MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|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
@Geoff: You are right. – Mark Sapir Jul 28 '13 at 20:36
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
up vote 0 down vote accepted

the details of our computation can now be found in

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.