This is a continuation of this [question](https://mathoverflow.net/questions/138025/centralizer%20of%20the%20order%202%5Ek%20cyclic%20permutation%20matrix%20over%20F_2), 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.](https://mathoverflow.net/questions/138025/centralizer%20of%20the%20order%202%5Ek%20cyclic%20permutation%20matrix%20over%20F_2) 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](http://www.theoryofgroups.ir/?_action=articleInfo&article=2643&vol=252).



centralizer of a n-cyclic permutation matrix over F_2 in GL(n,2)
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
$$ (\mathbb{F}\_2 C\_n)^\* \cong 
 (\mathbb{F}\_2 C\_m)^\* \times \big((\mathbb{F}\_2C\_{2^k})^\*\big)^m.$$


