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Could we show [itex]d\mid c[/itex]?

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- #51

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Could we show [itex]d\mid c[/itex]?

- #52

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In this method, ##a^2=9x^2, b^2=9y^2## .

Could we show [itex]d\mid c[/itex]?

This implies ##c^2## is a multiple of 3. Then you can continue with the same method I did. Using contradiction to prove that solutions don't exist.

However, I'm not sure how to prove that d divides c (if I'm not wrong, d is the gcd of a,b).

- #53

Math_QED

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I'm satisfied with your explanations. Well done!Sure. First I'll explain why x is even. Since l is less than equal to m and n, we can divide both the sides by ##2^{2l}## . That in any scenario gives either x is even or x and y are even or all x,y,z are even (when l=m=n) by taking mod 4 both sides. A similar reasoning is followed for z.

As to why it is okay to consider only two cases instead of three (by including the case of m being smaller), considering the case for x is the same as considering the case for y. If it hadn't been the same, I would have to consider 3 cases. I don't know how to explain this one any better. You could think that if (a,b,c) satisfy the equation, then (b,a,c) will also satisfy.

In fact, there is a shorter solution.

You got to ##a^2+b^2 +c^2\equiv 0 \bmod 4##, implying that ##2|a,b,c##. Now, if we take a minimal positive solution ##(a,b,c)## this procedure yields the solution ##(a/2,b/2,c/2)##, contradicting minimality.

I think this is in the lines of what @nuuskur was trying.

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- #54

fresh_42

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... plus ##f'(x)\neq 0## for all ##x##.I was thinking about it in a "give a counter-example" fashion. The question then is rather easy don't you think? I could put forward any function ##f## where ##f(0) \neq f(1)## for example.

If you have three conditions which are needed to draw a single conclusion, then the violation of any of the conditions together with the opposite of the conclusion is a counterexample. The best solution would have been to list three examples, violating one condition after the other. And a fourth example where the conclusion holds without any of the conditions.

- #55

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Indeed, formally it's known as method of infinite descent - stems from induction (equivalent, too, I think). By the way, I revised #33.I think this is in the lines of what @nuuskur was trying.

- #56

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[tex]

(gh)^{-1} = g^{-1}h^{-1} = (hg)^{-1} \iff gh = hg.

[/tex]

Suppose [itex]G[/itex] is abelian then for every [itex]g,h\in G[/itex]

[tex]

gh \mapsto (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1}.

[/tex]

Now, pick [itex]g\in G[/itex] and write [itex]g= h^{-1}\sigma (h)[/itex], then

[tex]

\sigma (g) = \sigma (h^{-1}\sigma (h)) = (\sigma (h))^{-1} (h^{-1})^{-1} = (h^{-1}\sigma (h))^{-1} = g^{-1}.

[/tex]

By 7a [itex]G[/itex] is abelian.

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- #57

Math_QED

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Correct! I will give feedback on the measure theory question soon! Bit busy now!

[tex]

(gh)^{-1} = g^{-1}h^{-1} = (hg)^{-1} \iff gh = hg.

[/tex]

Suppose [itex]G[/itex] is abelian then for every [itex]g,h\in G[/itex]

[tex]

gh \mapsto (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1}.

[/tex]

Now, pick [itex]g\in G[/itex] and write [itex]g= h^{-1}\sigma (h)[/itex], then

[tex]

\sigma (g) = \sigma (h^{-1}\sigma (h)) = (\sigma (h))^{-1} (h^{-1})^{-1} = (h^{-1}\sigma (h))^{-1} = g^{-1}.

[/tex]

By 7a [itex]G[/itex] is abelian.

- #58

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Could we show [itex]d\mid c[/itex]?

[tex]

a = 3 ^{k_1} \ldots p_n^{k_n} \quad b = 3^{l_1}\ldots p_n^{l_n} \quad c = 3^{r_1} \ldots p_n ^{r_n}

[/tex]

where the powers are non-negative. Then [itex](a,b) = 3 ^{m_1} \ldots p_n ^{m_n}[/itex], where [itex]m_j = \min \{k_j,l_j\}[/itex]. The goal is to show [itex]m_j\leq r_j[/itex]. The initial equality can be written as

[tex]

3^{2m_1} \ldots p_n^{2m_n} \left ( 3^{2k_1 - 2m_1} \ldots p_n^{2k_n-2m_n} + 3^{2l_1 - 2m_1} \ldots p_n^{2l_n-2m_n}\right ) = 3^{2r_1+1}\ldots p_n^{2r_n}

[/tex]

Suppose, for a contradiction, some [itex]m_j>r_j[/itex]. The case [itex]j\neq 1[/itex] would run into [itex]2m_j + k = 2r_j[/itex], where [itex]k\geq 0[/itex], which is impossible. So it must be that [itex]j=1[/itex]. But then [itex]2m_1 + k = 2r_1 + 1[/itex] for some [itex]k\geq 0[/itex]. By assumption [itex]m_1 \geq r_1+1[/itex] so

[tex]

2m_1 + k = 2r_1 + 1 \Rightarrow 2r_1 + 1 - k \geq 2r_1 + 2 \Rightarrow k\leq -1,

[/tex]

a contradiction. Thus [itex](a,b) \mid c[/itex].

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- #59

Math_QED

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What happens when we take ##X =\mathbb{N}, A_1=\{0,1\},A_2=\{0\}##?Let [itex]\Sigma[/itex] be an infinite sigma algebra on a set [itex]X[/itex], choose [itex]\emptyset\subset A_1\subset X[/itex] and pick for every [itex]n\in\mathbb N[/itex]

[tex]

A_{n+1} \in \Sigma \setminus \left ( \left\lbrace\emptyset, X\right\rbrace \cup \left\lbrace\bigcup_{k=1}^mA_k\mid m\leq n\right\rbrace\cup\left\lbrace\bigcap _{k=1}^m A_k^c\mid m\leq n\right\rbrace\right ).

[/tex]

[itex]\Sigma[/itex] is infinite, so we can do this. Put [itex]S_n := \bigcup _{k=1}^nA_k\in\Sigma,n\in\mathbb N[/itex]. Observe that all [itex]S_n\subset X[/itex].

The sequence [itex]S_n^c,n\in\mathbb N,[/itex] is strictly decreasing. Put

[tex]

T_n := S_n^c\setminus S_{n+1}^c\in \Sigma, \quad n\in\mathbb N.

[/tex]

The [itex]T_n[/itex] are pairwise disjoint. The map

[tex]

\mathcal P(\mathbb N) \to \Sigma, \quad A\mapsto \begin{cases} \emptyset, &A=\emptyset \\ \bigcup_{x\in A}T_x, &\emptyset\subset A\subset\mathbb N \\ X, &A=\mathbb N \end{cases}

[/tex]

is injective. Note that if [itex]A\subset\mathbb N[/itex] then [itex]\bigcup _{x\in A}T_x \subseteq S_1^c \subset X[/itex].

Then ##S_1=S_2 = A_1## and ##T_1=\emptyset## so again injectivity is violated.

- #60

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This problem is of the devil mhh, the following is a cheapshotWhat happens when ...

[tex]

\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x

[/tex]

is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct [itex]\emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N[/itex] and put [itex]S_n := \bigcup _{k=1}^n A_k\in\Sigma[/itex]. Then [itex]T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N,[/itex] is a sequence of pairwise disjoint non-empty subsets.

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- #61

Math_QED

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Now we are going somewhere. Or course, the question is why such a sequence exists. How can we construct one?This problem is of the devil mhh, the following is a cheapshot

Take a sequence of pairwise disjoint nonempty subsets [itex]S_n\in\Sigma, n\in\mathbb N[/itex]. Then

[tex]

\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x

[/tex]

is injective, because of disjointedness.

- #62

Math_QED

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Again, it can happen that ##T_i=\emptyset## for some i. See my last counterexample.This problem is of the devil mhh, the following is a cheapshot

[tex]

\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x

[/tex]

is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct [itex]\emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N[/itex] and put [itex]S_n := \bigcup _{k=1}^n A_k\in\Sigma[/itex]. Then [itex]T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N,[/itex] is a sequence of pairwise disjoint non-empty subsets.

- #63

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You're right, we need a change of gears ..Again, it can happen that ##T_i=\emptyset## for some i. See my last counterexample.

[tex]

\Sigma = \sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ).\tag{E}

[/tex]

On the one hand we have by definition

[tex]

\Sigma _S \cup \Sigma _{S^c} \subseteq \Sigma \Rightarrow\sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ) \subseteq \sigma (\Sigma) = \Sigma.

[/tex]

Conversely, take [itex]A\in \Sigma[/itex], then

[tex]

A = A\cap (S\cup S^c) = A\cap S \cup A\cap S^c \in \sigma \left ( \Sigma _S \cup \Sigma _{S^c}\right ).

[/tex]

Suffices to show the following: if [itex]S[/itex] has property [itex](P)[/itex], then there exists a partition [itex]S = T\dot{\cup}T'[/itex] such that [itex]T,T'\in\Sigma _S[/itex] are non-empty and at least one of them has property [itex](P)[/itex]. Since [itex]X[/itex] has property [itex](P)[/itex] by assumption, we can repeatedly apply this fact and obtain a strictly decreasing sequence of non-empty proper subsets.

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- #64

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For the sake of convenience/not cluttering denote

[tex]

\frac{d^n}{dz^n} f(z) =: f_n(z) =: f_n \quad\mbox{and}\quad f^n(z) := (f(z)) ^n =: f^n.

[/tex]

Some preliminaries. By the chain and product rules

[tex]

\begin{align*}

(gf)_1 &= g_1(f)f_1 \\

(gf)_2 &= g_2(f)f_1^2 + g_1(f)f_2 \\

(gf)_3 &= g_3(f)f_1^2 + 3g_2(f)f_1f_2 + g_1(f)f_3

\end{align*}

[/tex]

Checking ahead in wiki - the Schwarzian in general would have to be:

[tex]

S_{gf} = (S_g \circ f) \cdot f_1^2 + S_f

[/tex]

Now it's pretty straightforward:

[tex]

\begin{align*}

S_{gf} &= \frac{(gf)_3}{(gf)_1} - \frac{3}{2} \left (\frac{(gf)_2}{(gf)_1}\right )^2 \\

&= \frac{g_3(f)f_1^3 + 3g_2(f)f_1f_2 + g_1(f)f_3}{g_1(f)f_1} \\

&-\frac{3}{2} \cdot \frac{g_2^2(f)f_1^4 + 2g_1(f)g_2(f)f_1^2f_2 + g_1^2(f)f_2^2}{g_1^2(f)f_1^2} \\

\end{align*}

[/tex]

Post-reduction:

[tex]

\begin{align*}

\frac{g_3(f)}{g_1(f)}f_1^2 + \frac{f_3}{f_1} - \frac{3}{2} \left (\frac{g_2^2(f)}{g_1^2(f)}f_1^2 + \frac{f_2^2}{f_1^2}\right )

\end{align*}

[/tex]

Re-arrange:

[tex]

\begin{align*}

S_{gf} &= \left (\frac{g_3(f)}{g_1(f)} - \frac{3}{2} \left ( \frac{g_2(f)}{g_1(f)}\right )^2\right ) f_1^2 + \left (\frac{f_3}{f_1} - \frac{3}{2} \left (\frac{f_2}{f_1}\right )^2\right ) \\ &= (S_g \circ f) f_1^2 + S_f

\end{align*}

[/tex]

The Schwarzian is said to have negative derivative if [itex]f'(z) \neq 0[/itex] implies [itex]S_f (z) <0[/itex], thus if both Schwarzians are negative, then by definition the sum is a negative Schwarzian.

[tex]

\frac{d^n}{dz^n} f(z) =: f_n(z) =: f_n \quad\mbox{and}\quad f^n(z) := (f(z)) ^n =: f^n.

[/tex]

[tex]

\begin{align*}

(gf)_1 &= g_1(f)f_1 \\

(gf)_2 &= g_2(f)f_1^2 + g_1(f)f_2 \\

(gf)_3 &= g_3(f)f_1^2 + 3g_2(f)f_1f_2 + g_1(f)f_3

\end{align*}

[/tex]

Checking ahead in wiki - the Schwarzian in general would have to be:

[tex]

S_{gf} = (S_g \circ f) \cdot f_1^2 + S_f

[/tex]

Now it's pretty straightforward:

[tex]

\begin{align*}

S_{gf} &= \frac{(gf)_3}{(gf)_1} - \frac{3}{2} \left (\frac{(gf)_2}{(gf)_1}\right )^2 \\

&= \frac{g_3(f)f_1^3 + 3g_2(f)f_1f_2 + g_1(f)f_3}{g_1(f)f_1} \\

&-\frac{3}{2} \cdot \frac{g_2^2(f)f_1^4 + 2g_1(f)g_2(f)f_1^2f_2 + g_1^2(f)f_2^2}{g_1^2(f)f_1^2} \\

\end{align*}

[/tex]

Post-reduction:

[tex]

\begin{align*}

\frac{g_3(f)}{g_1(f)}f_1^2 + \frac{f_3}{f_1} - \frac{3}{2} \left (\frac{g_2^2(f)}{g_1^2(f)}f_1^2 + \frac{f_2^2}{f_1^2}\right )

\end{align*}

[/tex]

Re-arrange:

[tex]

\begin{align*}

S_{gf} &= \left (\frac{g_3(f)}{g_1(f)} - \frac{3}{2} \left ( \frac{g_2(f)}{g_1(f)}\right )^2\right ) f_1^2 + \left (\frac{f_3}{f_1} - \frac{3}{2} \left (\frac{f_2}{f_1}\right )^2\right ) \\ &= (S_g \circ f) f_1^2 + S_f

\end{align*}

[/tex]

The Schwarzian is said to have negative derivative if [itex]f'(z) \neq 0[/itex] implies [itex]S_f (z) <0[/itex], thus if both Schwarzians are negative, then by definition the sum is a negative Schwarzian.

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- #65

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Consider the map

[tex]

\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.

[/tex]

[tex]

\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)

[/tex]

[tex]

\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.

[/tex]

Due to injectivity, the cycles are disjoint, thus there are precisely [itex]m[/itex] such cycles, which implies [itex]\varphi (a)[/itex] is an odd permutation. Consider the signum morphism [itex]\varepsilon : \mbox{Sym}(G) \to \{-1,1\}[/itex] where [itex]\varepsilon (\sigma ) = 1[/itex] iff [itex]\sigma [/itex] is an even permutation. We have [itex]\varepsilon \varphi : G\to \{-1,1\}[/itex] a surjective morphism, because [itex]e\mapsto 1[/itex] ([itex]\varphi (e)[/itex] has no inversions). By first isomorphism theorem

[tex]

G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}

[/tex]

which implies [itex]|\mbox{Ker} (\varepsilon\varphi)| = m[/itex]. Also [itex]\mbox{Ker} (\varepsilon\varphi) \triangleleft G[/itex].

summer's almost over ..

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- #66

Math_QED

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Correct! Well done! Could you maybe give some insight in why you thought about that action?Suppose GG is of order 2m2m, where m>1m>1 is odd.

Consider the map

[tex]

\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.

[/tex]

By Sylow's first theorem there exists a∈Ga∈G of order two. Note that (g,ga),g∈G,(g,ga),g∈G, are cycles in the permutation φ(a)φ(a), becausethe map φ(g)φ(g) is injective, because gx=gygx=gy implies x=yx=y and for a fixed h∈Gh∈G we have φ(g)(g−1h)=gg−1h=hφ(g)(g−1h)=gg−1h=h. Take g,h∈Gg,h∈G, then

[tex]

\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)

[/tex]

[tex]

\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.

[/tex]

Due to injectivity, the cycles are disjoint, thus there are precisely mm such cycles, which implies φ(a)φ(a) is an odd permutation. Consider the signum morphism ε:Sym(G)→{−1,1}ε:Sym(G)→{−1,1} where ε(σ)=1ε(σ)=1 iff σσ is an even permutation. We have εφ:G→{−1,1}εφ:G→{−1,1} a surjective morphism, because e↦1e↦1 (φ(e)φ(e) has no inversions). By first isomorphism theorem

[tex]

G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}

[/tex]

which implies |Ker(εφ)|=m|Ker(εφ)|=m. Also Ker(εφ)◃GKer(εφ)◃G.

summer's almost over ..

- #67

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Disclaimer: This spoiler may not be a spoiler at all, as I am no sure if the answer is correct :) Too long since I last used calculus to solve anything other than the most basic and even longer since I last worked on solving a physics problem :)Summary:1. - 2. posed and moderated by @QuantumQuest

3. - 8. posed and moderated by @Math_QED

9. - 10. posed and moderated by @fresh_42

keywords: calculus, abstract algebra, measure theory, mechanics, dynamical systems

QuestionsThree identical airplanes start at the same time at the vertices of an equilateral triangle with side length ##L##. Let's say the origin of our coordinate system is the center of the triangle. The planes fly at a constant speed ##v## above ground in the direction of the clockwise next airplane. How long will it take for the planes to reach the same point, and which are the flight paths?

9.

If the solution is incorrect, please explain which assumption or step is incorrect.

Suppose the current distance between 2 planes is ##a## and ##\delta## is the small distance traveled by each of them, along the direction of edges of the current triangle, in a very small instant of time. Let ##b## denote the distance between the 2 planes after this instant. Using trigonometry, I found that the following equation determines the value of ##b##:

##b^2 = a^2 - 3 \delta (a - \delta)##

If ##v## is the constant velocity and ##\Delta t## is the small time instant, then ##\delta = v \Delta t##, yielding ##b^2 = a^2 - 3 v \Delta t (a - v\Delta t)##

If ##D^2## is used as the variable denoting the squared length of side of equilateral triangle, then we see that ##\Delta D^2 = -3 v \Delta t (D - v\Delta t)##.

$$

\lim_{\Delta t \rightarrow 0} {\frac {\Delta D^2} {\Delta t}} = -3 v D \Rightarrow

\frac {dD^2} {dt} = -3 v D \Rightarrow 2 D \frac {dD} {dt} = - 3 v D \Rightarrow \\

\frac {dD} {dt} = - \frac {3} {2} v

$$

So the cumulative change in the value of ##D## over a period of time ##T## would be

$$

\int_0^T - \frac {3} {2} v \, dt = - \frac {3} {2} v T

$$

When the planes meet, the value of D would be 0 whereas its initial value is ##L##, so the cumulative change in ##D## has to be ##-L##. Substituting this in the previous equation gives the value of ##T## to be

$$

-L / (- \frac {3} {2} v) = \frac {2L} {3v}

$$

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I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some [itex]\varphi : G\to\mbox{Sym}(G)[/itex] such that I would get [itex]\varepsilon \varphi : G\to \{-1,1\}[/itex] surjective and apply the first isomorphism theorem. Of course, getting a preimage for [itex]1[/itex] is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in [itex]\mbox{Sym}(G)[/itex] which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order [itex]2m[/itex] gave a clue. A way to get such cycles is to put [itex](g,ga)[/itex] where [itex]a[/itex] is of order two and at that point the definition of [itex]\varphi[/itex] became self-evident.Correct! Well done! Could you maybe give some insight in why you thought about that action?

- #69

Math_QED

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Yes, using actions to find kernels is something that works when all other things seem to fail. You are right btw that this exercise generalises (via induction)I think P6 generalises via induction to the case [itex]2^rm[/itex].

I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some [itex]\varphi : G\to\mbox{Sym}(G)[/itex] such that I would get [itex]\varepsilon \varphi : G\to \{-1,1\}[/itex] surjective and apply the first isomorphism theorem. Of course, getting a preimage for [itex]1[/itex] is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in [itex]\mbox{Sym}(G)[/itex] which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order [itex]2m[/itex] gave a clue. A way to get such cycles is to put [itex](g,ga)[/itex] where [itex]a[/itex] is of order two and at that point the definition of [itex]\varphi[/itex] became self-evident.

The generalisation is:

If ##G## is a group of order ##2^nm## with ##m## odd and ##G## has a cyclic Sylow 2-subgroup, then ##G## has a normal subgroup of order ##m##.

- #70

fresh_42

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This seems to be correct and is similar to what @nuuskur didDisclaimer: This spoiler may not be a spoiler at all, as I am no sure if the answer is correct :) Too long since I last used calculus to solve anything other than the most basic and even longer since I last worked on solving a physics problem :)

If the solution is incorrect, please explain which assumption or step is incorrect.

Suppose the current distance between 2 planes is ##a## and ##\delta## is the small distance traveled by each of them, along the direction of edges of the current triangle, in a very small instant of time. Let ##b## denote the distance between the 2 planes after this instant. Using trigonometry, I found that the following equation determines the value of ##b##:

##b^2 = a^2 - 3 \delta (a - \delta)##

If ##v## is the constant velocity and ##\Delta t## is the small time instant, then ##\delta = v \Delta t##, yielding ##b^2 = a^2 - 3 v \Delta t (a - v\Delta t)##

If ##D^2## is used as the variable denoting the squared length of side of equilateral triangle, then we see that ##\Delta D^2 = -3 v \Delta t (D - v\Delta t)##.

$$

\lim_{\Delta t \rightarrow 0} {\frac {\Delta D^2} {\Delta t}} = -3 v D \Rightarrow

\frac {dD^2} {dt} = -3 v D \Rightarrow 2 D \frac {dD} {dt} = - 3 v D \Rightarrow \\

\frac {dD} {dt} = - \frac {3} {2} v

$$

So the cumulative change in the value of ##D## over a period of time ##T## would be

$$

\int_0^T - \frac {3} {2} v \, dt = - \frac {3} {2} v T

$$

When the planes meet, the value of D would be 0 whereas its initial value is ##L##, so the cumulative change in ##D## has to be ##-L##. Substituting this in the previous equation gives the value of ##T## to be

$$

-L / (- \frac {3} {2} v) = \frac {2L} {3v}

$$

https://www.physicsforums.com/threads/math-challenge-august-2019.975478/page-2#post-6214607

My solution is slightly different, but of course not basically.

The side length of the triangle at ##t=0## is ##L(0)=L.## For the position ##\vec{r}(t)## of the first airplane we have ##|\vec{r}(0)|=r(0)=\dfrac{2}{3}L\cos \dfrac{\pi}{6}=\dfrac{L}{\sqrt{3}}.## The distance between the airplanes are the same at any point in time, because of the symmetry, i.e. the airplanes will always mark the vertices of an equilateral triangle with its center at the origin. Thus the angle between the velocity ##\vec{v}(t)## and the position ##\vec{r}(t)## is always

$$

\sphericalangle (\vec{v}(t),\vec{r}(t))=\psi(t)=\psi(0)=\psi =\pi - \dfrac{\pi}{6}

$$

Thus we have

\begin{align*}

\vec{v}(t)&=\dot{\vec{r}}(t)\\

\vec{r}(t))\vec{v}(t)&=\vec{r}(t)\dot{\vec{r}}(t)\\

r\cdot v \cdot \cos \psi &=\dfrac{1}{2}\dfrac{d}{dt}(\vec{r}(t)\vec{r}(t)\\

r\cdot v \cdot \cos \psi &=\dfrac{1}{2}\dfrac{dr^2}{dt}\\

r\cdot v \cdot \cos \psi &=r\dfrac{dr}{dt}\\

\dfrac{dr}{dt}&= -v \dfrac{\sqrt{3}}{2}\\

r(t)&= \dfrac{L}{\sqrt{3}}-v\dfrac{\sqrt{3}}{2}t

\end{align*}

Hence ##r(t_f)=0 ## implies ##t_f=\dfrac{2L}{3v}.##

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Ok, now that we know..You're right, we need a change of gears ..

[tex]

\Sigma = \sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ).\tag{E}

[/tex]

On the one hand we have by definition

[tex]

\Sigma _S \cup \Sigma _{S^c} \subseteq \Sigma \Rightarrow\sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ) \subseteq \sigma (\Sigma) = \Sigma.

[/tex]

Conversely, take [itex]A\in \Sigma[/itex], then

[tex]

A = A\cap (S\cup S^c) = A\cap S \cup A\cap S^c \in \sigma \left ( \Sigma _S \cup \Sigma _{S^c}\right ).

[/tex]

Suffices to show the following: if [itex]S[/itex] has property [itex](P)[/itex], then there exists a partition [itex]S = T\dot{\cup}T'[/itex] such that [itex]T,T'\in\Sigma _S[/itex] are non-empty and at least one of them has property [itex](P)[/itex]. Since [itex]X[/itex] has property [itex](P)[/itex] by assumption, we can repeatedly apply this fact and obtain a strictly decreasing sequence of non-empty proper subsets.

Proof of fact.Suppose [itex]S\in\Sigma[/itex] has property [itex](P)[/itex]. Then [itex]\Sigma _S[/itex] is an infinite sub sigma algebra. Pick [itex]T \in \Sigma _S \setminus \{\emptyset, S\}[/itex] Write [itex]S= T\cup (S\cap T^c)[/itex]. By (E) at least one of the respective sub sigma algebras must be infinite.

[tex]

\mathcal P(\mathbb N) \setminus \{ \emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} A_x,

[/tex]

is injective, thus an infinite sigma algebra must be at least of the cardinality of the continuum.

- #72

Math_QED

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Will give feedback tomorrow. Currently busy with abstract algebra and measure theory needs a change of mindset :P.Ok, now that we know..

[tex]

\mathcal P(\mathbb N) \setminus \{ \emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} A_x,

[/tex]

is injective, thus an infinite sigma algebra must be at least of the cardinality of the continuum.

- #73

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The fundamental theorem of abelian groups says that for any finite abelian group

[tex]

A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}

[/tex]

Hyperlinking breaks my raw text for some reason. I'll be using Sylow's theorems which I will refer to as theorems 1, 2 and 3 respectively. (https://en.wikipedia.org/wiki/Sylow_theorems)

In the abelian case there are [itex]\mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3[/itex].

**Quick remark**. Being a Sylow [itex]p-[/itex]subgroup is invariant with respect to conjugating, thus if [itex]H[/itex] was a unique Sylow [itex]p-[/itex]subgroup, we would have [itex]gHg^{-1} = H, g\in G,[/itex] i.e [itex]H[/itex] would be normal.

**Fact.** Suppose a non-abelian group [itex]G[/itex] has order [itex]pq[/itex] (primes) with [itex]q\equiv 1 \pmod{p}[/itex], then there are precisely [itex]q[/itex] Sylow [itex]p-[/itex]subgroups in [itex]G[/itex].

**Proof of fact. **Let [itex]n_p,n_q[/itex] be the numbers of Sylow [itex]p,q-[/itex]subgroups respectively. By assumption [itex]q>p[/itex]. By theorem 3 [itex]n_q \equiv 1 \pmod{q}[/itex] and by theorem 1 [itex]n_q\mid p[/itex], which forces [itex]n_q = 1[/itex], thus we have a (normal) subgroup of order [itex]q[/itex], which makes it cyclic, therefore abelian.

For [itex]n_p[/itex] we thus have two choices: [itex]n_p = 1,q[/itex]. If [itex]n_p = 1[/itex], then [itex]G\cong \mathbb Z_p \oplus \mathbb Z_q[/itex] would be abelian, thus it must be that [itex]n_p = q[/itex].

Our non-abelian group [itex]G[/itex] therefore contains three Sylow [itex]2-[/itex]subgroups, call them [itex]H_1,H_2,H_3[/itex], which by theorem 2 are all conjugate to each other. We have [itex]\mbox{Sym}(H_1,H_2,H_3) \cong S_3[/itex]. We show that [itex]G\cong S_3[/itex]. Define

[tex]

\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}

[/tex]

(also called conjugating)

Firstly, if [itex]H[/itex] is a Sylow [itex]p-[/itex]subgroup, then conjugating it gives another Sylow [itex]p-[/itex]subgroup, hence it must hold that [itex]gH_ig^{-1} = H_j[/itex] for some [itex]j[/itex].

The map [itex]\varphi (g)[/itex] is injective, because if [itex]gH_ig^{-1} = gH_jg^{-1}[/itex], then

[tex]

h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.

[/tex]

The argument is symmetrical, thus [itex]H_i=H_j[/itex]. It is surjective due to finiteness.

Take [itex]g,h\in G[/itex], then

[tex]

\varphi (gh) (H_j) = (gh)H_j(gh)^{-1} = g(hH_jh^{-1})g^{-1} = g (\varphi (h)(H_j)) g^{-1} = \varphi (g) (\varphi (h)(H_j)) = (\varphi (g) \varphi (h))(H_j).

[/tex]
To show it's an isomorphism, it suffices to show it's injective (because [itex]|G| = |S_3| = 6[/itex]). Suppose [itex]\varphi (g) = \mbox{id}[/itex] i.e [itex]gH_jg^{-1}=H_j[/itex], then by definition [itex]g\in N(H_j)[/itex] the normaliser of [itex]H_j[/itex]. If [itex]N(H_j) = G[/itex], then [itex]H_j[/itex] would be normal, which would then make it equal to its conjugates, contradicting the fact that there are three Sylow [itex]2-[/itex]subgroups.

Therefore, all [itex]N(H_j) = H_j[/itex] and [itex]g\in H_1\cap H_2\cap H_3 = \{e\}[/itex] i.e [itex]g=e[/itex], which makes [itex]\varphi[/itex] an isomorphism.

[tex]

A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}

[/tex]

Hyperlinking breaks my raw text for some reason. I'll be using Sylow's theorems which I will refer to as theorems 1, 2 and 3 respectively. (https://en.wikipedia.org/wiki/Sylow_theorems)

For [itex]n_p[/itex] we thus have two choices: [itex]n_p = 1,q[/itex]. If [itex]n_p = 1[/itex], then [itex]G\cong \mathbb Z_p \oplus \mathbb Z_q[/itex] would be abelian, thus it must be that [itex]n_p = q[/itex].

Our non-abelian group [itex]G[/itex] therefore contains three Sylow [itex]2-[/itex]subgroups, call them [itex]H_1,H_2,H_3[/itex], which by theorem 2 are all conjugate to each other. We have [itex]\mbox{Sym}(H_1,H_2,H_3) \cong S_3[/itex]. We show that [itex]G\cong S_3[/itex]. Define

[tex]

\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}

[/tex]

(also called conjugating)

The map [itex]\varphi (g)[/itex] is injective, because if [itex]gH_ig^{-1} = gH_jg^{-1}[/itex], then

[tex]

h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.

[/tex]

The argument is symmetrical, thus [itex]H_i=H_j[/itex]. It is surjective due to finiteness.

Take [itex]g,h\in G[/itex], then

[tex]

\varphi (gh) (H_j) = (gh)H_j(gh)^{-1} = g(hH_jh^{-1})g^{-1} = g (\varphi (h)(H_j)) g^{-1} = \varphi (g) (\varphi (h)(H_j)) = (\varphi (g) \varphi (h))(H_j).

[/tex]

Therefore, all [itex]N(H_j) = H_j[/itex] and [itex]g\in H_1\cap H_2\cap H_3 = \{e\}[/itex] i.e [itex]g=e[/itex], which makes [itex]\varphi[/itex] an isomorphism.

Last edited:

- #74

DEvens

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- #75

Math_QED

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A character table does not determine a group uniquely. If I recall correctly, ##D_8## and ##Q_8## are groups with the same character table that are not isomorphic.

Do you maybe mean Cayley table?

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