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College Mathematics

4 questions from Sylow’s theorem: Qn 4

Prove that if |G| = 8000 then G is not simple .

SOLUTION

If \( |G| = 2^3 \times 10^3 = 2^6 \times 5^3 \\ consider \ , \ n_5 = (5k+1) | 2^6 \\ n_5 = 1 , 16 \\ n_2 = (2k +1) | |G| \\ \Rightarrow n_2 = 5 \ , 25 ,\ 125 \) .

Let , H and K are two Sylow – 5- subgroups

\( |H \cap K | | |H| = 5^3 \\ \Rightarrow |H \cap K | = 1 \, 5 , \ 25 \)

If \( |H \cap K| = 1 \\ \Rightarrow |HK| = \frac {|H||K|}{|H \cap K |} = (125)^2 > |G| \)

\( \\ \Rightarrow \Leftarrow \\ \)

If \( H \cap K = 5 \ \\ \ If \ n_5 = 16 \ and \ n_5 = 16 \nequiv 1 \ (mod 11^2) \) .

So there are Sylow- p – subgroups H and K are \( H \cap K \) is of index p in both H and K . Hence normal in each .

So \( N_H(H \cap K) = H ; \ N_K(H \cap K ) = K \\ \Rightarrow | N_G(H \cap K )| > 125 [as \ H \cap K \subset N_G(H \cap K) ] \\ \Rightarrow 125| \ |N_G(H \cap K) \ and \ |N_G (H \cap K )| > 125 \) .

By the …………………………….. \( |N_G(H \cap K)| = 125 \times k \)

althogh this will not be required .

\(N_G(H \cap K ) = 125 \times 8 \rightarrow for \ this \ case \)

\( [ G: N_G(H \cap K ) = 8 \ but \ |G| \not| 8! \ \ \Rightarrow G \ is \ not \ simple \\ N_G(H \cap K) = 125 \times 2^4 \rightarrow for \ this \ case ] \\ [ G:N_G(H \cap K) = 4 \ but \ |G| \not| 4! \Rightarrow G \ is \ not \ simple \\ N_G(H \cap K) = 125 \times 2^5 \rightarrow for \ this \ case ] \\ [G:N_G(H \cap K) =2 \Rightarrow N_G(H \cap K) … G \Rightarrow G \ is \ not \ simple \\ |N_G(H \cap K) |= 125 \times 2^6 \\ =|G| \rightarrow for \ this \ case \\ \\ \\ \\ \\ H \cap K …G \\ \Rightarrow G \ is \ not \ simple \)

Categories
College Mathematics

4 questions from Sylow’s theorem: Qn 3

Prove that if |G| = 2376 then G is not simple .

SOLUTION

\( |G| = 2376 = 2^3 \times 3^3 \times 11 \)

If \( n_{11} = 12 \\ \\ Let \ , H \in Syl_{11}(G) \ then \ consider \ \ N_G (H) ; [ G : N_G(H) ] \\ n_{11} = 12 \\ \Rightarrow | N_G(H) | = \frac {2376}{12} = 198 \\ \\ \Rightarrow [ N_G(H) : C_G9H) ] \mid |Aut H | =10 \\ \Rightarrow |C_G(H) | =99 \ or \ 198 \)

So , \( 9 \mid |C_G(H) | \) in either case .

So , \( C_G(H) \) has a Sylow 3- subgroups P (say) \(\Rightarrow \) P commutes every element of H [ \( as P \leq C_G(H) \) ]

Now , consider ,

\( H \leq C_G(P) \leq N_G(P) \leq G \Rightarrow |H| = 11 \mid |N_G(P) | \) .

We again have that this P Sylow -3 -subgroup of \( C_G(H) \) is a subgroup of a Sylow -3 -subgrou Q (say) of G .

Now , \( [ Q : P ] = 3 \Rightarrow P \leq Q \\ \Rightarrow Q \leq N_G(P) \\ \Rightarrow |Q| = 27 | N_G(P) \) .

So , we have a subgroup R (say) \( N_G(P) \)

which is divisible by 27 \( \Rightarrow divisible \ by \ lcm(11 ,27) = 297 \) .

Now \( [G : R ] \leq 8

Now , observe that |G| | 8 .

\( \Rightarrow |G| | k! \ \ \forall \ k= 1(1) 8 \\ \Rightarrow G \ can’t \ be \ simple \)

Categories
College Mathematics

4 questions from Sylow’s theorem: Qn 2

I have come up with the Question no. 2 of four problems related to Sylow’s theorem with high difficulty level. Let’s take a look at the problem and understand the concept.

Let P be a Sylow p- group of a finite group G and let H be a subgroup of G containing \( N_{G}(P) \) . Prove that \( H = N_{G}(H) \).

Solution

Let \( P \in Syl_{P}(G) \ and H \leq G \ such \ that \ N_{G}(P) \subset H \)

Claim : Frattinis Argument :

If G is a finite group with normal subgroup h and if \( P \in Syl_{P}(H) \) then G is not great notation can be confirmed in \( \backslash[G ,E] \) .

\( G’ = H N_{G}(P) \)

where \( N_{G} \) is the normalizer of P in G’ .

& \( H N_{G}(P) \) means the product of group subsets .

PROOF :

By Sylow’s 2nd theorem we know that any two Sylow subgroups are conjugate to each other and here we are considering P to be a Sylow P subgroup of H .

So they are conjugate in H . ………………………………….(1)

Now , for any \(g \in G \ or \ gHg^{-1} = H \ \ [as \ H \leq G \ given] \Rightarrow gPg^{-1} \subset H \ but \ gPg^{-1 } is \ also \ a \ group . \\ \\ and |gPg^{-1}| = |P| \)

So \( gPg^{-1} \) is also a sylow – p -subgroup in H .

so \( \exists h \in H \ such \ that gPg^{-1} = hPh^{-1} \ \ [by \ (1)] \Rightarrow h^{-1}gPg^{-1}h = P \Rightarrow h^{-1}g \in N_{G} (P) \Rightarrow g \in HN_{G}(P) \\ \Rightarrow G \subset HN_{G}(P) = H \)

Now clearly \( HN_{G}(P) \subset G \) .

So , we have \( N_{\epsilon} = HN_{G}(P) = H \)

hence the proof of the claim .

Now come back to our question ,

If we consider \( G’ = N_{G}(H) \) So , \( H \leq G’ \) .

and \( P \leq H \) where P is a Sylow p-group of G then it is na Sylow p- group of G’ as well .

So by Frattinis Argument we have that.

Make sure to visit the Question no. 1 to have the your concepts clear.

Categories
College Mathematics

4 questions from Sylow’s theorem: Qn 1

Prove that if |G| = 616 then G is not simple .

SOLUTION

\( |G| = 616 = 2^3 \times 7 \times 11 \)

Consider the 11 – sylow subgroup of G .

\( n_{11} \mid |G| = 616 \ \ n_{11} = number \ of \ sylow_ {11} subgroup \\ \Rightarrow (11k + 1) \mid 56 \ \Rightarrow n \in \{ 1 \ , 56 \} \)

k = 0 , 5

\( n_{7} \ is \ the \ number \ of \ sylow_ 7 \ subgroup \\\Rightarrow n_{7} = 97k + 1)|88 \ \ k = 0 , 1 , 3 \rightarrow n_{7} \in {1 , 8 , 22}\)

\( n_{2} =( 2k + 1) | 77 \Rightarrow k = 0 , 3 , 5 , 38 \\ n_{2} \in \{ 1 , 7 , 11 , 77 \} \)

Suppose that \( n_{7} \neq 1 and \ n_{11} \neq 1 \) then for any \( H_{1} , H_{2} \in Syl_{11} (G) \\ k_1 , k_2 \in Syl_7 (G) \)

we have \( |H_1 \cap H_2 | \mid |H_1| , |H_2| \\ \Rightarrow H_1 \cap H_2 = \{ e \} \ any \ k_1 \cap k_2 = \{ e \} \ \ and |H_1 \cap k_1 | \ \mid |H_1| and |k_1 | \\ \Rightarrow |H_1 \cap k_1 | = 1 \Rightarrow H_1 \cap k_1 = \{ e \} \) \)

So we have \( 56 \times (11 – 1) = 560 \) elements of order 11 .

at last \( 8 \times (7 -1) = 48 \) elements of order 7 .

As 616 – 560 – 48 = 8

\( \Rightarrow \) G has only one \( sylow_2 \) subgroup & \( \Rightarrow \) That \( sylow_1 \) subgroup would be normal in G . \( \\ \\ \\ \\ \Rightarrow \) G is not simple .

Categories
Research Arithmetic Dynamics

Arithmetical Dynamics: Part 6

Arithmetical dynamics is the combination of dynamical systems and number theory in mathematics.

Again, we are here with the Part 6 of the Arithmetical Dynamics Series. Let’s get started….

Consider fix point of \( R(z) = z^2 – z \) .

Which is the solution of $$ R(z) = z \\ \Rightarrow z^2 – z =z \\ \Rightarrow z^2 – 2z =0 \\ \Rightarrow z(z -2) =0 $$

Now , consider the fix point of \( R^2(z) \) . \( \\ \) $$ i.e. R^2(z) = R . R(z) = R(z^2 -z)\\ \Rightarrow (z^2 -z)^2 – z^2 +z =z \\ \Rightarrow z^4 -2z^3 = 0 \\ \Rightarrow z^3(z- 2) =0 $$

So , every solution of \( R^2(z) =z \) is asolution of \( R(z) =z \) .

Here comes the question of existence of periodic point .

I. N . Baker proved that ,

Theorem:

Let P be a polynomial of degree at least 2 and suppose that P has no periodic points of period n . Then n=2 and P is to \( z \rightarrow z^2 – z \) .

Theorem:

Let \( R , \ \ (\frac {P}{Q}) \) be a rational function of degree $$ d = max \{ degree(P) , degree(Q) \} , \ where \ d \geq 2 . $$

Make sure you visit the previous part of this Arithmetical Dynamics Series.

Categories
Research Arithmetic Dynamics

Arithmetical Dynamics: Part 5

Arithmetical dynamics is the combination of dynamical systems and number theory in mathematics. The basic objective of Arithmetical dynamics is to explain the arithmetic properties with regard to underlying geometry structures.

Again, we are here with the Part 5 of the Arithmetical Dynamics Series. Let’s get started….

And suppose that R has no periodic points of period n . Then (d, n) is one of the pairs \( (2,2) ,(2,3) ,(3,2) ,(4 ,2) \) , each such pair does arise from some R in this way .

The example of such pair is $$ 1. R(z) = z +\frac {(w-1)(z^2 -1)}{2z} ; it \ has \ no \ points \ of \ period \ 3 .$$

$$ 2. If R(z) = \frac {z^3 +6}{3z^2} , then \ R \ has \ no \ points \ of \ period \ 2 \\ \\ \\ \\ R^2(z) = z \Rightarrow \frac {(\frac {z^3 +6}{3z^2})^3 + 6}{ 3 (\frac {z^3 + 6}{3z^2})^2 } =z \Rightarrow \frac {(z^3 +6)^3+(27 \times 6) z^6}{z^2 \times 3^2 \times (z^3 + 6)^2} =z $$

$$ 3. \ If \ R(z) = \frac {-z(1+ 2z^3)}{1-3z^3} \ then \ R \ no \ points \ of \ period \ 2 .$$

Make sure you visit the previous part of this Arithmetical Dynamics Series.

Categories
Research Arithmetic Dynamics

Arithmetical Dynamics: Part 0

Arithmetical dynamics is the combination of dynamical systems and number theory in mathematics.

We are here with the Part 0 of the Arithmetical Dynamics Series. Let’s get started….

Rational function \( R(z)= \frac {P(z)}{Q(z)} \) ; where P and Q are polynimials . There are some theory about fixed points .

Theorem:

Let \( \rho \) be the fixed point of the maps R and g be the Mobius map . Then \( gRg^{-1} \) has the same number of fixed points at \( g(\rho) \) as \( R \) has at \( \rho \).

Theorem :

If \( d \geq 1 \) , a rational map of degree d has previously \( d+1 \) fixed points in .

To each fixed point \( \rho \) of a rational maps R , we associate a complex number which we call the multiplier \( m(R , \rho) \) of R at \( \rho \) .

$$ m(r, \rho) = \{ R^{,}(\rho) ; \ if \ \rho \neq \propto \ and \ \frac {1}{R^{,}(\rho)} ; \ if \ \rho = \propto $$

Now, we dive into classification of fixed points and this is purely local matter , it applies to any analytic function and in particular , to the local inverse(when it exists ) of a rational map .

Make sure you visit the Introduction to Arithmetical Dynamics post of this Series.

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Research Arithmetic Dynamics

Arithmetical Dynamics: Part 4

We are here with the Part 4 of the Arithmetical Dynamics Series. Let’s get started….

Arithmetical dynamics is the combination of dynamical systems and number theory in mathematics.

\( P^m(z) = z \ and \ P^N(z)=z \ where \ m|N \Rightarrow (P^m(z) – z) | (P^N(z)-z) \)

The proof of the theorem in Part 0 :

Let , P be the polynomials satisfying the hypothesis of theorem 6.2.1 .

Let , \( K = \{ z \in C | P^N(z) =z \} \\ \)

and let \( M =\{ m \in Z : 1 \leq m \leq N , m|N \} \\ \)

then each \( z \in K \) is a fixed point of \( P^m \) for some \( m \in M \) and we let m(z) be the minimal such m . \( \\ \)

The proof depends on establishing the inequalities , $$ d^{N-1} (d-1) \leq \sum_{k} [\mu (N, z) – \mu (m(z) ,z) ] \\ \leq N(d-1) $$ ………………………………..\( (1) \) where \( \mu (n, w) \) is the no of fixed points of \( P^n \) at w .$$ (1) \Rightarrow d^{N-1} \leq N \\ therefore , N= 1 + (N-1) \leq 1 + (N – 1)(d – 1) \leq [1 + (d-1)]^{N-1} = d^{N – 1} \leq N $$

Make sure you visit the Arithmetical Dynamics Part 3 post of this Series before the Arithmetical Dynamics Part 4.

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Research Arithmetic Dynamics

Arithmetical Dynamics: Part 3

We are here with the Part 3 of the Arithmetical Dynamics Series. Let’s get started….

Arithmetical dynamics is the combination of dynamical systems and number theory in mathematics.

Theory:

Let \( \{ \zeta_1 , ……., \zeta_m \} \) be a ratinally indifferent cycle for R and let the multiplier of \( R^m \) at each point of the cycle be \( exp \frac {2 \pi i r}{q} \) where \( (r,q) =1 \) . Then \( \exists \ k \in Z \) and\( mkq \) distinct component \( F_1 , F_2 , ….. , F_{mkq} \) s.t. at each \( \zeta_j \) there are exactly \( kq \) of these component containing a petal of angle \( \frac {2 \pi}{kq} \ at \ \zeta \) .

Further R acts as a permutation J on \( F_1 , F_2 , ….. , F_{mkq} \) where J is a composition of k disjoint cycles of length mqJ a petal based at \( \zeta_j \) maps under R to a petal based at \( \zeta_{j+1} \)

Petal theorem :

As there are \( K_j \) such cycles of components for the rationally indifferent cycle \( c_j \) , we see that there are at least \( \sum_{j} \) critical points of P in \( C \) thus \( \sum k_j \leq d-1 \Rightarrow \) we can take the uppper bound to be \( N(d-1) \)

Make sure you visit the Arithmetical Dynamics Part 2 post of this Series before the Arithmetical Dynamics Part 3.

Categories
Research Arithmetic Dynamics

Arithmetical Dynamics: Part 2

We are here with the Part 2 of the Arithmetical Dynamics Series. Let’s get started….

Arithmetical dynamics is the combination of dynamical systems and number theory in mathematics.

The lower bound calculation is easy .

But for the upper bound , observe that each \( z \in K \) lies in some cycle of length m(z) and we these cycles by \( C_1 , C_2 …..,C_q \) . Further , we denote the length of the cycle by \( m_j \) , so , if \( z \in C_j \) then , \( m(z)= m_j \) $$ \sum_{j=1}^{q} \sum_{z \in C_j} [\mu(N,z) – \mu(m_j ,z)] $$ we can confine our attention xparatly .

Now , \( \mu(N,z) = \mu(m_j -, z) \) whenever \( z \in C_j \rightarrow \) rationally indifferent .

So , nonzero contribution comes from rationally different cycles , \( C_j \) .

Theorem:

  1. If m|n , then \( R^n \) has no fixed at \( \zeta_j \) .
    1. If m|n but \( m_q \not | n \) , then \( R^n \) has our fixed point at \( \zeta \) .
    2. If \( m_q |n \) then \( R^n \) has fixed point .

Make sure you visit the Arithmetical Dynamics Part 1 post of this Series before the Arithmetical Dynamics Part 2.