Math Challenge - July 2021

[fresh_42]

Summary:: Group Theory, Lie Algebras, Number Theory, Manifolds, Calculus, Topology, Differential Equations.
1. (solved by @Infrared ) Suppose that $G$ is a finite group such that for each subgroup $H$ of $G$ there exists a homomorphism $\varphi \,:\, G \longrightarrow H$ such that $\varphi(h)=h$ for all $h \in H$. Show that $G$ is a product of groups of prime order.2. (solved by @StoneTemplePython and by @julian ) Let $G$ be a finite group that operates on a set $X$. Then the number of orbits is
$$
|X/G|=\dfrac{1}{|G|}\sum_{g\in G}|X^g|
$$
where $X^g=\{\,x\in X\, : \,g.x=x\,\}$ are the fixed points in $X\,.$3. (solved by @martinbn ) Prove that there is a Lie algebra isomorphism $\mathfrak{g} \hookrightarrow \mathfrak{gl(g)}$ if $\mathfrak{g}$ is a semisimple Lie algebra. Is this also a necessary condition?4. (solved by @julian ) If $n>1$ is a square-free natural number, prove for all $k>1$
$$
\sum_{d|n} \sigma(d^{k-1})\,\varphi(d)=n^k
$$
Remark: $\varphi$ is Euler's phi-function and $\sigma(m)$ the sum of divisors of $m$.5. (solved by @martinbn ) Show that
$$
M:=\{\,x\in \mathbb{R}^3\,|\,x_1+x_2+x_3=0\, , \,x_1^2+2x_2^2+x_3^2-2x_2(x_1+x_3)=9\,\} \subseteq \mathbb{R}^3
$$
is a manifold, and determine the tangent space $T_pM$ and the normal space $N_pM$ at $p=(2,-1,-1)\in M\,.$6. (solved by @Infrared ) Two persons $P$ and $Q$ play the following game:
$P$ starts by selecting exactly one real value for $a,b,$ or $c$ in the equation
$$
x^3+ax^2+bx+c=0
$$
Then $Q$ does the same for one of the remaining coefficients before $P$ finally chooses the last value. $P$ wins if and only if the equation has three different real roots. Is there a winning strategy for one of the players?7. What are the composition factors of $\rm GL(2,\mathbb{F}_{19})\,?$8. (solved by @Infrared ) Show that any group $G$ of order $70$ has always a normal subgroup of order $5$.9. (solved by @Infrared ) Let $X,Y,Z$ be topological spaces, $X$ covering compact (not necessarily Hausdorff), and $Z$ Hausdorff. Let $g\, : \,X\longrightarrow Y$ be continuous, and $h\, : \,X \longrightarrow Z$ surjective and continuous.
Show that the following statements are equivalent:

1. $g(x)=g(x')$ for all $x,x'\in X$ with $h(x)=h(x').$
2. There is a continuous function $f\, : \,Z\longrightarrow Y$ with $g=f\circ h\,.$
3. There is a unique continuous function $f\, : \,Z\longrightarrow Y$ with $g=f\circ h\,.$

10. (solved by @The Fez ) Given $y'''=y''+y'-y\,.$ Determine a fundamental system, and solve the initial value problem $y(0)=1\, , \,y'(0)=0\, , \,y''(0)=3\,.$

High Schoolers only (until 26th)
11. (solved by @Not anonymous ) Assume we have put a Cartesian coordinate system on France and got the following positions:
Paris $(0,0)$, Lyon $(3,-8)$ and Marseille $(4,-12)$. Look up the definitions and calculate the distance between Lyon and Marseille according to

• the Euclidean metric.
• the maximum metric.
• the French railway metric.
• the Manhattan metric.
• the discrete metric.

12. (solved by @kshitij ) Consider the ellipse in the first quarter of a Cartesian coordinate system
$$
\dfrac{(x-2)^2}{4}+\left(y-1\right)^2=1
$$
and rotate it such that the coordinate axes are always tangents to the ellipse.
Which locus describes the center of the ellipse during a full rotation?

13. (solved by @kshitij and @Not anonymous ) Maximize $f(x,y,z)=4x^2y^2-(x^2+y^2-z^2)^2$ under the conditions $x+y+z=c$ and $x,y,z > 0.$14. (solved by @kshitij , shorter solution possible) Let $\triangle ABC$ be a triangle and $s$ a straight which intersects all three sides (or their prolongations), say $X \in BC\, , \,Y\in CA\, , \,Z\in AB$ are the intersection points. Prove
$$
\overline{AZ}\cdot \overline{BX}\cdot \overline{CY}=\overline{AY}\cdot \overline{BZ}\cdot \overline{CX}
$$


15. (solved by @Shreya ) Otto von Guericke, who invented the air pump, led an experiment in Berlin in 1654. Two groups of eight horses tried in vain to pull apart two bronze hemispheres between which a vacuum was created. Assume that the radius $R$ of the hemispheres is so thin that we can neglect the difference between inner and outer radius.

Show that the force required to tear apart the hemispheres is $F=\pi R^2 \cdot \Delta p$ where $\Delta p$ is the difference of air pressure within and outside the hemispheres. Next assume $R=30\,\rm cm$, the inner pressure of $0.1\,\rm bar$ and the outer pressure of $1.013\,\rm bar\,$ Which force had each group of horses to apply in order to separate the hemispheres?

P.s.: These are easy questions this month. Please solve them in detail and with an explanation.

 

[Office_Shredder]

6 is stupidly yes by Zermelo's theorem. If A doesn't have a winning strategy, then it means that for every choice of first coefficient, B can pick a coefficient such that A can't force a win. This means B can pick a coefficient to force a win.

I assume you want to know what the strategy is as part of it?

 

[fresh_42]

6 is stupidly yes by Zermelo's theorem. If A doesn't have a winning strategy, then it means that for every choice of first coefficient, B can pick a coefficient such that A can't force a win. This means B can pick a coefficient to force a win.

I assume you want to know what the strategy is as part of it?

Well, yes, and if possible with proof. And who are A and B? I also have an understanding problem. Did you say that B wins, or is that a contradiction to the premise that A loses? I am lost in translation.

 

[Office_Shredder]

Well, yes, and if possible with proof. And who are A and B? I also have an understanding problem. Did you say that B wins, or is that a contradiction to the premise that A loses? I am lost in translation.

Whoops, that was supposed to be P and Q. If P doesn't have a winning strategy, it means for every first choice P makes, Q can keep P from winning. The only way Q can do that is by picking a coefficient such that P can't win on their third turn. But that means P has to lose, so Q has to be able to force a win.

I haven't even constructed which player wins, I'm just proving one of them has a winning strategy for someone

Is there a winning strategy for one of the players?

The answer is yes, but not in a satisfying way (yet)

 

[fresh_42]

The answer is yes, but not in a satisfying way (yet)

Indeed. And Ben-Gurion does not count!

 

[The Fez]

Problem 10:

Characteristic equation becomes $$\lamda^3 - \lamda^2 - \lamda + 1 = 0$$ with roots of 1, 1, -1. So the fundamental equation is $$F(x) = C_1e^-x + C_2e^x + C_3xe^x$$ Solving for initial conditions gives you $$F(x) = -e^-x + e^x + 1.5xe^x$$

 

[The Fez]

Sorry, evidently can't spell lambda. And my LaTeX sucks. Hopefully you get it. first term should be e^-x.

 

[The Fez]

And I can't read either. I had F(0) = 0. With the actual stated problem C₁ = 1, C₂ = 0, and C₃ = 1. So $$F(x) = e^{-x} + xe^x$$

 

[kshitij]

Spoiler: Problem 12

It is given that the coordinate axes are tangent to the ellipse and we know that these axes are perpendicular to each other thus, their point of intersection i.e., the origin will lie on the director circle of the ellipse.

We know that, the radius of the director circle is $\sqrt{a^2+b^2}$, where $a,b$ are semi transverse and conjugate axes of the ellipse, which in our case is $\sqrt5$ and the center of the director circle is same as the center of the ellipse.

Now, when we rotate the ellipse, the radius of the director circle will be unaffected and the director circle will always pass through origin even after rotation as the coordinate axes remain tangent to the ellipse while rotating.

So, the situation is something like this,

We can clearly see that the locus of the centers of the director circles will also be a circle whose radius is equal to that of the director circle as its radius is the distance from origin to the center of director circles which is equal to radius of the director circles as they pass through origin.

And we know that the center of both the ellipse and the director circle is same, so the locus of center of ellipse is $$x^2+y^2=5$$

Spoiler: Problem 13

\begin{align*}
&4x^2y^2-(x^2+y^2-z^2)^2\\
&(2xy-x^2-y^2+z^2)(2xy+x^2+y^2-z^2)\\
&(z^2-(x-y)^2)((x+y)^2-z^2)\\
&(z-x+y)(z+x-y)(x+y-z)(x+y+z)\\
&c(c-2x)(c-2y)(c-2z)\\
A.M &\geq G.M\\
\frac{c+(c-2x)+(c-2y)+(c-2z)}{4}&\geq\left(c(c-2x)(c-2y)(c-2z)\right)^\frac1 4\\
\frac{4c-2c}{4}&\geq\left(c(c-2x)(c-2y)(c-2z)\right)^\frac1 4\\
\frac{c}{2}&\geq\left(c(c-2x)(c-2y)(c-2z)\right)^\frac1 4\\
\left(c(c-2x)(c-2y)(c-2z)\right)&\leq \frac{c^4}{16}\\
4x^2y^2-(x^2+y^2-z^2)^2&\leq \frac{(x+y+z)^4}{16}
\end{align*}

Spoiler: Problem 14

Using sine law in triangle CXY,
$$\begin{align}
\dfrac{\sin D}{\overline{CX}}=\dfrac{\sin (\pi-(C+D))}{\overline{CY}}\nonumber\\
\dfrac{\sin D}{\overline{CX}}=\dfrac{\sin (C+D)}{\overline{CY}}\nonumber\\
\dfrac{\overline{CY}}{\overline{CX}}=\dfrac{\sin (C+D)}{\sin D}
\end{align}$$
Using sine law in triangle AZY,
$$\begin{align}
\dfrac{\sin D}{\overline{AZ}}=\dfrac{\sin (A-D)}{\overline{AY}}\nonumber\\
\dfrac{\overline{AZ}}{\overline{AY}}=\dfrac{\sin (D)}{\sin (A-D)}
\end{align}$$
Finally on using sine law in triangle BZX,
$$\begin{align}
\dfrac{\sin (C+D)}{\overline{BZ}}=\dfrac{\sin (A-D)}{\overline{BX}}\nonumber\\
\dfrac{\overline{BX}}{\overline{BZ}}=\dfrac{\sin (A-D)}{\sin (C+D)}
\end{align}$$
On multiplying equations 1,2 and 3,
$$\begin{align*}
\dfrac{\overline{AZ}}{\overline{AY}} \cdot \dfrac{\overline{BX}}{\overline{BZ}} \cdot \dfrac{\overline{CY}}{\overline{CX}}&=\dfrac{\sin (D)}{\sin (A-Y)} \cdot \dfrac{\sin (A-D)}{\sin (C+D)} \cdot \dfrac{\sin (C+D)}{\sin D}\\
\dfrac{\overline{AZ}}{\overline{AY}} \cdot \dfrac{\overline{BX}}{\overline{BZ}} \cdot \dfrac{\overline{CY}}{\overline{CX}}&=1\\
\overline{AZ}\cdot \overline{BX}\cdot \overline{CY}&=\overline{AY}\cdot \overline{BZ}\cdot \overline{CX}
\end{align*}$$
Hence proved.

 

[fresh_42]

And I can't read either. I had F(0) = 0. With the actual stated problem C₁ = 1, C₂ = 0, and C₃ = 1. So $$F(x) = e^{-x} + xe^x$$

You could have been a bit more detailed so that we could actually read what you thought, but ok.

To demonstrate what I mean:

The characteristic polynomial $p\left(\dfrac{d}{dx}\right)(y)=0$ is given by $p(x)=x^3-x^2-x+1=(x-1)^2(x+1)$ with a double root at $c_1=1$ and a simple root at $c_2=-1\,.$ With $D=\dfrac{d}{dx}$ we verify
\begin{align*}
(D-1)(D-1)(e^x)&=(D-1)(e^x-e^x)=D(0)-0=0\\
(D-1)(D-1)(xe^x)&=(D-1)(xe^x+e^x)=(e^x+xe^x)-(xe^x+e^x)=0\\
(D+1)(e^{-x})&=-e^{-x}+e^{-x}=0
\end{align*}
so we get three linear independent solutions and a basis by the fundamental system
$$
\{\,\varphi_1=e^x\, , \,\varphi_2=xe^x\, , \,\varphi_3=e^{-x}\,\}
$$
With the initial values for $y=\sum_{k=1}^3 a_k\varphi_k(x)=a_1e^x+a_2xe^x+a_3e^{-x}$
\begin{align*}
y(0)=1 &\Longrightarrow a_1+a_3=0\\
y'(0)=0&\Longrightarrow a_1+a_2-a_3=0\\
y''(0)=3&\Longrightarrow a_1+2a_2+a_3=3
\end{align*}
which results in $(a_1,a_2,a_3)=(0,1,1)$ and $y(x)=xe^x+e^{-x}\,.$It was an easy question, and it doesn't take too long to write it down in a way that newcomers can understand, too.

 

[fresh_42]

It is given that the coordinate axes are tangent to the ellipse and we know that these axes are perpendicular to each other thus, their point of intersection i.e., the origin will lie on the director circle of the ellipse.

Which was exactly why I posted the question. Prove it! And do we get the entire circle?

We know that, the radius of the director circle is $\sqrt{a^2+b^2}$, where $a,b$ are semi transverse and conjugate axes of the ellipse, which in our case is $\sqrt5$ and the center of the director circle is same as the center of the ellipse.

Now, when we rotate the ellipse, the radius of the director circle will be unaffected and the director circle will always pass through origin even after rotation as the coordinate axes remain tangent to the ellipse while rotating.

So, the situation is something like this,
View attachment 285325
We can clearly see that the locus of the centers of the director circles will also be a circle whose radius is equal to that of the director circle as its radius is the distance from origin to the center of director circles which is equal to radius of the director circles as they pass through origin.

And we know that the center of both the ellipse and the director circle is same, so the locus of center of ellipse is $$x^2+y^2=5$$

 

[fresh_42]

\begin{align*}
&4x^2y^2-(x^2+y^2-z^2)^2\\
&(2xy-x^2-y^2+z^2)(2xy+x^2+y^2-z^2)\\
&(z^2-(x-y)^2)((x+y)^2-z^2)\\
&(z-x+y)(z+x-y)(x+y-z)(x+y+z)\\
&c(c-2x)(c-2y)(c-2z)\\
A.M &\geq G.M\\
\frac{c+(c-2x)+(c-2y)+(c-2z)}{4}&\geq\left(c(c-2x)(c-2y)(c-2z)\right)^\frac1 4\\
\frac{4c-2c}{4}&\geq\left(c(c-2x)(c-2y)(c-2z)\right)^\frac1 4\\
\frac{c}{2}&\geq\left(c(c-2x)(c-2y)(c-2z)\right)^\frac1 4\\
\left(c(c-2x)(c-2y)(c-2z)\right)&\leq \frac{c^4}{16}\\
4x^2y^2-(x^2+y^2-z^2)^2&\leq \frac{(x+y+z)^4}{16}
\end{align*}

Are there no equality signs where you live?

$4x^2y^2-(x^2+y^2-z^2)^2\leq c^4$ as well, so what?
Why is this the maximal value and not just an upper bound?
Do you know what it means geometrically and why?
When does equality hold? Does it hold?

 

[fresh_42]

Spoiler: Problem 14

View attachment 285328
Using sine law in triangle CXY,
$$\begin{align}
\dfrac{\sin D}{\overline{CX}}=\dfrac{\sin (\pi-(C+D))}{\overline{CY}}\nonumber\\
\dfrac{\sin D}{\overline{CX}}=\dfrac{\sin (C+D)}{\overline{CY}}\nonumber\\
\dfrac{\overline{CY}}{\overline{CX}}=\dfrac{\sin (C+D)}{\sin D}
\end{align}$$
Using sine law in triangle AZY,
$$\begin{align}
\dfrac{\sin D}{\overline{AZ}}=\dfrac{\sin (A-D)}{\overline{AY}}\nonumber\\
\dfrac{\overline{AZ}}{\overline{AY}}=\dfrac{\sin (D)}{\sin (A-D)}
\end{align}$$
Finally on using sine law in triangle BZX,
$$\begin{align}
\dfrac{\sin (C+D)}{\overline{BZ}}=\dfrac{\sin (A-D)}{\overline{BX}}\nonumber\\
\dfrac{\overline{BX}}{\overline{BZ}}=\dfrac{\sin (A-D)}{\sin (C+D)}
\end{align}$$
On multiplying equations 1,2 and 3,
$$\begin{align*}
\dfrac{\overline{AZ}}{\overline{AY}} \cdot \dfrac{\overline{BX}}{\overline{BZ}} \cdot \dfrac{\overline{CY}}{\overline{CX}}&=\dfrac{\sin (D)}{\sin (A-Y)} \cdot \dfrac{\sin (A-D)}{\sin (C+D)} \cdot \dfrac{\sin (C+D)}{\sin D}\\
\dfrac{\overline{AZ}}{\overline{AY}} \cdot \dfrac{\overline{BX}}{\overline{BZ}} \cdot \dfrac{\overline{CY}}{\overline{CX}}&=1\\
\overline{AZ}\cdot \overline{BX}\cdot \overline{CY}&=\overline{AY}\cdot \overline{BZ}\cdot \overline{CX}
\end{align*}$$
Hence proved.

Correct, but not the shortest possible way.

 

[StoneTemplePython]

It's interesting the difference that cosmetic changes can make. The fact that you wrote (2) with a sum over a group and divided by $\vert G\vert$ (as opposed to this named formula as being for $\vert G\vert \cdot \text{number of orbits}$) prompted me to think of a proof using Representation Theory-- and Perron-Frobenius to finish. I don't want to reinvent the wheel in terms of Rep Theory results though. Should I wait until a more standard proof is completed?

 

[fresh_42]

It's interesting the difference that cosmetic changes can make. The fact that you wrote (2) with a sum over a group and divided by $\vert G\vert$ (as opposed to this named formula as being for $\vert G\vert \cdot \text{number of orbits}$) prompted me to think of a proof using Representation Theory-- and Perron-Frobenius to finish. I don't want to reinvent the wheel in terms of Rep Theory results though. Should I wait until a more standard proof is completed?

We can have more than one solution. I mean, the Lemma has more than one name .

 

[StoneTemplePython]

I lost a little steam toward the end but hopefully didn't skip too many steps

Spoiler: Problem 2

note: in general any matrices are taken to have scalars in $\mathbb C$ and I assume $X$ is a finite set -- that's the standard setup for Burnside's Formula that I've seen though I don't think finiteness of $X$ was mentioned in the problem.

lemma:
if you have a group of permutation matrices $H$ such that
$\text{trace}\big(T\big) = 1$ with
$T:=\frac{1}{\vert H\vert}\sum_{P \in H}P$
$\implies T = \frac{1}{\vert H\vert} \mathbf {11}^T$
proof:
permutation matrices are doubly stochastic so
$T$ has $\mathbf 1$ as a left and right eigenvector but $T$ is idempotent so
$\text{rank} \big(T\big)=\text{trace} \big(T\big)=1\implies T=\frac{1}{\vert H\vert} \mathbf {11}^T$

easy (\base) case:
suppose $1=\dfrac{1}{|G|}\sum_{g\in G}|X^g|$
(it is probably prudent, though not necessary, to first interpret this in terms of the special case of the regular representation i.e. where $X=G$)
then collect the elements of $X$ in $\mathbf B$. Now for $g\in G$ we have

$g\mathbf B = \mathbf B P_g$
this defines a permutation (matrix) representation i.e. $P_g = \phi(g)$ for some surjective homomorphism $\phi : G\longrightarrow G' \subset GL_{\vert X\vert}(\mathbb C)$
edit note: changed $\in$ to $\subset$

So

$1=\dfrac{1}{|G|}\sum_{g\in G}|X^g|=\dfrac{1}{|G'|}\cdot\Big(\dfrac{1}{ \vert \ker \phi\vert }\sum_{g\in G}|X^g|\Big)=\dfrac{1}{|G'|}\cdot\Big(\dfrac{1}{ \vert \ker \phi\vert}\sum_{g\in G}\text{trace}\big(P_g\big)\Big)$

$=\dfrac{1}{|G'|}\cdot\Big(\dfrac{1}{ \vert \ker \phi\vert}\cdot\vert \ker \phi\vert\sum_{P_g\in G'}\text{trace}\big(P_g\big)\Big)=\text{trace}\big(\dfrac{1}{|G'|}\sum_{P_g\in G'}P_g\big)$
and application of the *lemma* tells us that $G$ acts transitively on $X$

general case:
$m=\dfrac{1}{|G|}\sum_{g\in G}|X^g|$
then by the same manipulations as in the easy case we know
$m=\dfrac{1}{|G'|}\sum_{P_g\in G'}\text{trace}\big(P_g\big) = \langle \chi_{\phi}, \chi_1\rangle$

where $\langle , \rangle$ denotes the standard (averaged) hermitian inner product used in complex representations for finite groups, and $\rho_1$ denotes the trivial representation that maps everything to the identity (with character $\chi_1$).

This implies that $G'$ is a direct sum of irreducibles where we know that with respect to the trivial representation, there are exactly $m$ copies. Thus the matrices in $G'$ have $m$ linearly independent eigenvectors in common with eigenvalue 1, the Perron root. This implies there are $m$ distinct communicating $G'$-left-invariant 'subspaces' in $\mathbf B$, i.e. if we had chosen our ordering from $X$ wisely in creating $\mathbf B$, then each $P_g$ would be block diagonal with $m$ blocks, where each block $m_i$ is the same size across matrices. (Perhaps should be tightened up in terms of wording and proof though the result easily follows from averaging and applying Perron-Frobenius to the averaged matrix.)

Note: the term $G'$-left-invariant 'subspaces' in $\mathbf B$, is terminology I made up. It should be clear that what this is, i.e. a collection of 1 or more orbits. To confirm each one of these $G'$-left-invariant 'subspaces' is in fact a single orbit, just recurse on $X'$ with the $k$ elements in $X$ that are in this subspace, and application of the 'easy case' proves that action of $G$ is transitive on $X'$ hence this is a single orbit.

 

[fresh_42]

I lost a little steam toward the end but hopefully didn't skip too many steps

Too many for me, far too many. What is $\mathbf{11}$, what $\mathbf{11}^T$ or $\mathbf{B}\,\rm ?$ Why is $\mathbf{T}$ idempotent? Why should I collect the elements of $X$ in $\mathbf{B}$ if they are already collected in $X\,\rm ?$ Does $\mathbf{T}$ even exist, except for $H=\left\{\begin{pmatrix} 1 \end{pmatrix}\right\}\subseteq \mathbb{M}_{1,1}(\mathbb{F}) \,\rm ?$ Shouldn't the number of permutated elements play a role? What is $P_g\,\rm ?$

Sorry, but I am completely lost.

 

[Office_Shredder]

Stone, once you have the base case doesn't the rest fall out by simply observing that the sum over all the fixed points can be broken up into the sum over all the fixed points in orbit 1, plus the sum of all the fixed points in orbit 2, etc, which since each sum yields 1 adds up to the number of orbits.

I guess you didn't prove that every transitive action yields a 1 on the left side (you proved a 1 yields a transitive orbit) but it feels like there must be an easier way to do it like that

Edit:
For 3, did you really mean a lie algebra isomorphism? I thought the general linear algebra was not semismple. Did you just mean an embedding?

 

[StoneTemplePython]

Too many for me, far too many. What is $\mathbf{11}$, what $\mathbf{11}^T$ or $\mathbf{B}\,\rm ?$ Why is $\mathbf{T}$ idempotent? Why should I collect the elements of $X$ in $\mathbf{B}$ if they are already collected in $X\,\rm ?$ Does $\mathbf{T}$ even exist, except for $H=\left\{\begin{pmatrix} 1 \end{pmatrix}\right\}\subseteq \mathbb{M}_{1,1}(\mathbb{F}) \,\rm ?$ Shouldn't the number of permutated elements play a role? What is $P_g\,\rm ?$

Sorry, but I am completely lost.

so $\mathbf 1$ is the ones vector and $\mathbf {11}^T$ is the ones matrix -- I use this notation all the time on this site and $\mathbf v \mathbf y^T$ is completely standard matrix vector notation. My approach to the lemma fairly closely follows that of Artin's Algebra's chapter on Representation Theory -- though after re-examining the text, not quite as closely as I thought. With some notational changes I am using $\mathbf B$ to mimic the regular representation development in that chapter. In particular $\mathbf B$ is ordered whereas a set is not. So pick an arbitrary labelling and if X has cardinality n, then $\mathbf B =\bigg[\begin{array}{c|c|c} x_1 & \cdots & x_{n-1} & x_{n}\end{array}\bigg]\implies g\mathbf B =\bigg[\begin{array}{c|c|c} gx_1 & \cdots & gx_{n-1} & gx_{n}\end{array}\bigg] = \mathbf B P_g$ where $P_g$ is a permutation matrix.

You’re asking what $P_g$ is but I explicitly addressed this in my post when I wrote

$g\mathbf B = \mathbf B P_g$
this defines a permutation (matrix) representation i.e. $P_g = \phi(g)$ for some surjective homomorphism $\phi : G\longrightarrow G' \in GL_{\vert X\vert}(\mathbb C)$

That said I definitely could say this differently by getting rid of $\mathbf B$ and just assert X to be an ordered set. But in any case the point is $P_g$ is the standard permutation matrix representation for g in G implied by the way g acts on the elements in our set X.

- - - -
There's a miscellaneous problem in either edition of that book's rep theory chapter that says "Let G be a finite subgroup of $GL_n(\mathbb C)$. Prove that if $\sum_g \text{trace} g = 0$ then $\sum_g g = \mathbf 0$. There's basically two proofs of this, one (after dividing by $\vert G\vert$) using idempotence or two be clever and back into Schur’s Lemma. What I did in the lemma is implied by that problem and I went the idempotence route.

Spoiler

$T^2=\Big(\frac{1}{\vert H\vert}\sum_{P\in H}P\Big)\Big(\frac{1}{\vert H\vert}\sum_{P'\in H}P'\Big)=\frac{1}{\vert H\vert^2}\sum_{P\in H}\Big(P\sum_{P'\in H}P'\Big)=\frac{1}{\vert H\vert^2}\sum_{P\in H}\vert H\vert \cdot T=\frac{1}{\vert H\vert^2}\cdot \vert H\vert^2 \cdot T=T$

missing detail:
$P\sum_{P'\in H}P'=\sum_{P'\in H}PP'=\sum_{P''\in H}P''=\sum_{P\in H}P=\vert H\vert \cdot T$

The does $T$ even exist question is strange. Of course it exists, you can always add square matrices of the same size (and we sum over all permutation matrices P in H) and since this is over a field of characteristic zero ($\mathbb C$) you can always divide by a positive integer (the cardinality of H).

I do remember how it can be quite exhausting to review answers to these challenges. I had guessed the sort of things in my post would be automatic for you though it seems not to be the case. Probably a good idea to put a pin in this.

 

[StoneTemplePython]

Stone, once you have the base case doesn't the rest fall out by simply observing that the sum over all the fixed points can be broken up into the sum over all the fixed points in orbit 1, plus the sum of all the fixed points in orbit 2, etc, which since each sum yields 1 adds up to the number of orbits.

I guess you didn't prove that every transitive action yields a 1 on the left side (you proved a 1 yields a transitive orbit) but it feels like there must be an easier way to do it like that

Edit:
For 3, did you really mean a lie algebra isomorphism? I thought the general linear algebra was not semismple. Did you just mean an embedding?

I'm not sure about your edit and e.g. what 3 refers to. The idea of my proof is stunningly simple but gets bogged down in details. The core idea is:
1.) look at the action of $G$ on the set $X$ and (after some ordering of $X$) this implies a standard permutation matrix representation $P_g$ for each $g \in G$. If we choose the ordering of $X$ nicely, then the orbits will be obvious -- the $P_g$'s will all be block diagonal in a nice way where each block is a permutation matrix.
2.) the identity after considering the permutation representation, we can interpret the sum via character theory and see $m=\dfrac{1}{|G|}\sum_{g\in G}|X^g|= \langle \chi_{\phi}, \chi_1\rangle$
which gives us the number of copies of Perron roots/vectors that are common to all $P_g$.
3.) In fact Perron-Frobenius theory tells us there are $m$ orthonormal real non-negative eigenvectors common for all $P_g\in G'$ but this completely determines the orbit structure.

More simply: the sum $m=\dfrac{1}{|G|}\sum_{g\in G}|X^g|$ counts common orthonormal Perron vectors associated with the permutation (matrix) representation of the group and hence counts orbits. I find this to be an extremely nice interpretation of Burnside's Formula.

your comment
"I guess you didn't prove that every transitive action yields a 1 on the left side"
I'm not certain what this means but I have a couple hunches and if either are right, it should be a 2-3 line proof. If you want to clarify here... (I may be too tired but I kind of think I see where you're going with this and I think your first paragraph is right -- it could dramatically streamline things.)

 

[Infrared]

I'll post a few:

1) Note that if $H$ is a proper and nontrivial subgroup of $G$, then by counting, $\ker(\varphi)$ is a proper, nontrivial, and normal subgroup of $G.$ Hence, $G$ can only be simple if it has no proper, nontrivial subgroups, meaning it must be cyclic of prime order.

Having dealt with the case that $G$ is simple, suppose now that $N$ is a normal, proper, and nontrivial subgroup of $G$. Consider the map $f:G\to N\times G/N$ given by $f(g)=(\varphi(g),gN).$ This is clearly a group homomorphism and its kernel is trivial since $gN$ is only trivial when $g\in N$, in which case $\varphi(g)=1$ implies $g=1.$ So it is an isomorphism (since domain and codomain have the same finite cardinality). By downward induction, we just need to check that $N$ and $G/N$ have the same defining property that $G$ does, as the previous paragraph is the base case.

If $K$ is a subgroup of $H$, then it is also a subgroup of $G$, so we have a map $\varphi:G\to K$ as in the problem statement, and this restricts to the desired map $H\to K$. Next, subgroups of $G/N$ are of the form $G/M$ for $M$ a normal group of $G$ containing $N.$ We have the map $\varphi:G\to N$, which restricts to $\varphi:M\to N$, which induces the desired map $G/M\to G/N.$

2) I know a proof so I won't spoil it, but it's a clever application of the Orbit-Stabilizer theorem and no other machinery needed :)

6. I think $P$ can win. A winning first move is to choose $a=0.$ If $Q$ then chooses a value for $b$, then $P$ considers the cubic $x^3+bx$ and selects any value $c$ such that the graph of $x^3+bx$ is not tangent to the horizontal $y=-c.$ On the other hand, if $Q$ chooses a value for $c$, then consider separately $c\neq 0$ and $c=0$. If $c=0$, then $P$ may choose $b=1$ since $x^3+x$ does not have a multiple root. If $c\neq 0$, then $P$ may choose $b=0$ since $x^3+c$ only has a multiple root when $c=0.$ [Edit: As pointed out later, I badly misread the problem. I give an actual solution later]

8. Let $n$ be the number of $5$-Sylow subgroups. By the Sylow theorems, $n$ must divide $70/5=14$, so is either $1,2,7$ or $14.$ Another Sylow theorem says that $n$ is $1$ mod $5$, leaving only the possibility $n=1$. If $H\subset G$ is a subgroup of order $5,$ then so is $gHg^{-1}$ and therefore $H=gHg^{-1}.$

9. Do you mean for $Y$ to be hausdorff instead of $Z$? If so, I think I have a solution.
1 implies 2: Assuming condition (1), given $z\in Z$, by surjectivity, we may choose $x\in X$ satisfying $h(x)=z$. Define $f(z)$ to be $g(x)$. This is well-defined because if $x,x'\in X$ both satisfy $h(x)=h(x')=z$, then $g(x)=g(x').$ It is clear that $g=f\circ h$. Finally we check continuity: let $F\subset Z$ be closed. Then $f^{-1}(F)=g(h^{-1}(F)).$ Since $h$ is continuous, $h^{-1}(F)$ is closed and hence compact (as $X$ is compact). Then $g(h^{-1}(F))$ is compact and hence closed, since compact subsets of Hausdorff spaces are closed.

2 implies 3: Suppose $g=f\circ h=f'\circ h$. Then $f=f'$ on the image of $h$, which is all of $Z$, so $f=f'.$

3 implies 1: Given the equation $g=f\circ h$, it is clear that $h(x)=h(x')$ implies $g(x)=g(x').$

 

[kshitij]

Which was exactly why I posted the question. Prove it! And do we get the entire circle?

Do you want me to derive the equation of director circle?

Let $y=mx+c$ be a tangent to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, so on solving these two we get,
\begin{align*}
b^2x^2+a^2(mx+c)^2&=a^2b^2\\
(b^2+a^2m^2)x^2+(2a^2mc)x+a^2(c^2-b^2)&=0
\end{align*}
For $y=mx+c$ to be a tangent to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, discriminant of the above quadratic equation must be zero (to ensure that the line cuts the ellipse at only one point),
\begin{align*}
4a^4m^2c^2-4a^2(b^2+a^2m^2)(c^2-b^2)&=0\\
4a^2(a^2m^2c^2-b^2c^2+b^4-a^2m^2c^2+a^2b^2m^2)&=0\\
4a^2b^2(-c^2+b^2+a^2m^2)&=0\\
c^2=a^2m^2+b^2\\
c=\pm \sqrt{a^2m^2+b^2}
\end{align*}
So, the equation of tangent to the ellipse is $$y=mx \pm \sqrt{a^2m^2+b^2}$$
Let the locus of the director circle is $P(h,k)$

Now, as the tangent passes through $P(h,k)$, we can write,
\begin{align*}
&y=mx \pm \sqrt{a^2m^2+b^2}\\
&k=mh \pm \sqrt{a^2m^2+b^2}\\
&k-mh= \pm \sqrt{a^2m^2+b^2}\\
&k^2+m^2h^2-2khm=a^2m^2+b^2\\
&(a^2-h^2)m^2+(2hk)m+(b^2-k^2)=0
\end{align*}
This is a quadratic equation in $m$ so its roots represents slope of tangents passing through $P(h,k)$ and by the definition of director the circle, they are perpendicular,
\begin{align*}
m_1m_2&=-1\\
\dfrac{b^2-k^2}{a^2-h^2}&=-1\\
h^2+k^2&=a^2+b^2\\
x^2+y^2&=a^2+b^2
\end{align*}
which is the locus of $P(h,k)$

And do we get the entire circle?

And yes, we don't get the entire circle as the ellipse is rotating only in the first quadrant, we get an arc of the circle $y=\sqrt{5-x^2}$ from $x=1$ to $x=2$ because when the major axis of the ellipse is horizontal, the $x$ coordinate of the center is $x=2$ (maximum case as on rotating the $x$ coordinate of the center decreases) and when the major axis of ellipse is vertical, the $x$ coordinate of the center is $x=1$ (the other maximum case as the ellipse cannot cross the coordinate axes, the distance from the center of the ellipse to the y-axis is equal to the length of semi-minor axis of the ellipse)

 

[kshitij]

Are there no equality signs where you live?

$4x^2y^2-(x^2+y^2-z^2)^2\leq c^4$ as well, so what?
Why is this the maximal value and not just an upper bound?
Do you know what it means geometrically and why?
When does equality hold? Does it hold?

You can ignore this attempt, as the question doesn't say that $x,y,z\gt 0$ so we cannot use $A.M \geq G.M$

 

[fresh_42]

I'll post a few:

1) Note that if $H$ is a proper and nontrivial subgroup of $G$, then by counting, $\ker(\varphi)$ is a proper, nontrivial, and normal subgroup of $G.$ Hence, $G$ can only be simple if it has no proper, nontrivial subgroups, meaning it must be cyclic of prime order.

Having dealt with the case that $G$ is simple, suppose now that $N$ is a normal, proper, and nontrivial subgroup of $G$. Consider the map $f:G\to N\times G/N$ given by $f(g)=(\varphi(g),gN).$ This is clearly a group homomorphism and its kernel is trivial since $gN$ is only trivial when $g\in N$, in which case $\varphi(g)=1$ implies $g=1.$ So it is an isomorphism (since domain and codomain have the same finite cardinality). By downward induction, we just need to check that $N$ and $G/N$ have the same defining property that $G$ does, as the previous paragraph is the base case.

If $K$ is a subgroup of $H$, then it is also a subgroup of $G$, so we have a map $\varphi:G\to K$ as in the problem statement, and this restricts to the desired map $H\to K$. Next, subgroups of $G/N$ are of the form $G/M$ for $M$ a normal group of $G$ containing $N.$ We have the map $\varphi:G\to N$, which restricts to $\varphi:M\to N$, which induces the desired map $G/M\to G/N.$

Right.

6. I think $P$ can win. A winning first move is to choose $a=0.$ If $Q$ then chooses a value for $b$, then $P$ considers the cubic $x^3+bx$ and selects any value $c$ such that the graph of $x^3+bx$ is not tangent to the horizontal $y=-c.$ On the other hand, if $Q$ chooses a value for $c$, then consider separately $c\neq 0$ and $c=0$. If $c=0$, then $P$ may choose $b=1$ since $x^3+x$ does not have a multiple root. If $c\neq 0$, then $P$ may choose $b=0$ since $x^3+c$ only has a multiple root when $c=0.$

I read that as a hint since I want a rigorous proof. Just a remark: What if $Q$ chooses $b=0\,\rm ?$

8. Let $n$ be the number of $5$-Sylow subgroups. By the Sylow theorems, $n$ must divide $70/5=14$, so is either $1,2,7$ or $14.$ Another Sylow theorem says that $n$ is $1$ mod $5$, leaving only the possibility $n=1$. If $H\subset G$ is a subgroup of order $5,$ then so is $gHg^{-1}$ and therefore $H=gHg^{-1}.$

Yep.

9. Do you mean for $Y$ to be hausdorff instead of $Z$? If so, I think I have a solution.

No typo. $Z$ is required to be Hausdorff, not $Y$.

1 implies 2: Assuming condition (1), given $z\in Z$, by surjectivity, we may choose $x\in X$ satisfying $h(x)=z$. Define $f(z)$ to be $g(x)$. This is well-defined because if $x,x'\in X$ both satisfy $h(x)=h(x')=z$, then $g(x)=g(x').$ It is clear that $g=f\circ h$. Finally we check continuity: let $F\subset Z$ be closed. Then $f^{-1}(F)=g(h^{-1}(F)).$ Since $h$ is continuous, $h^{-1}(F)$ is closed and hence compact (as $X$ is compact). Then $g(h^{-1}(F))$ is compact and hence closed, since compact subsets of Hausdorff spaces are closed.

Yes, but if we only have $Z$ Hausdorff and not $Y$, then there is a bit more to do. We need the surjectivity of $h$ again and that $h$ is closed!

2 implies 3: Suppose $g=f\circ h=f'\circ h$. Then $f=f'$ on the image of $h$, which is all of $Z$, so $f=f'.$

3 implies 1: Given the equation $g=f\circ h$, it is clear that $h(x)=h(x')$ implies $g(x)=g(x').$

 

[fresh_42]

You can ignore this attempt, as the question doesn't say that $x,y,z\gt 0$ so we cannot use $A.M \geq G.M$

We may assume them to be positive since they all only occur squared in the statement. Your proof was basically ok, just a bit too sloppy (cp. my questions in post #12).

 

[fresh_42]

Do you want me to derive the equation of director circle?

Yes, but I thought of a geometric proof which would have been shorter (therefore the picture).

Let $y=mx+c$ be a tangent to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, so on solving these two we get,
\begin{align*}
b^2x^2+a^2(mx+c)^2&=a^2b^2\\
(b^2+a^2m^2)x^2+(2a^2mc)x+a^2(c^2-b^2)&=0
\end{align*}
For $y=mx+c$ to be a tangent to the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, discriminant of the above quadratic equation must be zero (to ensure that the line cuts the ellipse at only one point),
\begin{align*}
4a^4m^2c^2-4a^2(b^2+a^2m^2)(c^2-b^2)&=0\\
4a^2(a^2m^2c^2-b^2c^2+b^4-a^2m^2c^2+a^2b^2m^2)&=0\\
4a^2b^2(-c^2+b^2+a^2m^2)&=0\\
c^2=a^2m^2+b^2\\
c=\pm \sqrt{a^2m^2+b^2}
\end{align*}
So, the equation of tangent to the ellipse is $$y=mx \pm \sqrt{a^2m^2+b^2}$$
Let the locus of the director circle is $P(h,k)$

Now, as the tangent passes through $P(h,k)$, we can write,
\begin{align*}
&y=mx \pm \sqrt{a^2m^2+b^2}\\
&k=mh \pm \sqrt{a^2m^2+b^2}\\
&k-mh= \pm \sqrt{a^2m^2+b^2}\\
&k^2+m^2h^2-2khm=a^2m^2+b^2\\
&(a^2-h^2)m^2+(2hk)m+(b^2-k^2)=0
\end{align*}
This is a quadratic equation in $m$ so its roots represents slope of tangents passing through $P(h,k)$ and by the definition of director the circle, they are perpendicular,
\begin{align*}
m_1m_2&=-1\\
\dfrac{b^2-k^2}{a^2-h^2}&=-1\\
h^2+k^2&=a^2+b^2\\
x^2+y^2&=a^2+b^2
\end{align*}
which is the locus of $P(h,k)$

And yes, we don't get the entire circle as the ellipse is rotating only in the first quadrant, we get an arc of the circle $y=\sqrt{5-x^2}$ from $x=1$ to $x=2$ because when the major axis of the ellipse is horizontal, the $x$ coordinate of the center is $x=2$ (maximum case as on rotating the $x$ coordinate of the center decreases) and when the major axis of ellipse is vertical, the $x$ coordinate of the center is $x=1$ (the other maximum case as the ellipse cannot cross the coordinate axes, the distance from the center of the ellipse to the y-axis is equal to the length of semi-minor axis of the ellipse)

 

[fresh_42]

so $\mathbf 1$ is the ones vector and $\mathbf {11}^T$ is the ones matrix -- I use this notation all the time on this site ...

I apologize that I am not familiar with your nomenclature. I guess I wouldn't have asked if it had been $\mathbf{1}\cdot \mathbf{1}^\tau$. The ambiguous use of $T$ is of course totally ok.

... and $\mathbf v \mathbf y^T$ is completely standard matrix vector notation. My approach to the lemma fairly closely follows that of Artin's Algebra's chapter on Representation Theory -- though after re-examining the text, not quite as closely as I thought. With some notational changes I am using $\mathbf B$ to mimic the regular representation development in that chapter. In particular $\mathbf B$ is ordered whereas a set is not. So pick an arbitrary labelling and if X has cardinality n, then $\mathbf B =\bigg[\begin{array}{c|c|c} x_1 & \cdots & x_{n-1} & x_{n}\end{array}\bigg]\implies g\mathbf B =\bigg[\begin{array}{c|c|c} gx_1 & \cdots & gx_{n-1} & gx_{n}\end{array}\bigg] = \mathbf B P_g$ where $P_g$ is a permutation matrix.

You’re asking what $P_g$ is but I explicitly addressed this in my post when I wrote

$g\mathbf B = \mathbf B P_g$

And a note that this was the defining equation and not a conclusion was obviously not necessary. Sorry for not guessing it.

this defines a permutation (matrix) representation i.e. $P_g = \phi(g)$ for some surjective homomorphism $\phi : G\longrightarrow G' \in GL_{\vert X\vert}(\mathbb C)$

I would ask what $G'$ means instead of guessing $G':=\operatorname{im} \phi ,$ but you are obviously annoyed by my lack of fantasy. And sure, it is not necessary to mention that $G'$ isn't $[G,G]$ in a proof in group theory. Or is it, and I missed the point again?

That said I definitely could say this differently by getting rid of $\mathbf B$ and just assert X to be an ordered set. But in any case the point is $P_g$ is the standard permutation matrix representation for g in G implied by the way g acts on the elements in our set X.

- - - -
There's a miscellaneous problem in either edition of that book's rep theory chapter that says "Let G be a finite subgroup of $GL_n(\mathbb C)$. Prove that if $\sum_g \text{trace} g = 0$ then $\sum_g g = \mathbf 0$. There's basically two proofs of this, one (after dividing by $\vert G\vert$) using idempotence or two be clever and back into Schur’s Lemma. What I did in the lemma is implied by that problem and I went the idempotence route.

Spoiler

$T^2=\Big(\frac{1}{\vert H\vert}\sum_{P\in H}P\Big)\Big(\frac{1}{\vert H\vert}\sum_{P'\in H}P'\Big)=\frac{1}{\vert H\vert^2}\sum_{P\in H}\Big(P\sum_{P'\in H}P'\Big)=\frac{1}{\vert H\vert^2}\sum_{P\in H}\vert H\vert \cdot T=\frac{1}{\vert H\vert^2}\cdot \vert H\vert^2 \cdot T=T$

missing detail:
$P\sum_{P'\in H}P'=\sum_{P'\in H}PP'=\sum_{P''\in H}P''=\sum_{P\in H}P=\vert H\vert \cdot T$

The does $T$ even exist question is strange. Of course it exists, you can always add square matrices of the same size (and we sum over all permutation matrices P in H) and since this is over a field of characteristic zero ($\mathbb C$) you can always divide by a positive integer (the cardinality of H).

I thought of an example like $T=\dfrac{1}{3}\cdot \begin{bmatrix} 1 & 1 & 0&\ldots\\1&1&0&\ldots\\0&0&1&\ldots\\ &&&\ddots \\ 0&0&\ldots&1 \end{bmatrix}$ where I could make the matrix arbitrary large. This example doesn't fulfill $\operatorname{trace}T =1$. Hence this condition restricts the number of possible groups $H$ significantly, and the question whether some groups remain is not trivial. You have always the identity in $H$ which can be made arbitrary large, regardless how large $|H|$ is. Sorry for asking.

I do remember how it can be quite exhausting to review answers to these challenges. I had guessed the sort of things in my post would be automatic for you though it seems not to be the case. Probably a good idea to put a pin in this.

 

[kshitij]

We may assume them to be positive since they all only occur squared in the statement. Your proof was basically ok, just a bit too sloppy (cp. my questions in post #12).

But I applied $A.M \geq G.M$ on $$(z-x+y)(z+x-y)(x+y-z)(x+y+z)$$
$(z-x+y),(z+x-y),(x+y-z)$ and $(x+y+z)$ aren't necessarily positive right?

Also to answer your questions in post #12, I have no idea about the geometric meaning of $A.M \geq G.M$
All I knew was that if we had $n$ positive real numbers, $a_1,a_2, ... ,a_n$ we can use $A.M \geq G.M$ and the equality holds if $a_1=a_2= ... =a_n$

 

[kshitij]

I think P can win. A winning first move is to choose a=0. If Q then chooses a value for b, then P considers the cubic x3+bx and selects any value c such that the graph of x3+bx is not tangent to the horizontal

I read that as a hint since I want a rigorous proof. Just a remark: What if Q chooses b=0?

I might be wrong, but I thought that when $a=0$ then, if the value of $b$ is chosen to be positive, then the equation $f(x)=x^3+ax^2+bx+c=0$ should have only one solution because we can see that
$f'(x)=3x^2+2ax+b$
Now to have two different values of $x$ where $f'(x)=0$ we should have (using discriminant >0)
$a^2>3b$
So, for any positive value of $b$ and $a=0$, $f'(x)>0$ so $f(x)$ will be strictly increasing function.

If P chooses any value of $a$ in the first move, Q can accordingly choose $b$ and P will never win as the function will be always increasing and have only one real root!

Infact, I thought that the winning strategy for P would be choosing the value of $b$ as a negative number in the first move, so if Q chooses a value of $a$ in the next move P can simply shift the x-axis up or down accordingly so as to intersect the curve at three points by choosing the value of $c$ and if Q chose the value of $c$ in the second move to make the curve lie above axis or below such that it doesn't cut it at three distinct points then P can choose the value of $a$ accordingly to shift the curve either up or down because $a$ is the coefficient of $x^2$ so it should have a greater say in the position of the curve in the plane and it is also the last move so Q cannot do anything after this.

 

[Infrared]

I read that as a hint since I want a rigorous proof. Just a remark: What if $Q$ chooses $b=0\,\rm ?$

Can you clarify what is not rigorous? I don't see why $Q$ choosing $b=0$ is not covered by what I wrote. I wrote that $P$ chooses any value of $b$ such that the graph $x^3+bx$ is not tangent to $y=-c.$ Hence if $b=0$ any nonzero value of $c$ works as the line $y=0$ is the only horizontal that's tangent to the graph of $x^3.$ This procedure works because $p(x)$ having a multiple root at $x=a$ is equivalent to the graph of $p$ being tangent to the $y$-axis at $x=a.$

No typo. $Z$ is required to be Hausdorff, not $Y$.

OK, my apologies. I had an idea of how I thought you wanted the solution to look and shouldn't have changed the problem. Anyway, put a relation on $X$ by $x\sim x'$ if $h(x)=h(x')$ and consider the induced quotient map $h':(X/\sim)\to Z.$ This is a continuous bijection, and since the map $h$ is closed as you mentioned (because a closed subset of $X$ is compact, and its image in $Z$ is compact in a hausdorff space; thus closed), the map $h'$ is too, making it a homeomorphism. Then if $g'$ is the induced map $(X/\sim)\to Y$, the map $f=(h')^{-1}\circ g'$ is continuous and clearly satisfies $f\circ h=g.$

 

[kshitij]

Can you clarify what is not rigorous? I don't see why Q choosing b=0 is not covered by what I wrote. I wrote that P chooses any value of b such that the graph x3+bx is not tangent to y=−c. Hence if b=0 any nonzero value of c works as the line y=0 is the only horizontal that's tangent to the graph of x3. This procedure works because p(x) having a multiple root at x=a is equivalent to the graph of p being tangent to the y-axis at

I don't understand what you're saying?

How can the equation $y=x^3+c$ have three different real roots for any value of $c$?

 

[fresh_42]

But I applied $A.M \geq G.M$ on $$(z-x+y)(z+x-y)(x+y-z)(x+y+z)$$
$(z-x+y),(z+x-y),(x+y-z)$ and $(x+y+z)$ aren't necessarily positive right?

Also to answer your questions in post #12, I have no idea about the geometric meaning of $A.M \geq G.M$
All I knew was that if we had $n$ positive numbers, $a_1,a_2, ... ,a_n$ we can use $A.M \geq G.M$ and the equality holds if $a_1=a_2= ... =a_n$

Sure. That gives you an upper bound. But why is it the maximum, and not way greater than the maximum?

But I applied $A.M \geq G.M$ on $$(z-x+y)(z+x-y)(x+y-z)(x+y+z)$$
$(z-x+y),(z+x-y),(x+y-z)$ and $(x+y+z)$ aren't necessarily positive right?

Also to answer your questions in post #12, I have no idea about the geometric meaning of $A.M \geq G.M$
All I knew was that if we had $n$ positive real numbers, $a_1,a_2, ... ,a_n$ we can use $A.M \geq G.M$ and the equality holds if $a_1=a_2= ... =a_n$

I added the condition $x,y,z > 0$ to save us a zoo of cases. That leaves us with fewer cases. Sorry for this inconvenience.

I meant the geometrical meaning of $f(x,y,z)$ not the meaning of mean values.

 

[kshitij]

Sure. That gives you an upper bound. But why is it the maximum, and not way greater than the maximum?

I added the condition $x,y,z > 0$ to save us a zoo of cases. That leaves us with fewer cases. Sorry for this inconvenience.

I don't know about why is that maximum, if equality holds then they should be maximum. I think I don't understand what you're trying to say?

Also even if $x,y,z \gt 0$, $x+y-z$ can still be less than zero and we cannot use A.M G.M inequality the way I did, if $x,y,z$ were sides of a triangle then its different though.

 

[kshitij]

I meant the geometrical meaning of f(x,y,z) not the meaning of mean values.

I just noticed that $$c(c-2x)(c-2y)(c-2z)$$
Looks similar to the formula for area of a triangle!
Is that what you meant, is geometrical meaning of $f(x,y,z)$ is that is 16 times the square of the area of a triangle with sides $x,y,z$! because area of a triangle with sides $x,y,z$ will be
$$\sqrt{\dfrac{c}{2}\left(\dfrac{c}{2}-x\right)\left(\dfrac{c}{2}-y\right)\left(\dfrac{c}{2}-y\right)}$$

 

[fresh_42]

Can you clarify what is not rigorous? I don't see why $Q$ choosing $b=0$ is not covered by what I wrote. I wrote that $P$ chooses any value of $b$ such that the graph $x^3+bx$ is not tangent to $y=-c.$ Hence if $b=0$ any nonzero value of $c$ works as the line $y=0$ is the only horizontal that's tangent to the graph of $x^3.$ This procedure works because $p(x)$ having a multiple root at $x=a$ is equivalent to the graph of $p$ being tangent to the $y$-axis at $x=a.$

You started with $P$ chooses $a=0$. And if $Q$ answers $b=0,$ then $P$ is left with $x^3+c$. So $a=0$ is no winning strategy for $P$.