i. m. soloveichik
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 Jan8 comment Kernel and image of a homomorphism $SL(2,5)\to S_5$ $S_5$ has a trivial center. Aug8 comment Proving that the intersection of a Sylow p-group with a normal subgroup is also a Sylow p-group Aug7 comment How prove $-\sqrt{2}\log(\cos x)\leq\sqrt{x\tan x-\sin^{2}x}$? Use the derivative Aug6 comment Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$ @JyrkiLahtonen That is not correct. For example the conjugates of $\sqrt{2}$ over the rationals are distinct but not linearly independent. Aug6 comment Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$ @PraphullaKoushik I should have said $1, s, \cdots, s^5$ are independent. Aug6 comment Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$ @PraphullaKoushik You can easily check that for $u=s+s^5$ then 1, u, u^2, u^3 are linearly independent over the rationals (using that $1, s, s^2, s^3, s^4, s^5, s^6$ are linearly independent over $Q$). Thus $u$ can't satisfy a rational polynomial of degree at most 3. It then follows easily that the degree 6 polynomial is irreducible. Aug6 comment Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$ @Praphulla Your polynomial is irreducible over $Q(\sqrt{-7})$ so multiply by its complex conjugate to get an irreducible polynomial over $Q$ Aug6 comment Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$ Multiply your cubic by its complex conjugate. Your cubic is $x^3+x^2+ \frac{1}{2}(3+\sqrt{7}I)x-1$ Aug1 comment Unique normal subgroups of every possible order @WilliamStagner That's proven in 39941. Aug1 comment Do two points determine a unique line in 4D space? @8bar If the two points are v, w then the plane has exactly the points $sv+tw$ for scalars $s,t$. You could find two vectors perpendicular to this plane and thereby determine this plane as the intersection of two 3-dimensional hyperplanes. Jul31 comment Groups of order $n^2$ that have no subgroup of order $n$ I checked n=31...44; all such groups have a subgroup of index n. Jul31 comment Characterization of solvable groups in terms of subgroups of certain orders? en.wikipedia.org/wiki/Hall_subgroup Jul31 comment Incentre and excentre of a triangle what does equivalent mean? Jul31 comment Largest Equilateral Triangle in a Polygon It is not in Math Reviews either. Can you post a pdf? Jul30 comment Groups of order $n^2$ that have no subgroup of order $n$ Must $n$ be even? Jul30 comment Largest Equilateral Triangle in a Polygon @NovaDenizen Thanks, so maybe make centroid of the triangle as far as possible from the vertices and inside the polygon. Jul30 comment Canonical embedding into dual space? What if it is an inner product space? Jul29 comment Structure of a group, $G$, of order $pq$ where $p, q$ are prime. How many Sylow p-subgroups?Sylow q-subgroups? Jul28 comment Finitely generated submodules use Smith Canonical form Jul26 comment Another Presentation of Certain Cyclic Groups Thanks for answering the other case too.