# Math Challenge - August 2019

• Challenge
• Featured
member 587159
This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets $S_n\in\Sigma, n\in\mathbb N$. Then
$$\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x$$
is injective, because of disjointedness.

Now we are going somewhere. Or course, the question is why such a sequence exists. How can we construct one?

member 587159
This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets $S_n\in\Sigma, n\in\mathbb N$. Then
$$\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x$$
is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct $\emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N$ and put $S_n := \bigcup _{k=1}^n A_k\in\Sigma$. Then $T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N,$ is a sequence of pairwise disjoint non-empty subsets.

Again, it can happen that ##T_i=\emptyset## for some i. See my last counterexample.

Again, it can happen that ##T_i=\emptyset## for some i. See my last counterexample.
You're right, we need a change of gears .. Call a subset $S\in \Sigma$ to have property $(P)$ iff the sub sigma algebra $\Sigma _S := \{S\cap A \mid A\in\Sigma\}$ is infinite. Note that
$$\Sigma = \sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ).\tag{E}$$
On the one hand we have by definition
$$\Sigma _S \cup \Sigma _{S^c} \subseteq \Sigma \Rightarrow\sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ) \subseteq \sigma (\Sigma) = \Sigma.$$
Conversely, take $A\in \Sigma$, then
$$A = A\cap (S\cup S^c) = A\cap S \cup A\cap S^c \in \sigma \left ( \Sigma _S \cup \Sigma _{S^c}\right ).$$
Suffices to show the following: if $S$ has property $(P)$, then there exists a partition $S = T\dot{\cup}T'$ such that $T,T'\in\Sigma _S$ are non-empty and at least one of them has property $(P)$. Since $X$ has property $(P)$ by assumption, we can repeatedly apply this fact and obtain a strictly decreasing sequence of non-empty proper subsets.

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

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For the sake of convenience/not cluttering denote

$$\frac{d^n}{dz^n} f(z) =: f_n(z) =: f_n \quad\mbox{and}\quad f^n(z) := (f(z)) ^n =: f^n.$$
Some preliminaries. By the chain and product rules

\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*}

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

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

Now it's pretty straightforward:

\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*}
Post-reduction:
\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*}
Re-arrange:
\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*}
The Schwarzian is said to have negative derivative if $f'(z) \neq 0$ implies $S_f (z) <0$, thus if both Schwarzians are negative, then by definition the sum is a negative Schwarzian.

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Suppose $G$ is of order $2m$, where $m>1$ is odd.

Consider the map
$$\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.$$
the map $\varphi (g)$ is injective, because $gx = gy$ implies $x=y$ and for a fixed $h\in G$ we have $\varphi (g) (g^{-1}h) = gg^{-1}h = h$. Take $g,h\in G$, then
$$\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)$$
By Sylow's first theorem there exists $a \in G$ of order two. Note that $(g,ga), g\in G,$ are cycles in the permutation $\varphi (a)$, because
$$\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.$$
Due to injectivity, the cycles are disjoint, thus there are precisely $m$ such cycles, which implies $\varphi (a)$ is an odd permutation. Consider the signum morphism $\varepsilon : \mbox{Sym}(G) \to \{-1,1\}$ where $\varepsilon (\sigma ) = 1$ iff $\sigma$ is an even permutation. We have $\varepsilon \varphi : G\to \{-1,1\}$ a surjective morphism, because $e\mapsto 1$ ($\varphi (e)$ has no inversions). By first isomorphism theorem
$$G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}$$
which implies $|\mbox{Ker} (\varepsilon\varphi)| = m$. Also $\mbox{Ker} (\varepsilon\varphi) \triangleleft G$.

summer's almost over .. Last edited:
• member 587159
member 587159
Suppose GG is of order 2m2m, where m>1m>1 is odd.

Consider the map
$$\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.$$
the 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
$$\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)$$
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), because
$$\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.$$
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
$$G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}$$
which implies |Ker(εφ)|=m|Ker(εφ)|=m. Also Ker(εφ)◃GKer(εφ)◃G.

summer's almost over .. Correct! Well done! Could you maybe give some insight in why you thought about that action?

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

Questions
9.
Three 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?

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 :)

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

I am unable to draw here the diagram I sketched on paper to derive the solution, but the fundamental observation is that if all the 3 planes fly at the same speed, then at any point in time till they meet, the positions of the 3 planes will continue to form an equilateral triangle and this triangle keeps shrinking in size as time progresses.

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

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

• archaic
I think P6 generalises via induction to the case $2^rm$.
Correct! Well done! Could you maybe give some insight in why you thought about that action?
I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some $\varphi : G\to\mbox{Sym}(G)$ such that I would get $\varepsilon \varphi : G\to \{-1,1\}$ surjective and apply the first isomorphism theorem. Of course, getting a preimage for $1$ is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in $\mbox{Sym}(G)$ which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order $2m$ gave a clue. A way to get such cycles is to put $(g,ga)$ where $a$ is of order two and at that point the definition of $\varphi$ became self-evident.

• member 587159
member 587159
I think P6 generalises via induction to the case $2^rm$.

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

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)

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##.

• nuuskur
Mentor
2021 Award
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 :)

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

I am unable to draw here the diagram I sketched on paper to derive the solution, but the fundamental observation is that if all the 3 planes fly at the same speed, then at any point in time till they meet, the positions of the 3 planes will continue to form an equilateral triangle and this triangle keeps shrinking in size as time progresses.

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

$$-L / (- \frac {3} {2} v) = \frac {2L} {3v}$$
This seems to be correct and is similar to what @nuuskur did

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}.##

You're right, we need a change of gears .. Call a subset $S\in \Sigma$ to have property $(P)$ iff the sub sigma algebra $\Sigma _S := \{S\cap A \mid A\in\Sigma\}$ is infinite. Note that
$$\Sigma = \sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ).\tag{E}$$
On the one hand we have by definition
$$\Sigma _S \cup \Sigma _{S^c} \subseteq \Sigma \Rightarrow\sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ) \subseteq \sigma (\Sigma) = \Sigma.$$
Conversely, take $A\in \Sigma$, then
$$A = A\cap (S\cup S^c) = A\cap S \cup A\cap S^c \in \sigma \left ( \Sigma _S \cup \Sigma _{S^c}\right ).$$
Suffices to show the following: if $S$ has property $(P)$, then there exists a partition $S = T\dot{\cup}T'$ such that $T,T'\in\Sigma _S$ are non-empty and at least one of them has property $(P)$. Since $X$ has property $(P)$ by assumption, we can repeatedly apply this fact and obtain a strictly decreasing sequence of non-empty proper subsets.

Proof of fact. Suppose $S\in\Sigma$ has property $(P)$. Then $\Sigma _S$ is an infinite sub sigma algebra. Pick $T \in \Sigma _S \setminus \{\emptyset, S\}$ Write $S= T\cup (S\cap T^c)$. By (E) at least one of the respective sub sigma algebras must be infinite.
Ok, now that we know..
Pick a strictly decreasing sequence $X \supset S_1 \supset S_2 \supset \ldots$ from the sigma algebra. Then put $A_n := S_n\setminus S_{n+1}, n\in\mathbb N$, which is a sequence of pairwise disjoint elements and
$$\mathcal P(\mathbb N) \setminus \{ \emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} A_x,$$
is injective, thus an infinite sigma algebra must be at least of the cardinality of the continuum.

member 587159
Ok, now that we know..
Pick a strictly decreasing sequence $X \supset S_1 \supset S_2 \supset \ldots$ from the sigma algebra. Then put $A_n := S_n\setminus S_{n+1}, n\in\mathbb N$, which is a sequence of pairwise disjoint elements and
$$\mathcal P(\mathbb N) \setminus \{ \emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} A_x,$$
is injective, thus an infinite sigma algebra must be at least of the cardinality of the continuum.
Will give feedback tomorrow. Currently busy with abstract algebra and measure theory needs a change of mindset :P.

• nuuskur and fresh_42
The fundamental theorem of abelian groups says that for any finite abelian group

$$A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}$$
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 $\mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3$.

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

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

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

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

Our non-abelian group $G$ therefore contains three Sylow $2-$subgroups, call them $H_1,H_2,H_3$, which by theorem 2 are all conjugate to each other. We have $\mbox{Sym}(H_1,H_2,H_3) \cong S_3$. We show that $G\cong S_3$. Define
$$\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}$$
(also called conjugating)
Firstly, if $H$ is a Sylow $p-$subgroup, then conjugating it gives another Sylow $p-$subgroup, hence it must hold that $gH_ig^{-1} = H_j$ for some $j$.

The map $\varphi (g)$ is injective, because if $gH_ig^{-1} = gH_jg^{-1}$, then
$$h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.$$
The argument is symmetrical, thus $H_i=H_j$. It is surjective due to finiteness.

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

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

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DEvens
Gold Member
Regarding question 5, a clarification question.

Would constructing all possible character tables be an acceptable answer? I can get my group theory text and swat this up, but it would take me a couple hours to refresh that stuff since I have not looked at it in some time.

member 587159
Regarding question 5, a clarification question.

Would constructing all possible character tables be an acceptable answer? I can get my group theory text and swat this up, but it would take me a couple hours to refresh that stuff since I have not looked at it in some time.

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?

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

$$A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}$$
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 $\mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3$.

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

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

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

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

Our non-abelian group $G$ therefore contains three Sylow $2-$subgroups, call them $H_1,H_2,H_3$, which by theorem 2 are all conjugate to each other. We have $\mbox{Sym}(H_1,H_2,H_3) \cong S_3$. We show that $G\cong S_3$. Define
$$\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}$$
(also called conjugating)
Firstly, if $H$ is a Sylow $p-$subgroup, then conjugating it gives another Sylow $p-$subgroup, hence it must hold that $gH_ig^{-1} = H_j$ for some $j$.

The map $\varphi (g)$ is injective, because if $gH_ig^{-1} = gH_jg^{-1}$, then
$$h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.$$
The argument is symmetrical, thus $H_i=H_j$. It is surjective due to finiteness.

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

Therefore, all $N(H_j) = H_j$ and $g\in H_1\cap H_2\cap H_3 = \{e\}$ i.e $g=e$, which makes $\varphi$ an isomorphism.
Typo
By theorem 3 $n_q \equiv 1 \pmod{q}$ and by theorem 3 $n_q\mid p$
As a continuation to #73.
In the abelian case there are
$$\mathbb Z_8 \qquad \mathbb Z_4 \oplus \mathbb Z _2 \qquad \mathbb Z_2 \oplus\mathbb Z_2\oplus\mathbb Z_2$$
Suppose $G$ is non-abelian of order $8$. If there is an element of order 8, then $G$ is cyclic, thus abelian. Suppose the maximal order is $2$ and pick $g,h\in G$. If $gh=e$, then they commute. Suppose $(gh)^2 = e$, then
$$gh = geh = g(ghgh)h = (gg)hg(hh) = ehge = hg$$
and again $G$ would be abelian. Thus we must have an element $a$ of order $4$. Its generated subgroup $\langle a \rangle =: H$ is of index $2$, therefore normal. Take $b\notin H$, then we have $bab^{-1}\in H$ due to normality. Notice that
$$(bab^{-1})^4 = bab^{-1}bab^{-1}bab^{-1}bab^{-1} = baaaab^{-1} = e.$$
If also $(bab^{-1}) ^2 = e$, then $baab^{-1} = e$ would lead to $aa=e$, contradicting the order of $a$.
Next, we find the possible values of $bab^{-1}$.
1. If $bab^{-1} = a^4 = e$, then $a=e$, which is impossible.
2. If $bab^{-1} = a$, then $ba = ab$, but this will make $G$ abelian. Indeed, pick $g,h\in G$. If they're both in $H$, then they commute. Suppose $g\notin H$ and write $g = ba^m$ and $h = a^n$, then
$$gh = ba^ma^n = ba^na^m = a^n ba^m = hg.$$
and if both reside outside $H$, then writing $g = ba^m$ and $h = ba^n$ would similarly lead to $gh=hg$. But our group is not abelian.
3. If $bab^{-1} = a^2$, then $(bab^{-1})^2 = e$ would contradict the order of $bab^{-1}$.
Thus, the only possibility is $bab^{-1}=a^3 =a^{-1}$.

There are two cases to consider.
1. $b$ is of order $2$, then we have $G \cong \langle a,b \mid \mbox{ord}(a) = 4, \mbox{ord}(b) = 2, bab^{-1} = a^{-1} \rangle$ this is the dihedral group $D_8$.
2. $b$ is of order $4$. We show $G$ is the dicyclic group $\mbox{Dic}_2$, where
$$\mbox{Dic}_2 \cong \langle x,y \mid x^{4} = e, x^2 = y^2, x^{-1} = y^{-1}xy \rangle.$$
Notice that
$$bab^{-1} = a^{-1} \Rightarrow b^{-1}ab = b^{-1}(ba^{-1}b^{-1})b = a^{-1}.$$
So we have left to show $a^2 = b^2$. Considering the natural projection $G\to G/H\cong\mathbb Z_2$, we have $b^2 \in H$. Consider possible values of $b^2$.
1. $b^2 = a^4 = e$ contradicts order of $b$.
2. $b^2 = a$ contradicts order of $a$
3. $b^2 = a^2$ is what we want.
4. $b^2 = a^{-1}$ contradicts order of $a^{-1}$ (hence order of $a$).
Thus $b^2 = a^2$ must hold.

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etotheipi

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Yes, but what is the calculation?

etotheipi
Yes, but what is the calculation?

Let the distance outside the town be d

The initial average speed before the change of heart is thus $$v_1 = \frac {d+10} {0.25 + \frac{d}{180}},$$
In a similar vein the average speed after the driver decides to lower his speeds is
$$v_2 = \frac {d+10} {0.5 + \frac{d}{160}},$$
We are told that the difference between these two speeds is 40kmh, so we solve the following equation for d
$$\frac {d+10} {0.25 + \frac{d}{180}} = \frac {d+10} {0.5 + \frac{d}{160}} + 40$$
Cleaning up that mess gives a quadratic
$$d^2 - 120d + 3600 = 0$$
which yields a single solution of d=60. Hence the total distance is d+10 = 70.

• fresh_42
13. David drives to work every working day by car. Outside towns he drives at an average speed of ##180\,\text{km/h}##. On the
##10\,\text{km}## in town, he drives at an average speed of ##40\,\text{km/h}##. As a result, he is often too fast and gets a ticket. Meanwhile he has realized that things can not go on like this and he decides to reduce his average speed by ##20\,\text{km/h}## in town as well as outside. How long is his way to work, if this reduces his average speed by ##40\,\text{km/h}## on total?

The total distance to work is 70 km, including the 10 km inside the town (which is given).

Let ##d## be the distance traveled outside the town when David goes to work. Then his original average speed would be $$s_1 = \frac {(d + 10)} {\frac {d} {180} + \frac {10} {40}}$$. After reducing the speeds inside and outside the towns by ##20 km/h## each, the new average speed becomes $$s_2 = \frac {(d + 10)} {\frac {d} {160} + \frac {10} {20}}$$

Given that the new average speed is lower than the previous average speed by ##40km/h##, we get $$s_1 - s_2 = 40 \Rightarrow \frac {(d + 10)} {\frac {d} {180} + \frac {10} {40}} - \frac {(d + 10)} {\frac {d} {160} + \frac {10} {20}} = 40$$

The above equation reduces to ##(d - 60)^2 = 0##, giving ##d = 60##

P.S.
Thanks for giving 1 simple question that I could solve confidently . But more thanks for the tough questions for which I couldn't not solve - I am keen to learn from the solutions posted by others and from the learning resources.

• fresh_42
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The total distance to work is 70 km, including the 10 km inside the town (which is given).

Let ##d## be the distance traveled outside the town when David goes to work. Then his original average speed would be $$s_1 = \frac {(d + 10)} {\frac {d} {180} + \frac {10} {40}}$$. After reducing the speeds inside and outside the towns by ##20 km/h## each, the new average speed becomes $$s_2 = \frac {(d + 10)} {\frac {d} {160} + \frac {10} {20}}$$

Given that the new average speed is lower than the previous average speed by ##40km/h##, we get $$s_1 - s_2 = 40 \Rightarrow \frac {(d + 10)} {\frac {d} {180} + \frac {10} {40}} - \frac {(d + 10)} {\frac {d} {160} + \frac {10} {20}} = 40$$

The above equation reduces to ##(d - 60)^2 = 0##, giving ##d = 60##

P.S.
Thanks for giving 1 simple question that I could solve confidently . But more thanks for the tough questions for which I couldn't not solve - I am keen to learn from the solutions posted by others and from the learning resources.
There are only a couple of rules to attack a problem, if there is no direct, computational approach:
• simplify it
• get rid of what disturbs
• find and use symmetries
E.g. problems with rational numbers are also problems with integers, What does it say in the integer case? Integer problems are often solved by looking at the remainders modulo one or more primes: If there is an integer solution, then there is a solution with remainders, too.
##x^2+y^2=z^2 \Longrightarrow x^2+y^2\equiv z^2 \mod 2 \Longrightarrow \{\,x,y,z\,\} \text{ contains at least one even number }## because not all ##x,y,z## can be odd, since this would give ##1+1=1## for the remainders.

If there are disturbing denominators, multiply them off.

If there are symmetries, find other symmetric expressions and investigate whether there are relations.

There are only a couple of rules to attack a problem, if there is no direct, computational approach:
• simplify it
• get rid of what disturbs
• find and use symmetries
E.g. problems with rational numbers are also problems with integers, What does it say in the integer case? Integer problems are often solved by looking at the remainders modulo one or more primes: If there is an integer solution, then there is a solution with remainders, too.
##x^2+y^2=z^2 \Longrightarrow x^2+y^2\equiv z^2 \mod 2 \Longrightarrow \{\,x,y,z\,\} \text{ contains at least one even number }## because not all ##x,y,z## can be odd, since this would give ##1+1=1## for the remainders.

If there are disturbing denominators, multiply them off.

If there are symmetries, find other symmetric expressions and investigate whether there are relations.

Hi @fresh_42 , thanks for the tips. In that example, when you say ##x^2 + y^2 \equiv z^2 \mod 2## did you mean to say ##(x^2 + y^2) \mod 2 \equiv z^2 \mod 2##?

2. Find the equation of a curve such that ##y''## is always ##2## and the slope of the tangent line is ##10## at the point ##(2,6)##.

Is the answer ##y = x^2 + 6x - 10##?

##y'' = 2 \Rightarrow y' = \int y'' \, dx = \int 2 \,dx = (2x + a)##, where ##a## is some constant
##y = \int (2x + a) \, dx = (x^2 + ax + b)## where ##b## is some constant

Since slope of tangent at ##x = 2## is 10, we get ##y'_{(x=2)} = 2 \Rightarrow 2 \times 2 + a = 10 \Rightarrow a = 6##

Since ##(2, 6)## is a point on the curve, ##y_{(x=2)} = 6 \Rightarrow (x^2 + 6x + b)_{(x=2)} = 6 \Rightarrow 4 + 12 + b = 10 \Rightarrow b = -6##

1. The maximum value of ##f## with ##f(x) = x^a e^{2a - x}## is minimal for which values of positive numbers ##a## ?

Is the answer ##a = e^{-2}##?

Here is the calculation I used - ##f'(x) = ax^{a-1} e^{2a - x} - x^a e^{2a - x} = x^{a-1} e^{2a - x} (a - x)##
At the maximal point, ##f'(x) = 0 \Rightarrow x=a## or ##x=0##. I have not worked out a rigorous proof for maximality at ##a##, but it can be seen that since when ##x > a##, ##f'(x) < 0##, ##f(a) > 0## (given ##a > 0##), ##x = a## must be the global maximum.

Therefore, maximum value of ##f## is ##a^a e^a##, which we will now denote by ##g(a)##, a function of variable ##a##. Obviously, for ##a > 0##, this expression will not have a finite maximum, since as ##a \rightarrow \inf##, ##f(a) \rightarrow \inf##. Only a finite minimum would be possible and at the minimal point ##g'(a) = 0##. To simplify differentiation, we may apply logarithm on the expression to find the minimum, which is valid for ##a > 0## since logarithm is a monotonically increasing function. ##h(a) = \ln g(a) = a \ln a + a##. At the minimal point, ##h' = 0 \Rightarrow 2 + \ln a = 0 \Rightarrow a = e^{-2}##

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Hi @fresh_42 , thanks for the tips. In that example, when you say ##x^2 + y^2 \equiv z^2 \mod 2## did you mean to say ##(x^2 + y^2) \mod 2 \equiv z^2 \mod 2##?
Yes, except the correct notation is ##a \equiv b \mod 2## or ##a=b \mod 2## The module automatically covers the entire equation. It is basically a function from ##\mathbb{Z}## to ##\mathbb{Z}/\mathbb{2Z}=\mathbb{Z}_2=\{\,0,1\,\}##. So ##\mod n## means to apply this function on both sides, otherwise it wouldn't make sense.

• Not anonymous
##2x^6+3y^6=z^3##

It suffices to prove that ##2x^3+3y^3=1## has no rational roots (expression obtained by dividing by ##z^3## ).

We see that ##\frac{3y^3}{2} = \frac{1}{2} -x^3##

We see that ##x^3= \frac{1}{2} - t^3## and ##y= \frac{2t^3}{3}##

If t is rational, then y is irrational. If t is irrational, then x is irrational. Hence there are no solutions for x and y in rationals.

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##2x^6+3y^6=z^3##

It suffices to prove that ##2x^3+3y^3=1## has no rational roots (expression obtained by dividing by ##z^3## ).
Prove it!

• Pi-is-3
1. The maximum value of ##f## with ##f(x) = x^a e^{2a - x}## is minimal for which values of positive numbers ##a## ?

Is the answer ##a = e^{-2}##?

Here is the calculation I used - ##f'(x) = ax^{a-1} e^{2a - x} - x^a e^{2a - x} = x^{a-1} e^{2a - x} (a - x)##
At the maximal point, ##f'(x) = 0 \Rightarrow x=a## or ##x=0##. I have not worked out a rigorous proof for maximality at ##a##, but it can be seen that since when ##x > a##, ##f'(x) < 0##, ##f(a) > 0## (given ##a > 0##), ##x = a## must be the global maximum.

Therefore, maximum value of ##f## is ##a^a e^a##, which we will now denote by ##g(a)##, a function of variable ##a##. Obviously, for ##a > 0##, this expression will not have a finite maximum, since as ##a \rightarrow \inf##, ##f(a) \rightarrow \inf##. Only a finite minimum would be possible and at the minimal point ##g'(a) = 0##. To simplify differentiation, we may apply logarithm on the expression to find the minimum, which is valid for ##a > 0## since logarithm is a monotonically increasing function. ##h(a) = \ln g(a) = a \ln a + a##. At the minimal point, ##h' = 0 \Rightarrow 2 + \ln a = 0 \Rightarrow a = e^{-2}##

One wrong assumption, x=a is not the global maximal for all a. If a is negative or an even natural number, then it is just a critical point. But the answer is correct because it is the global maxima of ##e^{-2}##.

Prove it!

Sorry, I thought it was sufficient to leave it at that.
Let ##x^2=a, y^2=b##.
Then we have ##3a^3+2b^3=z^3##

Let ##\frac{a}{z}=j ## and ## \frac{b}{z}=k##

If x,y,z are rational, then so are j and k.

Finally, if their are no rational solutions of j and k, then their are no rational solutions for x,y,z.

This is why it suffices to prove that ##3j^3+2k^3=1## has no rational solutions.

EDIT: I made one more mistake, I didn't mention that z is not equal to 0. If z=0, then the answers come to be (0,0,0) which is excluded.

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Sorry, I thought it was sufficient to leave it at that.
No. It is a bad habit and a common place to hide mistakes.

You have to show that a solution of the general version provides a solution for the restricted one. So let us assume we have a solution ##(r,s,t)## such that ##2r^6 +3s^6=t^3##.

From ##t=0## we obtain the trivial solution only, since the left hand side is nonnegative.
For ##t\neq 0## we get ##2 \left( \dfrac{r}{t} \cdot r \right)^3\cdot 3 \left( \dfrac{s}{t} \cdot s\right)^3=1\,.##

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