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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
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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 $|k|<F_i$. Nevertheless, trying to work things out along this idea seems to be a pain. So someone probably has a better idea.
Feb
3
comment The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
@LevBorisov: You are right, there are some details missing for which I don't have the time to check.
Feb
3
revised The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
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Feb
2
awarded  Nice Answer
Feb
2
revised The sequence $a_{n+1}=\left\lceil \frac{-1+\sqrt{5}}{2}a_{n}-a_{n-1} \right\rceil$ is periodic
added 720 characters in body