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Sean Eberhard's user avatar
Sean Eberhard's user avatar
Sean Eberhard's user avatar
Sean Eberhard
  • Member for 12 years, 11 months
  • Last seen this week
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What is the finite group $(\operatorname {PCO}^{\circ}_{2n})^{+}(q)$
I think this is right, at least if $p$ is odd, though I'm not familiar with Magma. I believe $\mathrm{PCO}_{2n}^\circ$ is more or the less the same thing as $\mathrm{PSO}_{2n}$, though people who know more than me don't like to write the latter thing for some reason. So $(\mathrm{PCO}_{2n}^\circ)^+(q)$ is probably just $\mathrm{PSO}^+(2n, q)$ in Magma.
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Combination of simple tensors - II
What is $S_k$? It's still not clear.
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Groups $P$ of order $p^5$ with $\Omega_1(P)=P$
To complement YCor's comment, for $p \ge 5$, a group of order $p^5$ must be a regular $p$-group, so $\Omega_1(P) = P$ is equivalent to saying that $P$ has exponent $p$.
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Order of abelian subgroup of the automorphism group of an abelian group
As for your last question, there are no other examples, but I may offend you by using representation theory again. We know that $B$ acts reguarly on the nonzero points of $V = \mathbb F_p^\ell$, in particular irreducibly, so by Schur's lemma, $K = \mathrm{End}_B(V)$ is a finite division ring, hence a field, and $B \le K^\times$. Since $K^\times$ is abelian and transitive on $V \setminus \{0\}$, $|K^\times| = |V| - 1$. This implies that $B = K^\times$ and $V \cong K$.
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Order of abelian subgroup of the automorphism group of an abelian group
@tomasz We have $f^i(A) \le f(p^{i-1} A) = p^{i-1} f(A) \le p^i A$ by induction. The assumption that $k \ge p$ is just used to ensure that $\binom{p^k}{i}$ is divisible by $p^k$ for $0 < i < k$.
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If two vertices induce isomorphic subgraphs when they are removed, are they conjugate?
For example I think you can take (the underlying graph of) the en.wikipedia.org/wiki/Snub_cube or the en.wikipedia.org/wiki/Snub_dodecahedron. A lot of groups have a "graphical regular representation", which is a Cayley graph whose automorphism group is just the group itself -- these will always give you an example.
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awarded
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A problem about the series $\sin(n^p)$
In case it's not obvious, I started here! I don't know if it's better to close this question or to post an answer with a signpost.
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Does a random matrix over $\mathbb{Z}_q$ map linearly independent vectors to statistically independent vectors?
Yes, this is true. Extend the sequence to a maximally linearly independent set. Then $r = m$ and we have a basis for the free $\mathbb Z_q$-module of rank $m$. Choices for the matrix $A$ correspond bijectively with choices of linear maps $\mathbb Z_q^m \to \mathbb Z_q^n$. It follows that a change of basis does not change the distribution.
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Number of Salem–Spencer subsets of $\{1,2,3,\dots ,n\}$
@SayanDutta The given argument shows only that $f(n)^{1/n} \le f(m)^{1/m}$ when $n$ is a multiple of $m$.
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Regular orbits for automorphisms of finite simple groups
Upgraded some n's to (n-1)'s so there are no issues with small n
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Density of a set of numbers whose prime factors are defined by congruences
Yes, also density zero. The relevant Euler product is $\prod_{p \in 12\mathbf Z - 1} (1-1/p) = 0$.
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Is the monster group maximal in SO(196883)?
Maybe it would be better to ask whether this is true generally for finite simple groups in their minimal-degree representation (with SO or SU or Sp as appropriate depending on the Schur indicator), and if not then for which simple groups is it true.
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