centralizer of a n-cyclic permutation matrix over F_2 in GL(n,2)

Question

This is a continuation of this question, where I talked about the case $n=2^k$. Let $C$ be the $n\times n$-permutation matrix over $\mathbb{F}_2$ of the $n$-cycle. We needed to know the explicit structure of its centralizer in $\mathrm{GL}_{n}(\mathbb{F}_2)$ (more precisely, of its factor $Z(n)$ over the subgroup generated by $C$), and we computed it. It's an abelian group, and so we were able to compute its decomposition into the sum of cyclic subgroups, as follows, where $n=2^km$, with $m$ odd. $$ Z(n)=\bigoplus_{i=2}^k (\mathbb{Z}/\mathbb{Z}_{2^{k+1-i}})^{2^{i-2}m}\oplus \mathbb{Z}/\mathbb{Z}_{2^{k}}^{m-1}\oplus Z(m). $$ The decomposition of $Z(m)$ is also explicit, and can be described in terms of factors of the $\mathbb{F}_2$-irreducible factors of the polynomial $x^m+1\in\mathbb{F}_2[x]$.

We wonder if this was already done. A comment to loc.cit. says that the case $n=2^k$ was done in in Prop. XI(5.7) of Bass's "Algebraic K-Theory". It is useful to note that the question is equivalent to describing the group of units in the $\mathbb{F}_2$-group algebra of $\mathbb{Z}/n\mathbb{Z}$. It appears that in general the structure of the group of units in the group algebra of $\mathbb{Z}/n\mathbb{Z}$ over a finite field is not known, as one can find fresh publications settling particular cases, cf. e.g. Int.J. Group Theory v.2, No. 4 (2013), 1-6.

Answer 1

I don't know if this is in the literature, but I claim that the structure of the unit group of $\mathbb{F}_q C_n$ in general can be derived using similar ideas as in the proof of Bass' Proposition XI.5.7. and some standard techniques:

Bass computes the structure of $(\mathbb{F}_p C_{p^k} )^* = \mathbb{F}_p^* \times (1+I)$, where $I$ is the augmentation ideal. He shows that $1+I$ is the direct product of the cyclic groups generated by $$ 1+d^i \quad (1\leq i < p^k, \quad p\nmid i),$$ where $d+1$ is a generator of the cyclic group $C_{p^k}$.

Now let $q=p^a$. Then again $(\mathbb{F}_qC_{p^k})^* = \mathbb{F}_q^* \times (1+I_q)$, where $I_q$ is the augmentation ideal in $\mathbb{F}_qC_{p^k}$. Let $B$ be a basis of $\mathbb{F}_q$ over $\mathbb{F}_p$ and $d\in I$ as above. By mimicking Bass' proof, I can show that now $1+I_q$ is the direct product of the cyclic groups generated by the elements $$ 1+bd^i\quad (1\leq i< p^k,\quad p\nmid i,\quad b\in B). $$ (I can add more details, if wanted.) Thus $(1+I_q) \cong (1+I_p)^a$.

Finally, let $n$ be arbitrary, $n=p^k m$ with $p\nmid m$, and $q=p^a$. Then $$ \mathbb{F}_q C_n \cong \mathbb{F}_q C_m \otimes_{\mathbb{F}_q} \mathbb{F}_qC_{p^k}.$$ The group algebra $\mathbb{F}_qC_m$ is a direct sum of fields of order $q^{e_j}$, since $p\nmid m$. (The $e_j$'s are the degrees of the irreducible factors of $x^m-1$ over $\mathbb{F}_q$.) It follows that $$ \mathbb{F}_q C_n \cong \bigoplus_j \mathbb{F}_{q^{e_j}} C_{p^k}. $$ We know the unit groups of the summands. It follows that $$ (\mathbb{F}_q C_n)^* \cong \prod_j ((\mathbb{F}_{q^{e_j}})^* \times (1+I_q)^{e_j}) \cong (\mathbf{F}_qC_m)^* \times (1+I_q)^m, $$ where the last isomorphism follows from rearranging factors and $\sum_j e_j = m$.

In particular, for your question we get $$ (\mathbb{F}_2 C_n)^* \cong (\mathbb{F}_2 C_m)^* \times \big((\mathbb{F}_2C_{2^k})^*\big)^m.$$