Category: Un pò di matematica
-
Let G be a finite group and φ: G—>G be an automorphism that fixes more than half of the elements of G. Prove that φ is the identity of G.
Let S={g ∈ G : φ(g)=g}. We know that by hypothesis |S|>|G|/2 and, by what was said previously, S is a subgroup of G (it is a non-empty subset of G and if x,y∈S we have that φ(x⁻¹y)=φ(x⁻¹)φ(y)=φ(x)⁻¹φ(y) )therefore|G|=|S|k for non-zero natural k. If k=1 I am done. In fact:|G|=|S|k>(|G|/2)k. Therefore if k>1 then k≥2…
-
Let G be a finite non-abelian group. Show that |Z(G)| ≤ |G|/4
If by contradiction |Z(G)|>(1/4)|G| then |G/Z(G)|<4 so G/Z(G) is cyclic, so G is abelian, absurd. ■
-
Is there an infinite group with a finite number of subgroups?
Suppose that G is an infinite group with a finite number of subgroups. We have two possibilities, G contains an infinite cyclic subgroup or it does not. If we are in the first case then this subgroup, being isomorphic to ℤ, has in turn an infinite number of subgroups, absurd. Then we must be in…
-
Let G be a finite group such that Aut(G)={e}. Prove that G is isomorphic to ℤ₂
Since Aut(G)={e} we would have that G is abelian (because we would have G/Z(G) ={e} and so G=Z(G) ) then the inversion x|—>x⁻¹ is an automorphism of G. If G has an element of order greater then 2 then the inversion is not trivial so G is a 2-elementary abelian groups and so G is…
-
Let G be a finite group. Is there always a finite group H such that G≅Aut(H)?
This is false in general. Indeed we will prove that if m is an odd positive integer then doesn’t exist a finite group H such that ℤm≅Aut(H). Indeed if such a group exists then Aut(H) is cyclic and not trivial, by the last post at link https://marcodamele.blog/2024/09/05/let-g-be-a-group-such-that-autg-is-cyclic-and-not-trivial-then-autg-is-finite-and-his-order-is-even/ we would have that Aut(H) is even which…
-
Solve in ℤ₅ the equation x⁷ + x³ – x +1= 0.
If x ∈ ℤ₅ such that x⁷+x³-x+1=0 then x(x⁶+x²-1)=-1 i.e. x(2x²-1)=4. Now (when I write x=m i mean the class of m modulo 5) x=0 —> x(2x²-1)=0≠4x=1 —> x(2x²-1)=1≠4x=2 —> x(2x²-1)=14=4x=3 —> x(2x²-1)=1≠4x=4 —> x(2x²-1)=124=4 so the solutions are the class of 2 and the class of 4 modulo 5.
-
Let G be a group such that Aut(G) is cyclic and not trivial. Then Aut(G) is finite and his order is even.
Since Aut(G) is cyclic then Inn(G) is cyclic so G/Z(G) is cyclic which implies G being abelian. Let us first show that Aut(G) has an element of order 2. Consider the map: f: x ∈ G |—> x⁻¹ ∈ G we have (G being abelian) f(xy)=(xy)⁻¹=y⁻¹x⁻¹=x⁻¹y⁻¹=f(x)f(y) so f is a group morphism. Now f is…
-
Let f ∈ ℚ[X] irreducible, then f doesn’t have multiple roots in ℂ
To prove the assertion suppose f=a₀+a₁X+…+anXⁿ and let α be a multiple root of f. Consider g=an⁻¹f . Then g is irreducible, monic and has α has a multiple root (which make, by the way, g the minimal polynomial over ℚ of α) . Then α has to be a root of g’ (the derivative…
-
There is no non-constant f ∈ ℤ[X]: f(x) is prime for every integer x
Suppose by contradiction that there exists a non-constant polynomial f with coefficients in ℤ such that f(x) is prime for every integer x. Then there exists a polynomial g ∈ ℤ[X] such that f(x)=xg(x)+p for every integer x where p is prime. Then we have for every integer x that: f(xp)=xpg(xp)+p=p(xg(xp)+1) it follows that for…
Marco Damele 