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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?

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    $\begingroup$ That is a good start. Now how can we represent the permutations in $S_n$ as orthogonal real matrices? $\endgroup$
    – hardmath
    Aug 3 '18 at 19:31
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$S_{n}$ 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 $\langle \varphi(\sigma)(e_i), \varphi(\sigma)(e_j)\rangle~= ~\langle e_i, e_j\rangle$.

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$S_n$ 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.

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