Show that two matrices over a finite field are similar

Let p be a prime. Show that the following p \times p matrices are similar in \mathsf{Mat}_p(\mathbb{F}_p): A = \mathcal{C}_{x^p-1} (the companion matrix of x^p-1) and B = J_p(1) (the p \times p Jordan block with eigenvalue 1).

Recall that the characteristic polynomial of \mathcal{C}_{x^p-1} is x^p-1 (as we showed here), and that over \mathbb{F}_p we have x^p-1 = (x-1)^p (as can be deduced from this precious exercise).

We claim that in fact x^p-1 is the minimal polynomial of A. Recall (from this previous exercise) that \mathsf{Sym}(p) (the symmetric group on p objects) is embedded in \mathsf{GL}_p(\mathbb{F}_p) by letting a permutation \sigma act index-wise on the elements of an ordered basis (i.e. \sigma \cdot e_i = e_{\sigma(i)}). Now A is in the image of this representation, and in fact is the matrix representing the permutation whose cycle decomposition is \sigma = (1\ 2\ 3\ \ldots\ p-1\ p). In particular, A^{p-1} = A^{-1} is not the identity transformation, so the minimal polynomial of A is indeed x^p-1.

So the elementary divisors of A are just (x-1)^p, and so A is similar to the Jordan block B.

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