# Any finite group is a subgroup of an orthogonal group [duplicate]

Prove that any finite group of order $n$ is isomorphic to a subgroup of $\mathbb{O}(n)$, the group of $n\times n$ orthogonal real matrices.

Attempt:

Let $G$ be a group of order $n$. Then $G$ is isomorphic to a subgroup of the symmetric group $S_n$.

But how to go further?

• That is a good start. Now how can we represent the permutations in $S_n$ as orthogonal real matrices? Aug 3 '18 at 19:31

$$S_{n}$$ acts on $$\mathbb{R}^n$$ by the equation $$\sigma . e_i= e_{\sigma(i)},$$ where $$\lbrace e_i \vert i= 1,2,...,n\rbrace$$ is the standard basis of $$\mathbb{R}^n$$ and $$\sigma \in S_n$$. Therefore we have a group morphism $$\varphi : S_n \rightarrow GL_n(\mathbb{R})$$ defined by $$\varphi(\sigma)(e_i)= e_{\sigma(i)}.$$ It is easy to check that $$\varphi$$ is one-one. Note that $$\varphi(S_n) \subset \mathbb{O}(n)$$, for $$\langle \varphi(\sigma)(e_i), \varphi(\sigma)(e_j)\rangle~= ~\langle e_i, e_j\rangle$$.
$S_n$ is embedded in the set of n x n matrices as permutation matrices. A permutation matrix represents a permutation by having a 1 in the ith row and the jth column if the permutation sends i to j, and zero otherwise. (One can also embed them by switching "row" and "column" in the preceding sentence.) Permutation matrices are orthogonal, and this correspondence is a bijection that maps permutation composition to matrix multiplication, i.e. is a isomorphism.