Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $C$ be the $2^k\times 2^k$-permutation matrix over $\mathbb{F}_2$ of the $2^k$-cycle. We needed to know the structure of its centralizer in $\mathrm{GL}_{2^k}(\mathbb{F}_2)$, and we computed it - it was not too easy. It's an abelian group, and so we were able to compute the decomposition of the quotient of the centralizer over the subgroup generated by $C$ into the sum of cyclic subgroups, as follows. $$ \bigoplus_{i=2}^k (\mathbb{Z}/\mathbb{Z}_{2^{k+1-i}})^{2^{i-2}}. $$

We wonder if this was already done. (We also did this for more general case of other primes, not only 2, formulas are similar).

Update: see centralizer of a n-cyclic permutation matrix over F_2 in GL(n,2) for a follow-up question.

share|improve this question
@Mark: That argument works in characteristic zero, but not here. The only eigenvalue in this case is 1. –  Geoff Robinson Jul 28 '13 at 20:28
The matrix is similar to a matrix in Jordan normal form with a single Jordan block and eigenvalue 1,so its centralizer is conjugate to the centralizer of that matrix in Jordan form. –  Geoff Robinson Jul 28 '13 at 20:31
The Jordan form matrix $J$ has centralizer which is all linear combinations of $J-I$ ( there are $2^{k}-1$ such powers) in which the identity ( ie 0-th power) occurs with non-zero coefficient. –  Geoff Robinson Jul 28 '13 at 20:34
You're calculating the group of units of the group algebra ${\mathbb F}_2C$ of a cyclic group $C$ of order $2^k$ (your matrix describes how a generator of $C$ acts on the regular representation, and so the centralizer is the group of units in $\operatorname{End}_{{\mathbb F}_2C}({\mathbb F}_2C)\cong\mathbb{F}_2C$). In this guise, it's calculated in Prop. XI(5.7) of Bass's Algebraic K-Theory. –  Jeremy Rickard Jul 29 '13 at 7:29
There is a paper by S. H. Murray, Conjugacy classes in Maximal Parabolic Subgroups of General Linear Groups, J. Algebra 233, 135-155 (2000). In Section 4 (Centralizers in general linear groups) he works out the centralizers of arbitrary elements in $GL_n(k)$. –  Tim Dokchitser Jul 31 '13 at 8:21
show 5 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.