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- Newsgroups: sci.math
- Subject: Re: 1992 Putnam problems and unofficial solutions
- Message-ID: <1992Dec18.142423.18674@husc3.harvard.edu>
- From: elkies@ramanujan.harvard.edu (Noam Elkies)
- Date: 18 Dec 92 14:24:22 EST
- References: <1g0bmsINNh22@agate.berkeley.edu> <a_rubin.723836042@dn66>
- Organization: Harvard Math Department
- Summary: yes, the n^2 bound in B-6 is sharp
- Nntp-Posting-Host: ramanujan.harvard.edu
- Lines: 29
-
- In article <a_rubin.723836042@dn66>
- a_rubin@dsg4.dse.beckman.com (Arthur Rubin) writes:
- ">Problem B6
- "
- ">Let M be a set of real n by n matrices such that
- ">(i) I \in M, where I is the n by n identity matrix;
- ">(ii) if A \in M and B \in M, then either AB \in M or -AB \in M, but not both;
- ">(iii) if A \in M and B \in M, then either AB = BA or AB = -BA;
- ">(iv) if A \in M and A \noteq I, there is at least one B \in M such that
- "> AB = -BA.
- "
- ">Prove that M contains at most n^2 matrices.
- "
- "But can it actually contain n^2 matrices?
- "
- "I KNOW this is NOT part of the problem as stated, but I am a little curious.
-
-
- Yes it can: let n be a power of 2; let G be the extraspecial group of
- order 2n^2 which has a *real* (even rational) n-dimensional representation
- with nontrivial central character; and let M consist of half of the images
- of G under that representation, using one of each pair of opposite matrices
- and of course choosing the identity out of the pair {I,-I}. Indeed it
- readily follows from the posted solution that G ( = M union -M ) is an
- extraspecial 2-group, and thence that the above construction is the only
- way to attain equality.
-
- --Noam D. Elkies (elkies@zariski.harvard.edu)
- Dept. of Mathematics, Harvard University
-