Homomorphisms into the General Linear Group

Question

Let $G$ be a finite group of order $n \geq 2$.

I want to prove that there always exists an injective homomorphism $\varphi:G \to GL_n(\mathbb R)$. Can you help?

Answer 1

By Cayley's theorem, we have an injective homomorphism, $$G\hookrightarrow S_n.$$ We have a natural action of $S_n$ on $\mathbb{R}^n$ by permuting the indices via: $$S_n \times \mathbb{R}^n \to \mathbb{R}^n, \ (\sigma,(x_1,...,x_n))\mapsto (x_{{\sigma}^{-1}(1)},...,x_{{\sigma}^{-1}(n)}).$$ This action induces a homomorphism $$\psi :S_n \to \mbox{Aut}(\mathbb{R}^n)=GL_n(\mathbb{R}), \ \psi(\sigma)((x_1,...,x_n))= (x_{{\sigma}^{-1}(1)},...,x_{{\sigma}^{-1}(n)}).$$

In fact, every group action induces a group homomorphism and vice-versa: Given an action $$\rho: G \times X \to X, \ (g,x)\mapsto g.x,$$ one can get a homomorphism from $G$ via $$\phi: G \to Aut(X) , \ \phi(g)(x):=g.x$$ I used the term "$Aut(\mathbb{R}^n)$" earlier for this reason. But in your case, since $\mbox{Aut}(\mathbb{R}^n)=GL_n(\mathbb{R})$, you might directly go to $GL_n(\mathbb{R}).$

You can easily check that the map thus defined is indeed a homomorphism. The claim follows.

I hope this helps!

Answer 2

Let $V$ denote the real vector space with basis $\{e_g:\, g\in G\}$. The homomorphism $\varphi:G\to GL(V)$ by $\varphi(g)e_h = e_{gh}$ is clearly injective.

(This construction is called the (real) regular representation of $G$. Sometimes that name is also given to the corresponding map $G \to GL(W)$ for the subspace $W = \{\sum v_g e_g\in V:\, \sum v_g = 0\}$.)

Answer 3

Consider $n\times n$ matrices labelled by the elements of $G$ (there are $n$ of them).

Consider for every $g\in G$ the matrix $A^{(g)}= (a^{(g)}_{hk})_{h,k \in G}$ with elements $$a^{(g)}_{hk} = \delta_{h, gk}$$
Let's check that $$A^{(g_1)} \cdot A^{(g_2)} = A^{(g_1 \cdot g_2)}$$ Indeed we have $$(A^{(g_1)} \cdot A^{(g_2)})_{hk} = \sum_{l} a^{(g_1)}_{hl} \cdot a^{(g_2)}_{lk}=\delta_{h, g_1 l} \cdot \delta_{l, g_2 k} = \delta_{h, g_1 g_2 k} = A^{(g_1 g_2)}_{hk} .$$

Hence we have a morphism from $G$ to $GL(n,\mathbb{Z})$.

Note that for $g\ne e$ the matrix $A^{(g)}$ has all the diagonal elements $0$ so it it not the identity matrix. We conclude that the morphism is injective.

This morphism corresponds to the left regular representation like in the answer of @anomaly:. One can also consider the right regular representation $g\colon \ e_h \mapsto e_{hg^{-1}}$