Peter Mueller
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 Apr 27 comment Character Values for Alternating Groups of degree $\geq 7$ Do you mean non-real when you say complex? A Magma calculation indicates that actually a stronger property might hold: For all alternating groups of degree $\le35$, there are at most $2$ non-rational values in each row and column. Actually, it suffices to show the assertion either for the rows, or for the columns, by using Brauer's Theorem for the Galois action on the irreducible characters and the conjugacy classes. Apr 27 comment Character Values for Alternating Groups of degree $\geq 7$ Apparently I'm missing the point -- why does that answer the question? Apr 26 revised Normalizers in symmetric groups added 84 characters in body Apr 26 revised Normalizers in symmetric groups Answered Stefan Kaohl's question Apr 17 awarded Nice Answer Mar 1 comment maximal order of elements in GL(n,p) @Anurag: No, if $p$ is prime then $AGL(1,p)$ contains an element of order $p$. If $q$ is a power of the prime $p$, then it is not hard to see that the elements in $AGL(n,q)$ have order at most $p(q^n-1)$. Feb 26 answered Applications of the Cayley-Hamilton theorem Feb 23 answered What can we say about the differences between roots of a polynomial with large Galois group? Feb 21 revised Maximal set of non commuting elements in a conjugacy class of $S_8$ added 710 characters in body Feb 20 answered Maximal set of non commuting elements in a conjugacy class of $S_8$ Feb 20 comment Maximal set of non commuting elements in a conjugacy class of $S_8$ @DerekHolt: I was wrong when I thought I had an argument that $224$ can't be achieved. Feb 20 comment Maximal set of non commuting elements in a conjugacy class of $S_8$ @Maryam: If I'm not wrong, then I have a clique of size 222 now. Feb 19 comment Maximal set of non commuting elements in a conjugacy class of $S_8$ Lovasz's theta upper bound shows that such a clique has size $\le224$, while mcqd (sicmm.org/konc/maxclique) quickly finds a clique of size $200$. Feb 12 comment Splitting subspaces and finite fields @Lev Borisov: This approach cannot work: Take for instance $m=2$, $n=3$ and let $T$ be the quadratic extension of $R$. Then the elements of $T$ are not in the value set you described, but $T\setminus R$ is not a subset of $S$. Feb 3 revised The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic added 7 characters in body Feb 3 comment The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic Yes, one certainly has to use the identity $\omega F_i=F_{i-1}-(-\omega)^i$, together with bad rational approximately of $\omega$. This implies for instance $\lceil(\omega(\pm F_i+k)\rceil=\pm F_{i-1}+\lceil\omega k\rceil$ for each nonzero integer $k$ with \$|k|