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Let $H(n) \subset O(n)$ be the set of mirror symmetries in $\mathbb{R}^n$ with respect to $(n - 1)$-planes containing the origin. One can see that for any $a, b, c \in H(2)$ we have $abcabc = id$, that is, there is a non-trivial relation in $H(2)$.

Questions: are there similar non-trivial relations in $H(n)$ for $n > 2$ (the answer is likely no, but the proof eludes me)? Does the answer depend on the underlying field (say, with $\mathbb{Z}_p^n$ instead of $\mathbb{R}^n$)?

Motivation: suppose that we want to solve the word problem in the group generated by elements of a set $A$ under relations $a ^ 2 = 1$ for any $a \in A$. One way to do this is to assign a random element $\varphi(a)$ of $H(n)$ (represented by an $n \times n$ matrix) to each $a \in A$, and check that $\varphi(a_1) \ldots \varphi(a_n) = id$ for a word $a_1 \ldots a_n$. The relation for $n = 2$ described above implies that for some words (e.g. $abcabc$) this method fails with probability $1$. The question is, are there similar examples for $n > 2$, and if no, why not? Note that the word problem is very simple to solve for this group, yet there are situations where this "matrix hashing" approach is useful, e.g., when there are modification queries happening to the word. In this situation, finite fields are preferrable to $\mathbb{R}$ since they don't introduce floating-point errors.

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    $\begingroup$ You might like to look at this: en.wikipedia.org/wiki/Tits_alternative . The group $O(2)$ is virtually abelian. For larger $n$, $O(n)$ contains a free subgroup, and in particular satisfies no law. $\endgroup$
    – HJRW
    Jun 28, 2018 at 14:00
  • $\begingroup$ And of course the answer depends on the underlying field. If you use a finite field, the orthogonal groups are finite and therefore satisfy a lot of nontrivial relations, for example relations like $x^{|G|} =1$ $\endgroup$ Jun 28, 2018 at 15:18

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