Every finite subgroup of symmetric $n\times n$ matrices with nonzero determinant is isomorphic to some $(\mathbb{Z}/2\mathbb{Z})^m$, $0\leq m \leq n$

Question

Suppose $G$ is such a subgroup. The products of symmetric matrices are still symmetric matrices, and we have $AB=(AB)^T=B^TA^T=BA$, so any such subgroup is a commuting family of symmetric matrices and can be simultaneously diagonalized by orthogonal matrices. The $(\mathbb{Z}/2\mathbb{Z})^m$ part reminds me of Sylvester law, and I suppose we can scale the length of the columns and rows to make any $A$ congruent to some $diag\{1,1,..,-1,-1,...\}$, (no $0$'s since det $A \neq 0$), and then map $-1$ to $1$ and $1$ to $0$ in the appropriate entry in $(\mathbb{Z}/2\mathbb{Z})^n$. But I don't think we can do the scaling simultaneously for every matrix in $G$.

Answer 1

You don't need to do any scaling. You already know that all matrices in $G$ are simultaneously diagonalizable. So you fix one such diagonalization, say it's given by a matrix $R$; the eigenvalues of any $A\in G$ are $\{-1,1\}$ (proof at the end). Then you map $G\to \{-1,1\}^n$ by $$ A\longmapsto \operatorname{diag}\{R^{-1}AR\}\subset\{-1,1\}^n. $$ Now you further map, as you say, to $(\mathbb Z/2\mathbb Z)^n$. This way $G$ is isomorphic to a subgroup of $(\mathbb Z/2\mathbb Z)^n$.

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Proof that $A\in G$ has eigenvalues in $\{-1,1\}$

As was already said above, we may assume that $A$ is diagonal, and the elements $a_1,\ldots,a_n$ of the diagonal are the eigenvalues of $A$. The $m^{\rm th}$ power of $A$ is still in $G$ and has eigenvalues $$ a_1^m,\ldots,a_n^m. $$ As $G$ is finite, these values have to cycle, and so $a_k\in\{-1,1\}$ is the only option, since since symmetric matrices have real eigenvalues and $0$ is disallowed by $A$ being invertible.