TITLE: Any finite group is a subgroup of an orthogonal group
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: $S_n$ 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.