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?

  • $\begingroup$ @DietrichBurde I would prefer in the answer the condition that a group is cyclic be used .. $\endgroup$
    – Widawensen
    Aug 31, 2023 at 9:10
  • $\begingroup$ 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. $\endgroup$ Aug 31, 2023 at 9:24

1 Answer 1


$A^k=I$ 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.

  • $\begingroup$ 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 $1$? $\endgroup$
    – Widawensen
    Aug 31, 2023 at 10:33
  • 2
    $\begingroup$ @Widawensen Consider the companion matrices of those cyclotomic polynomials with three or more nonzero terms. If their determinants are $-1$, negate their last columns. $\endgroup$
    – user1551
    Aug 31, 2023 at 11:50
  • $\begingroup$ @user1551 Very useful answer, thank you. So it seems that such companion matrices are very illustrative for this kind of problem. $\endgroup$
    – Widawensen
    Aug 31, 2023 at 12:17
  • 1
    $\begingroup$ From mine as well, but I did not find it useful to mention it, and I forgot to delete a last matrix copied-pasted from my first example. Thanks to you, I just deleted it. $\endgroup$ Aug 31, 2023 at 13:20
  • 1
    $\begingroup$ @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 $k$. $\endgroup$
    – user1551
    Aug 31, 2023 at 13:35

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