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Let $p$ be a prime number and $k$ be a positive integer such that $(k, p) = 1$. Let $m$ be the order of $p$ in $\mathbb{Z}/k\mathbb{Z}^×$. How many similarity classes of invertible matrix A ∊ ${\rm GL}_n(\mathbb{Z}/p^m\mathbb{Z})$ exist such that $A^k=I_n$?

Lemma: The matrices $A$ and $B$ are similar if and only if they have the same invariant factors.

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  • $\begingroup$ Let $(p_1,p_2,....,p_r)$ the invariant factors of A , then we have $p_r(X)=m_A(X)$ the minimal plynomial of A. Since $A^k=I_n$, then $p_r(X)$ devise the plynomial $X^k-1 $ ∊ $\mathbb{Z}/p^m\mathbb{Z}[X]$ $\endgroup$
    – N. SNANOU
    Commented Jul 30, 2016 at 0:57
  • $\begingroup$ Not sure what do from here on. $\endgroup$
    – N. SNANOU
    Commented Jul 30, 2016 at 13:43

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For any positive integer $k$ coprime to $p$, and any positive integers $n,m$ there is a bijection between conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ and conjugacy classes of elements of order $k$ in ${\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z}).$ This is because ${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})$ has a normal $p$-subgroup $U$ such that${\rm GL}_{n}(\mathbb{Z}/p^{m}\mathbb{Z})/U \cong {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$. ( It is also necessary to invoke the Schur-Zassenhaus Theorem).

Later edit: The matrix $A \in {\rm GL}_{n}(\mathbb{Z}/p\mathbb{Z})$ will have order $k$ if and only if the minimum polynomial $f(x) \in \left(\mathbb{Z}/p\mathbb{Z}\right) [x]$ of $A$ is multiplicity free, and the lcm of the multiplicative order of the roots of $f(x)$ ( in the algebraic closure of $\mathbb{Z}/p\mathbb{Z}$) is $k$.

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  • $\begingroup$ yes, thank you very much. Have you any idea about conjugacy classes of elements of order $k$ in ${\rm GL}_n(\mathbb{Z}/p\mathbb{Z})$?. $\endgroup$
    – N. SNANOU
    Commented Aug 7, 2016 at 13:30
  • $\begingroup$ This is now a question about rational canonical form. The answer is in some ways a routine computation though it gets a little messy. $\endgroup$
    – Geoff Robinson
    Commented Aug 7, 2016 at 13:49
  • $\begingroup$ Yes because a matrix is similar to exactly one rational canonical form, then it suffices to see the decomposition of the plynomial $X^k-1$ in $\mathbb{Z}/p\mathbb{Z}[X]$ that I am not sure what do about it. $\endgroup$
    – N. SNANOU
    Commented Aug 7, 2016 at 20:08
  • $\begingroup$ A splitting field for $X^{k}-1$ is ${\rm GF}(p^{m})$, where $m$ is defined as in the question. But there is still work to do. $\endgroup$
    – Geoff Robinson
    Commented Aug 7, 2016 at 20:22
  • $\begingroup$ I have added a note in the answer. $\endgroup$
    – Geoff Robinson
    Commented Aug 7, 2016 at 20:38

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