There's$\let\geq\geqslant\DeclareMathOperator{\GL}{GL}$ this exercise in my algebra course book:

Let $p$ be a prime and $A\neq I$ an $n\times n$ matrix over $\mathbb Z$ such that $A^p=I$. Prove that $n\geq p-1$.

It's in a section about reducibility of polynomials over $\mathbb Z$. (Gauss' lemma, Eisenstein's criterion etc.) We are also allowed to use some basic properties of UFD's, PID's or Euclidean Domains. The only solution (if it is correct(?)) I could think of is the following:

Let $q$ be a prime which is a primitive root modulo $p$ (which exists by Dirichlet's theorem on AP's). Reducing modulo $q$, $A$ has order $p$ in $\GL(n,q)$. Since $\GL(n,q)$ has order $(q^n-1)(q^n-q)(q^n-q^2)\cdots(q^n-q^{n-1})$, we should have $p\mid q^k-1$ for some $k\in\{1,\ldots,n\}$. Since $q$ is a primitive root modulo $p$, we have $n\geq p-1$.

I don't think this is the intended approach. This has hardly anything to do with polynomials, and Dirichlet's theorem has never been mentioned in our course.
Is there a more elementary way of proving this?

  • 2
    $\begingroup$ Hint: By Cayley-Hamilton, the minimal polynomial of $A$ has degree at most $n$. $\endgroup$ – Erick Wong Jan 8 '15 at 17:22

Suppose $A$ is such a matrix. Since $A\neq I$ satisfies $A^p-I=0$, the minimal polynomial of $A$ is either $x^{p-1}+\cdots+x+1$ or $x^p-1$ as \begin{equation} x^p-1=(x-1)(x^{p-1}+\cdots+x+1) \end{equation} with the two polynomials in the right being irreducible.

(Note that both possibilities may occur. Indeed, if $A^{p-1}+\cdots+A+I=0$, then the matrix $B=\begin{bmatrix}A&0\\0&1\end{bmatrix}$ still satisfies $B^p=I$ while it's minimal polynomial is $x^p-1$ since \begin{equation} B^{p-1}+\cdots+B+I=\begin{bmatrix}O&0\\0&p\end{bmatrix}\neq0.) \end{equation}

Now the characteristic polynomial of $A$ of degree $n$ is a multiple of the minimal polynomial of $A$, which is of degree $p-1$ or $p$. This shows that $n\geq p-1$ and that $n\geq p$ if $1$ is an eigenvalue of $A$.


The polynomial $$p(x) = x^{p-1} +x^{p-2} + \ldots +1 $$ is irreducible over $\mathbb{Q}$ ( see cyclotomic polynomials ).

Thus $p(x)$ is the minimal polynomial of $A$ over $\mathbb{Q}$, because $$x^p-1 = (x-1)( x^{p-1} +x^{p-2} + \ldots +1)$$ and $A \neq I $ by hypotheses.

The roots of $p(x)$ are all differents, the minimal and the characteristic polynomial have the same roots, the degree of the characteristic polyomial is $n$ and thus $n \geq p-1$.

  • $\begingroup$ It's indeed important to note that we are working over $\mathbb Q$. Thanks! $\endgroup$ – punctured dusk Jan 8 '15 at 17:45

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