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None of these groups are $2$-primitive except for ${\rm Sp}(2d,2)$. For ${\rm PSL}(d,q)$ with $d>2$, the $2$-point stabilizer fixes two projective points, say $\langle v_1 \rangle$ and $\langle v_2 \ …

It misses the point to think of model category as a tool for proving that homotopy categories are equivalent. In the case of simplicial sets and topological spaces, that equivalence long preceded the …

One can build a distributional subsolution which is a Lipschitz extension of $u$ by making an extension with a positive jump in radial derivative across the boundary. Say $u$ is harmonic on $B_1$ and …

$R$ has a nonzero polynomial that induces the zero function if and only if there are ideals $I$, $J$ such that $I$ is nontrivial, $IJ=0$, and $R/J$ is a ring satisfying the following condition: There …

If the Sylow $p$-subgroup was not normal, then it is not hard to see that the only possibility would be $4|(p-1)$ with $p+1$ Sylow $p$-subgroups. Then $G$ would act $2$-transitively by conjugation on …

Not a complete answer, but a long comment. Using standard results for Riccati equations, one can parametrize all (symmetric and non-symmetric) solutions. One can rewrite the equation in the form $$ \m …

Then answer is yes. Since $F(L)$ is a regular language, it suffices to prove the following result: If $K$ is a regular language, then $R(K) = \{u \in A^\omega \mid F(u) \subseteq K \}$ is $\omega$ …

In an undirected graph with $m$ edges there can be as many as $\Theta(m^2)$ simple 4-cycles, so that's a reasonable time bound to aim for. And it's easy enough to achieve: set up a data structure that …

Zero. Indeed, if the intersection $Q_1 \cap Q_2$ of two quadrics is singular at $p_1$, then there is a quadric $Q$ in the pencil generated by $Q_1$ and $Q_2$ which is singular at $p_1$. On the other h …

There are some examples in convex analysis. For example, in the book of Borwein & Lewis (Convex Analysis and Nonlinear Optimization) page 225, they use the PVE to prove that every Chebyshev set (i.e. …

There is a clear and more specific answer (with reference moreover!) here, despite the different question: https://math.stackexchange.com/a/18496/84625 In short $\operatorname{Aut}\left(\left(\mathb …

Building upon what others have already said, the number of hyperbolic fixed points is indeed countable, but need not be finite. First, it follows from the definition of a hyperbolic fixed point $x$ t …

Every closed set $F$ on the unit circle is the set of singularities of some analytic function. Take a countable dense subset $z_k$ of $F$ and then choose positive $a_k$ so small that the series $$f(z) …

Alternatively, a finite group of odd order has no real-valued complex irreducible character other than the trivial character. ( Proof (well-known): if $\chi$ is such a character, we have $\chi(g) = \c …

In general, once you've proven an inequality like this in ${\bf R}$ it holds automatically in any Euclidean space (including ${\bf C}$) by averaging over projections. ("Inequality like this" = inequa …