, \ldots, g_3.$$ Write down the corresponding vectors as the columns of a matrix $$M_1 = \pmatrix{ e_7, e_2, \ldots, e_3}$$ Do the same for $$g_2$$ to produce $$M_2$$, and so on.

The map $$g_1 \mapsto M_1, g_2 \mapsto M_2, \ldots$$ will be a injective homomorphism of groups.

(Unless I've messed up and forgotten a transpose somewhere...)

You can test this out pretty quickly by trying it for $$\Bbb Z/3\Bbb Z$$, or even $$S_3$$.