# Orthogonality of cyclic finite group of matrices

Let $$M$$ be a cyclic finite subgroup of $$GL(n,\mathbb R)$$.
Is it necessary for matrices $$A$$ from $$M$$ to be orthogonal?

Indeed, because $$A^k=I$$ for some $$k$$ we have determinant of real matrix $$A$$ equal to $$-1$$ or $$1$$.

But generally there are matrices with such determinant which are not orthogonal. However I can't find other example of $$M$$ where matrices are not orthogonal.

How to prove that $$AA^T=I$$ or to find counterexample?

• @DietrichBurde I would prefer in the answer the condition that a group is cyclic be used .. Aug 31, 2023 at 9:10
• It suffices to look at finite subgroups (even compact ones). But anyway, this post is explaining that you may restrict to orthogonal matrices, but you could also use non-orthogonal ones, if I understand correctly. Aug 31, 2023 at 9:24

$$A^k=I$$ for some $$k\ge1$$ does not imply $$AA^T=I.$$
As a counterexample, you can take any non-orthogonal matrix $$A$$ such that $$A^2=I,$$ like $$A=\begin{pmatrix}1&1\\0&1 \end{pmatrix}\begin{pmatrix}1&0\\0&-1 \end{pmatrix}\begin{pmatrix}1&1\\0&1 \end{pmatrix}^{-1}=\begin{pmatrix}1&-2\\0&-1 \end{pmatrix}.$$ Similarly, the matrix $$B:=\begin{pmatrix}1&1\\0&1 \end{pmatrix}\begin{pmatrix}\cos\frac{2\pi}k&-\sin\frac{2\pi}k\\\sin\frac{2\pi}k&\cos\frac{2\pi}k \end{pmatrix}\begin{pmatrix}1&1\\0&1 \end{pmatrix}^{-1}$$ satisfies $$B^k=I$$ (and $$\det B=1$$) but is not orthogonal.
• So it would be a subgroup consisting of only two elements $A$ and $I$. But what if there would be more elements in groups, especially when $A$ is not equal to $A^{-1}$? Is it possible similar counterexample with determinant equal to $1$? Aug 31, 2023 at 10:33
• @Widawensen Consider the companion matrices of those cyclotomic polynomials with three or more nonzero terms. If their determinants are $-1$, negate their last columns. Aug 31, 2023 at 11:50
• @Widawensen No. E.g. the characteristic polynomial of the $B$ in Anne’s answer is not a product of cyclotomic polynomials except for a few values of $k$. Aug 31, 2023 at 13:35