Challenge Math Challenge - November 2018

[fresh_42]

But if $|a + b| = \max(|a|,|b|)$ for all integers a and b, then |0| = |1 - 1| = max(|1|,|-1|) = max(1,1) = 1, which is contrary to our definition of this value function or norm. So there must be some flaw in the proof of this statement.

Yes. I had to use $|a| \neq |b|$ for otherwise I couldn't have concluded, that $|a|<\max\{\,|a+b|,|a|\,\}=|a+b|\,.$
In your post #146 where it is needed, we have $|kp| =|k|\cdot |p| \leq |p| \lt |m|\,.$

 

[lpetrich]

So with $|a+b| = \max(|a|,|b|)$ if $b \neq a$, I can do all the positive integers. For n relatively prime to p, the smallest number where |p| < 1, I get n = kp + m, where m is nonzero and less than p:
$|n| = |kp+m| = \max(|kp|,|m|) = \max(|kp|,1) = 1$.

With that case solved, consider n having p as a prime factor: $n = k p^m$ for k is relatively prime to p. Then,
$|n| = |k p^m| = |k||p|^m = |p|^m$.

For fraction $x = n_1/n_2$ where $n_1 = k_1 p^{m_1}$ and likewise for $n_2$, then $|x| = |p|^{m_1 - m_2}$.

 

[fresh_42]

So with $|a+b| = \max(|a|,|b|)$ if $b \neq a$, I can do all the positive integers. For n relatively prime to p, the smallest number where |p| < 1, I get n = kp + m, where m is nonzero and less than p:
$|n| = |kp+m| = \max(|kp|,|m|) = \max(|kp|,1) = 1$.

With that case solved, consider n having p as a prime factor: $n = k p^m$ for k is relatively prime to p. Then,
$|n| = |k p^m| = |k||p|^m = |p|^m$.

For fraction $x = n_1/n_2$ where $n_1 = k_1 p^{m_1}$ and likewise for $n_2$, then $|x| = |p|^{m_1 - m_2}$.

That doesn't define the norm. In the end the definition should work for any rational number including powers of $p$, $0$ and without self references.
Plus you still must show $|a+b| \leq \max\{\,|a|,|b|\,\}\,.$

 

[julian]

Solution to problem 16.

The actual calculations are in the attached pdf file. I just outline in this post what was done.

We define a basis for $\mathfrak{su}(2,\mathbb{C})$ as:

$u_1 = i \sigma_1 = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} , \quad u_2 = i \sigma_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} , \quad u_3 = i \sigma_3 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$

with brackets:

$[ u_1 , u_2 ] = - 2 u_3 , \quad [ u_2 , u_3 ] = - 2 u_1 , \quad [ u_3 , u_1 ] = - 2 u_2 .$

We compute $[(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')]$ defined by $[(\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3) , (\alpha_1' u_1 + \alpha_2' u_2 + \alpha_3' u_3)]$. We easily find:

$[(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')] =$
$= 2 (\alpha_3 \alpha_2' - \alpha_2' \alpha_3) u_1 + 2 (\alpha_1 \alpha_3' - \alpha_3 \alpha_1') u_2 + 2 (\alpha_2 \alpha_1' - \alpha_1 \alpha_2') u_3 .$

We define an adjusted $\varphi$ by:

\begin{align*}
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3) &= \varphi (\alpha_1 (i \sigma_1) + \alpha_2 (i \sigma_2) + \alpha_3 (i \sigma_3)) \\
&= \varphi( (i \alpha_1) \sigma_1 + ( i\alpha_2) \sigma_2+ (i \alpha_3) \sigma_3) \\
\end{align*}

Then from

\begin{align*}
\varphi(\alpha_1 \sigma_1 +\alpha_2 \sigma_2+\alpha_3 \sigma_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
&= x(- i \alpha_1 a_3 + \alpha_2 a_3 - \alpha_3 a_1 )+\\
&+ x^2(2 i \alpha_1 a_5 + 2 \alpha_2 a_5 + 2 \alpha_3 a_2 )+\\
&+ y(i \alpha_1 a_1 + \alpha_2 a_1 + \alpha_3 a_3 )+\\
&+ y^2(2 i \alpha_1 a_5 -2 \alpha_2 a_5 -2 \alpha_3 a_4 )+\\
&+ xy(- i \alpha_1 a_2 - i \alpha_1 a_4 +\alpha_2 a_2 - \alpha_2 a_4 )
\end{align*}

we have:

\begin{align*}
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy) = \\
= \varphi( (i \alpha_1) \sigma_1 + ( i\alpha_2) \sigma_2+ (i \alpha_3) \sigma_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
&= x(\alpha_1 a_3 +i\alpha_2 a_3 - i\alpha_3 a_1 )+\\
&+ x^2(-2\alpha_1 a_5 +2 i\alpha_2 a_5 + 2i\alpha_3 a_2 )+\\
&+ y(- \alpha_1 a_1 + i \alpha_2 a_1 +i\alpha_3 a_3 )+\\
&+ y^2(-2\alpha_1 a_5 -2i\alpha_2 a_5 -2i\alpha_3 a_4 )+\\
&+ xy(\alpha_1 a_2 +\alpha_1 a_4 +i\alpha_2 a_2 -i\alpha_2 a_4 ) \qquad Eq (1)
\end{align*}
Part 16 a) Is done in pdf file where I prove

$[\tilde{\varphi} (\alpha_1 , \alpha_2 , \alpha_3) , \tilde{\varphi} (\alpha_1' , \alpha_2' , \alpha_3')] = \tilde{\varphi} ([(\alpha_1 , \alpha_2 , \alpha_3) , (\alpha_1' , \alpha_2' , \alpha_3')])$

when applied to $1,x,x^2,y,y^2$, and $xy$ in turn.16 b) Components that are transformed into linear combinations of themselves under repeated application of $\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 +\alpha_3 u_3)$ form an invariant subspace. Irreducible components are an invariant subspace which cannot be separated into smaller invariant subspaces. From Eq 1 we have:

\begin{align*}
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.1 = 0\\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.x= - x i \alpha_3 + y (- \alpha_1 + i \alpha_2) \\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.x^2 = 2 x^2 i \alpha_3 + xy (\alpha_1 + i \alpha_2) \\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.y = x (- \alpha_1 + i \alpha_2) + y i \alpha_3 \\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.y^2 = - 2 y^2 i \alpha_3 - xy (- \alpha_1 + i \alpha_2) \\
\tilde{\varphi} (\alpha_1 u_1 + \alpha_2 u_2 + \alpha_3 u_3)&.xy = 2 x^2 (- \alpha_1 + i \alpha_2) - 2 y^2 (\alpha_1 + i \alpha_2) .
\end{align*}

It is then obvious that the irreducible components are:

$\{ 1 \}$
$\{ x , y \}$
$\{ y^2 , xy , x^2 \} .$16 c) The corresponding vectors of maximum weight are

$1 ,$
$y ,$
$x^2$

respectively - see the pdf file.

p.s. Hello @fresh_42. I've done the calculation with the adjusted $\varphi$ and still think there is a slight typo in the question. In the original question you wrote:

\begin{align*}
\varphi(\alpha_1 \sigma_1 +\alpha_2 \sigma_2+\alpha_3 \sigma_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
&= \dots\\
&+ y(-i \alpha_1 a_1 - \alpha_2 a_1 +i\alpha_3 a_3 )+\\
&+ \dots
\end{align*}

but I think it should say:

\begin{align*}
\varphi(\alpha_1 \sigma_1 +\alpha_2 \sigma_2+\alpha_3 \sigma_3)&.(a_0+a_1x+a_2x^2+a_3y+a_4y^2+a_5xy)= \\
&= \dots\\
&+ y(i \alpha_1 a_1 + \alpha_2 a_1 +i\alpha_3 a_3 )+\\
&+ \dots
\end{align*}

Could you check my calculations in the pdf.

 

[fresh_42]

@julian
Great job, Julian!
Btw.: The Heisenberg algebra question in 15. is far easier with less computations. Only a bit thinking about centers is needed.

You are right, there is a sign error in the setup of $\varphi$ somewhere.
I took the basis and the representation theorem from
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/
which I now checked again, and which turned out to be correct.

I thought I had taken $x$ as the eigenvector of minimal weight $-1$ and $y$ for the maximal eigenvector. But with this setting my definition of $\varphi$ doesn't match up. Guess I deserved all these calculations to check again:

Pauli - matrices $\sigma_j \, \longrightarrow$ skew-Hermitian version $i \cdot \sigma_j \,\longrightarrow$ basis $\langle U,V,W\,|\,[U,V]=W\; , \;[V,W]=U\; , \;[W,U]=V \rangle$ because this is easiest to memorize $\longrightarrow$ standard basis $\langle H,X,Y\,|\,[H,X]=2X\; , \;[H,Y]=-2Y\; , \;[X,Y]=H \rangle$ in order to have the example $\mathfrak{sl}_2$ from the textbooks with CSA $\langle H \rangle \,\longrightarrow$ and all the way back to the Pauli matrices to make the question "physics" compatible

At least I haven't chosen a weird basis in $\mathbb{C}x \oplus \mathbb{C}y$ and something like $3x-5y$ the maximal vector.

These were the ladder structures I had in mind:
$\mathfrak{su}_2.\mathbb{C}=\{\,0\,\}$
\begin{align*}
(0,1) &\stackrel{X}{\longrightarrow} (1,0)\stackrel{X}{\longrightarrow} (0,0)\\
(1,0)&\stackrel{Y}{\longrightarrow}(0,1)\stackrel{Y}{\longrightarrow}(0,0)
\end{align*}
\begin{align*}
(0,0,1)&\stackrel{X}{\longrightarrow}(0,-i,0)\stackrel{X}{\longrightarrow}(2,0,0)\stackrel{X}{\longrightarrow}(0,0,0)\\
(1,0,0)&\stackrel{Y}{\longrightarrow} (0,-i,0) \stackrel{Y}{\longrightarrow} (0,0,2)\stackrel{Y}{\longrightarrow}(0,0,0)
\end{align*}

 

[julian]

A partial go a problem 21:

The deck transformations of the covering space $\tilde{X}$ constitute a group of homeomorphisms of that covering space (where the group operation is the usual operation of composition of homeomorphisms).

Definitions:

Definition of a homeomorphism:

A function $h: X \rightarrow Y$ between two topological spaces is a homeomorphism if it has the following properties:

$h$ is a bijection (one-to-one and onto),
$h$ is continuous,
the inverse function $h^{-1}$ is continuous ($h$ is an open mapping).

Proving the Group property of $\mathcal{D} (p)$:

Closure under composition of two deck transformations:

Say $p \circ h_1 = p$ and $p \circ h_2 = p$ then we wish to prove that $p \circ (h_1 \circ h_2) = p$. For any $\tilde{x} \in \tilde{X}$ we have

\begin{align*}
[p \circ (h_1 \circ h_2)] (\tilde{x}) &= p \big( h_1 \big( h_2 (\tilde{x}) \big) \big) \\
& = [p \circ h_1] \big( h_2 (\tilde{x}) \big) \\
& = p \big( h_2 (\tilde{x}) \big) \qquad \qquad \text{as } p \circ h_1 = p \\
& = [p \circ h_2] (\tilde{x}) \\
& = p (\tilde{x}) \qquad \qquad \qquad \text{as } p \circ h_2 = p
\end{align*}

hence $p \circ (h_1 \circ h_2) = p$.

Identity Homeomorphism is a deck transformation:

The identity map $id_{\tilde{X}} : \tilde{X} \rightarrow \tilde{X}$ is a homeomorphism and we obviously have $p \circ id_{\tilde{X}} = p$.

Inverse of a deck transformation is a deck transformation:

Say $p \circ h = p$. For any $\tilde{x} \in \tilde{X}$ we have:

\begin{align*}
p (\tilde{x}) &= [p \circ (h \circ h^{-1})] (\tilde{x}) \\
&= [p \circ h] \big( h^{-1} (\tilde{x}) \big) \\
& = p \big( h^{-1} (\tilde{x}) \big) \qquad \qquad \qquad \text{as } p \circ h = p \\
& = [p \circ h^{-1}] (\tilde{x}) \\
\end{align*}

hence $p \circ h^{-1} = p$.

Associativity of deck transformations:

Suppose that $h_1 : \tilde{X} \rightarrow \tilde{X}$, $h_2 : \tilde{X} \rightarrow \tilde{X}$, and $h_3 : \tilde{X} \rightarrow \tilde{X}$ are deck transformations. Then $p \circ [(h_3 \circ h_2) \circ h_1] = p \circ [h_3 \circ (h_2 \circ h_3)]$. Proof. For any $\tilde{x} \in \tilde{X}$ we have

\begin{align*}
[ p \circ [(h_3 \circ h_2) \circ h_1 ] ] (\tilde{x}) & = p \big( ( h_3 \circ h_2 ) \big( h_1 (\tilde{x}) \big) \big) \\
& = p \big( h_3 \big( h_2 \big( h_1 (\tilde{x}) \big) \big) \big) \\
& = p \big( h_3 \big( (h_2 \circ h_1) (\tilde{x}) \big) \big) \\
& = [ p \circ [h_3 \circ (h_2 \circ h_1)] ] (\tilde{x}) . \\
\end{align*}Proof that $h (\tilde{x}) = \tilde{x}$ for $\tilde{x} \in \tilde{X}$ implies $h = id_{\tilde{X}}$:

First we prove that if $p \circ g = p \circ h$, where $g$ and $h$ are deck transformations, and $g (\tilde{x}) = h(\tilde{x})$ for some point $\tilde{x} \in \tilde{X}$ then $g = h$. It then easily follows that if $h (\tilde{x}) = \tilde{x}$ for some point $\tilde{x} \in \tilde{X}$ then $h = id_{\tilde{X}}$.We prove that $A := \{ \tilde{x} \in \tilde{X} : g (\tilde{x}) = h (\tilde{x}) \}$ is both open and closed, thereby proving that $A = \tilde{X}$ because of the connectedness of $\tilde{X}$.

Both $g : \tilde{X} \rightarrow \tilde{X}$ and $h : \tilde{X} \rightarrow \tilde{X}$ are continuous maps as they are homeomorphisms. Let $\tilde{x} \in \tilde{X}$. There exists an open neighborhood $U \in X$ containing the point $p (g(\tilde{x}))$ with $p^{-1} (U)$ a disjoint union of open sets $V_\iota$ each of which is homeomorphically mapped onto $U$ by $p$ (for $\tilde{U} \in V_\iota$ we write $p | \tilde{U} : \tilde{U} \rightarrow U$ is a homeomorphism). One of these open sets contains $g (\tilde{x})$, denote it by $\tilde{U}$. Also one of these open sets contains $h (\tilde{x})$ (this is because $p \circ h = p \circ g$) Let us denote this open set by $\tilde{V}$. Let $N_{\tilde{x}} = g^{-1} (\tilde{U}) \cap h^{-1} (\tilde{V})$. Then $N_{\tilde{x}}$ is an open set in $\tilde{X}$ containing $\tilde{x}$ (open because $g$ and $h$ are continuous functions as they are homeomorphisms).

Consider the case when $\tilde{x} \in A$. Then $g (\tilde{x}) = h (\tilde{x})$, and therefore $\tilde{V} = \tilde{U}$. It follows from this that both $g$ and $h$ map the open set $N_{\tilde{x}}$ into $\tilde{U}$. We now use that $p \circ g = p \circ h$ and that $p | \tilde{U} : \tilde{U} \rightarrow U$ is a homeomorphism to prove that $g | N_{\tilde{x}} = h | N_{\tilde{x}}$. Take an arbitrary point $\tilde{x}' \in N_{\tilde{x}}$ other than the point $\tilde{x}$. Suppose that $g (\tilde{x}') \not= h (\tilde{x}')$, but we have $(p \circ g) (\tilde{x}') = (p \circ h) (\tilde{x}')$. This says two different points get mapped to the same point by $p | \tilde{U}$, but that contradicts that $p | \tilde{U}$ is injective. As such we must have that $g (\tilde{x}') = h (\tilde{x}')$ for all $\tilde{x}' \in N_{\tilde{x}}$. Thus $N_{\tilde{x}} \subset A$. We have thus shown that for each $\tilde{x} \in A$ there exists an open set set $N_{\tilde{x}}$ such that $\tilde{x} \in N_{\tilde{x}}$ and $N_{\tilde{x}} \subset A$. Thus $A$ is open.

Next we show that the set $\tilde{X} / A$ is open as well. So consider the case $\tilde{x} \in \tilde{X} / A$. In this case $\tilde{U} \cap \tilde{V} = \emptyset$ since $g (\tilde{x}) \not= h (\tilde{x})$ (having both $g (\tilde{x})$ and $h (\tilde{x})$ in $\tilde{U}$ together with $(p \circ g) (\tilde{x}) = (p \circ h) (\tilde{x})$ is in contradiction with the injectivity of $p | \tilde{U}$). Again define $N_{\tilde{x}} = g^{-1} (\tilde{U}) \cap h^{-1} (\tilde{V})$. But $g (N_{\tilde{x}}) \subset \tilde{U}$ and $h (N_{\tilde{x}}) \subset \tilde{V}$. Therefore $g (\tilde{x}') \not= h (\tilde{x}')$ for $\tilde{x}' \in N_{\tilde{x}}$, and thus $N_{\tilde{x}} \subset \tilde{X} / A$. We have then shown that for each $\tilde{x} \in \tilde{X} / A$ there exists an open set $N_{\tilde{x}}$ such that $\tilde{x} \in N_{\tilde{x}}$ and $N_{\tilde{x}} \subset \tilde{X} / A$. Thus $\tilde{X} / A$ is open.

The subset $A$ of $\tilde{X}$ is therefore both open and closed. As $A$ was assumed to be non-empty, we deduce that $A = \tilde{X}$, because $\tilde{X}$ is connected. Thus $g = h$, which was the required resultThe last step is to note that a function such that $h (\tilde{x}) = \tilde{x}$ for some point $\tilde{x} \in \tilde{X}$ and the identity function $id_{\tilde{X}}$ (satisfying $h (\tilde{x}) = id_{\tilde{X}} (\tilde{x})$ for this point $\tilde{x} \in \tilde{X}$) means that $h = id_{\tilde{X}}$.

 

[fresh_42]

A partial go a problem 21: ...

What do you mean by partial? It looks fine to me!

Only a few minor remarks:

• Subsequent application of functions is associative. Done. And $p\circ (h\circ g^{-1})=p$ does the other group properties in one step.
• For the second part, you could have started as you did with the pair $(h,g)$ but then switched to $g=1$. Just because it is less to write.
• $A$ is closed can be done shorter by using the fact that diagonals in Hausdorff spaces are closed. This is even equivalent to $T_2$.

 

[julian]

What do you mean by partial? It looks fine to me!

Only a few minor remarks:

• Subsequent application of functions is associative. Done. And $p\circ (h\circ g^{-1})=p$ does the other group properties in one step.
• For the second part, you could have started as you did with the pair $(h,g)$ but then switched to $g=1$. Just because it is less to write.
• $A$ is closed can be done shorter by using the fact that diagonals in Hausdorff spaces are closed. This is even equivalent to $T_2$.

I'm still looking into the issue of when homeomorphims form a group. I have proven that the general result that the composition of homeomorphisms is a homoemorphism and a result about an inverse function being a homeomorphism.

But I think there are subtle issues relating to the topology. For example the Identity Homeomorphism:

The identity map $id_X : X \rightarrow X$ is obviously bijective. However, the identity mapping $id_X : (X ,\tau) \rightarrow (X , \tau')$ is continuous if and only if $\tau' \subseteq \tau$. Thus if we topologize $X$ in such a way that this inclusion is proper the identity mapping in this direction will be continuous while its inverse (also the identity) will not. We will only have a continuous inverse if $\tau = \tau'$.

Does this mean that in order to have a group like structure do we have to consider only homeomorphims that preserve the topology $\tau$?

I have some intuition that deck transformation would preserve the topology of $\tilde{X}$ but haven't formed a rigorous proof yet.

 

[fresh_42]

Does this mean that in order to have a group like structure do we have to consider only homeomorphims that preserve the topology?

I don't quite understand. The definition was for homeomorphisms, so the question is obsolete. Whether $p\circ h = p$ forms a group if $h$ isn't homeomorph is another issue. I don't think so, since the bijection might be sufficient for the algebraic properties, but not for the topological. And isn't the whole issue of homeomorphisms, that they preserve the topologies?

I have some intuition that deck transformation would preserve the topology of $\tilde{X}$ but haven't formed a rigorous proof yet.

It's defined that way.

 

[julian]

I was trying to prove the continuity of the inverse of $h^{-1}$ by showing $( h^{-1} )^{-1} = h$ using bijective properties and that $h$ is continuous. Now I realize you can and should prove the continuity of the inverse $h^{-1}$ by topological arguments. Cleared up now.

 

[julian]

Is there a mistake in problem 20? It seems that $I = \langle \mathcal{I} \rangle$ is NOT a normal subgroup of $F$, and hence $F / I$ is not a group (meaning that $\mathcal{F} / \sim_\mathcal{I}$ does NOT admit a group structure).

It seems that it is $J = \langle \mathcal{J} \rangle$ that is a normal subgroup of $F$ and hence $F / J$ is a group (meaning $\mathcal{F} / \sim_\mathcal{J}$ that does admit a group structure).

To prove that $I$ is not a normal subgroup of $F$ we just have to give an example of when $f^{-1} i f \notin I$ for some $i \in I$ and some $f \in F$. An example of this is

\begin{align*}
[ u^{-1} \circ r \circ u ] (z) & = [ v \circ r \circ u ] (z) \\
& = v \big( \big( \frac{1}{2} (-1 + i \sqrt{3} z) \big)^{-1} \big) \\
& = v \big( - \frac{1}{2} (1 + i \sqrt{3}) \frac{1}{z} \big) \\
& = - \frac{1}{2} (1 + i \sqrt{3}) \big( - \frac{1}{2} (1 + i \sqrt{3}) \frac{1}{z} \big) \\
& = \frac{1}{2} (-1 + i \sqrt{3}) \frac{1}{z} \notin I
\end{align*}

(where we have used that $v (z)$ is the inverse of $u (z)$).

I'll give all details in my next post.

 

[fresh_42]

$r(u(z))=r\left( \left( \dfrac{1}{2}\left( -1+i\sqrt{3} \right)z \right) \right)= \dfrac{1}{2}\left( -1+i\sqrt{3} \right)z^{-1} \neq -\dfrac{1}{2}\left( 1+i\sqrt{3} \right)z^{-1}$ .
Sorry, I should have mentioned that.

 

[julian]

$r(u(z))=r\left( \left( \dfrac{1}{2}\left( -1+i\sqrt{3} \right)z \right) \right)= \dfrac{1}{2}\left( -1+i\sqrt{3} \right)z^{-1} \neq -\dfrac{1}{2}\left( 1+i\sqrt{3} \right)z^{-1}$ .
Sorry, I should have mentioned that.

So you are saying the maps act upon the $z$ in other words. That is the way I originally did the question and I still got that $I$ is not a normal subgroup! Let me redo the example I gave in my previous post:

\begin{align*}
[u^{-1} \circ r \circ u] (z) & = [v \circ r \circ u] (z) \\
& = v \left( \frac{1}{2} \left( - 1 + i \sqrt{3} \right) \frac{1}{z} \right) \\
& = \frac{1}{2} \left( - 1 + i \sqrt{3} \right) \frac{1}{- \frac{1}{2} \left( 1 + i \sqrt{3} \right) z} \\
& = - \frac{1}{2} (1 + i \sqrt{3}) z^{-1} \notin I
\end{align*}

so the conclusion is still that $I$ is not a normal subgroup.

 

[fresh_42]

So you are saying the maps act upon the $z$ in other words.

Yes. This happens if one "invents" problems rather than copy them from the internet, I guess.

I calculated:
\begin{align*}
\varphi(u^{-1})(r)=u^{-1}ru &= u^{-1}r\left(\dfrac{1}{2}\left( -1+i\sqrt{3} \right)z \right)\\
&= v \left(\dfrac{1}{2}\left( -1+i\sqrt{3} \right)z^{-1} \right)\\
&= \left(-\dfrac{1}{2}\left( 1+i\sqrt{3} \right) \right)\left(\dfrac{1}{2}\left( -1+i\sqrt{3} \right)z ^{-1} \right)\\
&= -\dfrac{1}{4}\left(1+i\sqrt{3} \right)\left(-1+i\sqrt{3} \right)z^{-1}\\
&= -\dfrac{1}{4} \cdot \left(-1 - 3 \right)z^{-1}\\
&= z^{-1}
\end{align*}
The goal is to consider the structure of a certain group with twelve elements.

 

[julian]

Yes. This happens if one "invents" problems rather than copy them from the internet, I guess.

I calculated:
\begin{align*}
\varphi(u^{-1})(r)=u^{-1}ru &= u^{-1}r\left(\dfrac{1}{2}\left( -1+i\sqrt{3} \right)z \right)\\
&= v \left(\dfrac{1}{2}\left( -1+i\sqrt{3} \right)z^{-1} \right)\\
&= \left(-\dfrac{1}{2}\left( 1+i\sqrt{3} \right) \right)\left(\dfrac{1}{2}\left( -1+i\sqrt{3} \right)z ^{-1} \right)\\
&= -\dfrac{1}{4}\left(1+i\sqrt{3} \right)\left(-1+i\sqrt{3} \right)z^{-1}\\
&= -\dfrac{1}{4} \cdot \left(-1 - 3 \right)z^{-1}\\
&= z^{-1}
\end{align*}
The goal is to consider the structure of a certain group with twelve elements.

But if the maps act upon the $z$ then you should have:

\begin{align*}
v \left( \frac{1}{2} \left( - 1 + i \sqrt{3} \right) \frac{1}{z} \right) & = \frac{1}{2} \left( - 1 + i \sqrt{3} \right) \frac{1}{v (z)} \\
& = - \frac{1}{2} (1 + i \sqrt{3}) z^{-1}
\end{align*}

 

[fresh_42]

But if the maps act upon the $z$ then you should have:

\begin{align*}
v \left( \frac{1}{2} \left( - 1 + i \sqrt{3} \right) \frac{1}{z} \right) & = \frac{1}{2} \left( - 1 + i \sqrt{3} \right) \frac{1}{v (z)} \\
& = - \frac{1}{2} (1 + i \sqrt{3}) z^{-1}
\end{align*}

I doesn't act on the map, it acts on $z$. And as we don't have rings or algebras, the factors simply remain factors. As said, I wanted to have a group with only twelve elements. The intended clue was, that although we can define equivalence relations for any subgroup, only the normal ones give a again a factor (or quotient) group.

 

[julian]

I'm confused. In post #162 you basically told me that I should not interpret $r (u (z))$ as meaning

$r (u (z)) = r \left( \frac{1}{2} (-1 + i \sqrt{3}) z \right) = \left( \frac{1}{2} (-1 + i \sqrt{3}) z \right)^{-1} = - \frac{1}{2} (1 + i \sqrt{3}) z^{-1}$

but should interpret as this instead:

$r (u (z)) = r \left( \frac{1}{2} (-1 + i \sqrt{3}) z \right) = \frac{1}{2} (-1 + i \sqrt{3}) z^{-1} = \frac{1}{2} (-1 + i \sqrt{3}) r (z) .$

 

[fresh_42]

I already apologized for not being clear. What we have is $\langle\mathcal{I} \rangle = V_4$ and $\langle\mathcal{J} \rangle=\mathbb{Z}_3$ and I wanted to combine them to a semidirect product, i.e. a group with $12$ elements, $A_4$ in this case, i.e. $\varphi\, : \,\mathbb{Z}_3 \longrightarrow \operatorname{Aut}(V_4)$ by conjugation $\varphi(w)(s)=wsw^{-1}$.

Beside the unfortunate parenthesis in post #162, which meat "only $z$" is affected, I don't see a problem.

The only failure was, that I should have added $s(c\cdot z) = c s(z)$ for $s\in \langle I \rangle$ and $w$ being a left multiplication by a constant factor for all $w\in \langle J \rangle$.

 

[julian]

I'm confused. In post #162 you basically told me that I should not interpret $r (u (z))$ as meaning

$r (u (z)) = r \left( \frac{1}{2} (-1 + i \sqrt{3}) z \right) = \left( \frac{1}{2} (-1 + i \sqrt{3}) z \right)^{-1} = - \frac{1}{2} (1 + i \sqrt{3}) z^{-1}$

but should interpret as this instead:

$r (u (z)) = r \left( \frac{1}{2} (-1 + i \sqrt{3}) z \right) = \frac{1}{2} (-1 + i \sqrt{3}) z^{-1} = \frac{1}{2} (-1 + i \sqrt{3}) r (z) .$

The point I'm making is that in post #162 you told me not to interpret $r ( u (z))$ as usual function composition! Well then how am I to interpret $a (b (z))$ generally?

 

[fresh_42]

The point I'm making is that in post #162 you told me not to interpret $r ( u (z))$ as usual function composition! Well then how am I to interpret $a (b (z))$ generally?

Well, it can be defined as a function, only that $s(cz^\varepsilon):=cs(z^\varepsilon)$ for $s\in \langle \mathcal{I}\rangle$ and $\varepsilon =\pm 1$, simply because $s(1)$, resp. $s(c)$ isn't defined, and I didn't say that it should be extended on $\mathbb{C}$, so I only failed to say, how else it has to be defined, if not as such an extension. The formulation $u := L_{\frac{1}{2}(-1+i\sqrt{3})}$ and $v:=L_{-\frac{1}{2}(1+i\sqrt{3})}$ where $L_c$ notes the left multiplication with $c$ would have made it clear what to do with $\langle \mathcal{J} \rangle$.
I was so focused on the group that I forgot to drop a few words on the functions.

 

[julian]

I am able to form a twelve dimensional group via ordinary function composition (maybe not the group you had in mind). And then able to prove that $J$ is a normal subgroup of this group. Not sure how anything is wrong. Could you have a look.

Determining the groups:

The group $I = \langle \mathcal{I} \rangle$:

From

\begin{align*}
(q \circ q) (z) & = -q (-z) = z = p (z) \\
(q \circ r) (z) & = q (z^{-1}) = - z^{-1} = s (z) = (r \circ q) (z) \\
(q \circ s) (z) & = q (-z^{-1}) = z^{-1} = r (z) = (s \circ q) (z) \\
(r \circ r) (z) & = r (z^{-1}) = z = p (z) \\
(s \circ s) (z) & = s (- z^{-1}) = z = p (z) \\
\end{align*}

we see that we have closure. We obviously have inverses to every element is they are involutions.

Associativity:

Associativity, and this will be true generally here, follows from

\begin{align*}
[(a \circ b) \circ c] (z) & = (a \circ b) \big ( c (z) \big) \\
& = a \big( b \big( c (z) \big) \big) \\
& = a \big( (b \circ c) (z) \big) \\
& = [a \circ (b \circ c)] (z) \\
\end{align*}

where $a$, $b$, and $c$ are any of the maps that we will encounter in the question.

therefore we have all the properties of a group. The group table for $I = \langle \mathcal{I} \rangle$ is:

\begin{array}{c|c|c|c}
& p & q & r & s \\
\hline p & p & q & r & s \\
\hline q & q & p & s & r \\
\hline r & r & s & p & q \\
\hline s & s & r & q & p \\
\end{array}

The group $J = \langle \mathcal{J} \rangle$:

Consider the compositions:

\begin{align*}
v \circ u & = v \big( - {1 \over 2} (- 1 + i \sqrt{3}) z \big) = - {1 \over 2} (1 + i \sqrt{3}) \cdot {1 \over 2} (-1 + i \sqrt{3}) \cdot z = z = 1 (z) \\
u \circ v & = u \big( - {1 \over 2} (1 + i \sqrt{3}) z \big) = {1 \over 2} (-1 + i \sqrt{3}) \cdot - {1 \over 2} (1 + i \sqrt{3}) z = z = 1 (z) \\
u \circ u & = u \big( {1 \over 2} (-1 + i \sqrt{3}) z \big) = {1 \over 2} (-1 + i \sqrt{3}) {1 \over 2} (-1 + i \sqrt{3}) z = - {1 \over 2} (1 + i \sqrt{3}) = v (z) \\
v \circ v & = v \big( - {1 \over 2} (1 + i \sqrt{3}) z \big) = - {1 \over 2} (1 + i \sqrt{3}) \cdot - {1 \over 2} (1 + i \sqrt{3}) z = {1 \over 2} (-1 + i \sqrt{3}) = u (z)
\end{align*}

We have closure under consecutive applications of the operations:

\begin{align*}
1 & = u \circ v = v \circ u \\
& u \\
& v \\
\end{align*}

which includes the identity map.

Inverses:

We have that $u$ is the inverse of $v$, and $v$ is the inverse of $u$.

Associativity:

Associativity has already been established, it is just due to the nature of composition.

The group table for $J = \langle \mathcal{J} \rangle$ is:

\begin{array}{c|c|c|c}
& 1 & u & v \\
\hline 1 & 1 & u & v \\
\hline u & u & v & 1 \\
\hline v & v & 1 & u \\
\end{array}

The group $F = \langle \mathcal{F} \rangle$:

The set of functions we obtain if we combine any of the functions in $\mathcal{I}$ and $\mathcal{J}$ by consecutive applications is denoted by $\mathcal{F}=\langle\mathcal{I},\mathcal{J} \rangle$. The group associated with $\mathcal{F}$ is denoted $F = \langle \mathcal{F} \rangle$.

In order to find $\mathcal{F}$ first consider the composite operations:

\begin{align*}
(r \circ u) (z) & = [u (z)]^{-1} = {2 \over (-1 + i \sqrt{3}) z} = - {1 \over 2} (1 + i \sqrt{3}) {1 \over z} \\
(r \circ v) (z) & = [v (z)]^{-1} = - {2 \over (1 + i \sqrt{3}) z} = {1 \over 2} (-1 + i \sqrt{3}) {1 \over z} \\
(s \circ u) (z) & = - [u (z)]^{-1} = - {2 \over (-1 + i \sqrt{3}) z} = - {1 \over 2} (1 + i \sqrt{3}) {1 \over z} \\
(s \circ v) (z) & = - [v (z)]^{-1} = {2 \over (1 + i \sqrt{3}) z} = - {1 \over 2} (-1 + i \sqrt{3}) {1 \over z} \\
\end{align*}

From the functions in $\mathcal{I}$ and $\mathcal{J}$ and the functions

\begin{align*}
(q \circ u) (z) & = - {1 \over 2} (-1 + i \sqrt{3}) z \quad \text{map denoted } qu \\
(q \circ v) (z) & = {1 \over 2} (1 + i \sqrt{3}) z \quad \text{map denoted } qv \\
(r \circ u) (z) & = - {1 \over 2} (1 + i \sqrt{3}) {1 \over z} \quad \text{map denoted } ru \\
(s \circ u) (z) & = {1 \over 2} (1 + i \sqrt{3}) {1 \over z} \quad \text{map denoted } su \\
(r \circ v) (z) & = {1 \over 2} (-1 + i \sqrt{3}) {1 \over z} \quad \text{map denoted } rv \\
(s \circ v) (z) & = - {1 \over 2} (-1 + i \sqrt{3}) {1 \over z} \quad \text{map denoted } sv \\
\end{align*}

one can verify the following "multiplication" table:

\begin{array}{c|c|c|c}
& 1 & q & r & s & u & v & qu & qv & ru & rv & su & sv \\
\hline 1 & 1 & q & r & s & u & v & qu & qv & ru & rv & su & sv \\
\hline q & q & 1 & s & r & qu & qv & u & v & su & sv & ru & rv \\
\hline r & r & s & 1 & q & ru & rv & su & sv & u & v & qu & qv \\
\hline s & s & r & q & 1 & su & sv & ru & rv & qu & qv & u & v \\
\hline u & u & qu & rv & sv & v & 1 & qv & q & r & ru & s & su \\
\hline v & v & qv & ru & su & 1 & u & q & qu & rv & r & sv & s \\
\hline qu & qu & u & sv & rv & qv & q & v & 1 & s & su & r & ru \\
\hline qv & qv & v & su & ru & q & qu & 1 & u & sv & s & rv & r \\
\hline ru & ru & su & v & qv & rv & r & sv & s & 1 & u & q & qu \\
\hline rv & rv & sv & u & qu & r & ru & s & su & v & 1 & qv & q \\
\hline su & su & ru & qv & v & sv & s & rv & r & q & qu & 1 & u \\
\hline sv & sv & rv & qu & u & s & su & r & ru & qv & q & v & 1 \\
\end{array}

where "multiplication" is composition of functions. This multiplication table forms a group multiplication table as we have closure under composition and inverses. As we have closure under composition we have

\begin{align*}
\mathcal{F} & = \big\{ z\stackrel{p}{\mapsto} z\; , \; z\stackrel{q}{\mapsto} -z\; , \;z\stackrel{r}{\mapsto} z^{-1}\; , \;z\stackrel{s}{\mapsto}-z^{-1} , \\
& \qquad z\stackrel{u}{\longmapsto}\frac{1}{2}(-1+i \sqrt{3})z\; , \;z\stackrel{v}{\longmapsto}-\frac{1}{2}(1+i \sqrt{3})z, \\
& \qquad z\stackrel{qu}{\longmapsto}- \frac{1}{2} (-1 + i \sqrt{3}) z \; , \; z\stackrel{qv}{\longmapsto} \frac{1}{2} (1 + i \sqrt{3}) z, \\
& \qquad z\stackrel{ru}{\longmapsto}- \frac{1}{2} (1 + i \sqrt{3}) \frac{1}{z} \; , \; z\stackrel{su}{\longmapsto} \frac{1}{2} (1 + i \sqrt{3}) \frac{1}{z} \\
& \qquad z\stackrel{rv}{\longmapsto} \frac{1}{2} (-1 + i \sqrt{3}) \frac{1}{z} \; , \; z\stackrel{sv}{\longmapsto}- \frac{1}{2} (-1 + i \sqrt{3}) \frac{1}{z}
\big\}
\end{align*}

and the above group multiplication table is for the group denoted by $F$.Verifying that $I = \langle \mathcal{I} \rangle$ is not a normal subgroup of $F = \langle \mathcal{F} \rangle$:Definition: Let $F$ be a group. A subgroup $N$ of $F$ is normal if $a^{-1} n a \in N$ for every $n \in N$ and every $a \in F$.

It appears that $I$ is a not a normal subgroup. We verify that $g^{-1} i g \not\in I$ for some $i \in I$ and for some $g \in F$. When $g \in I$ we obviously have $g^{-1} i g \in I$ as $I$ is itself a group. We check the other cases:

\begin{align*}
u^{-1} \circ q \circ u & = v \circ qu = q \\
v^{-1} \circ q \circ v & = u \circ qv = q \\
u^{-1} \circ r \circ u & = v \circ ru = rv \; \leftarrow \\
v^{-1} \circ r \circ v & = u \circ rv = vr \; \leftarrow \\
u^{-1} \circ s \circ u & = v \circ su = sv \; \leftarrow \\
v^{-1} \circ s \circ v & = u \circ sv = su \; \leftarrow \\
\end{align*}

and

\begin{align*}
qu^{-1} \circ q \circ qu & = qv \circ q \circ qu = q \\
qv^{-1} \circ q \circ qv & = qu \circ q \circ qv = q \\
qu^{-1} \circ r \circ qu & = qv \circ r \circ qu = rv \; \leftarrow \\
qv^{-1} \circ r \circ qv & = qu \circ r \circ qv = ru \; \leftarrow \\
qu^{-1} \circ s \circ qu & = qv \circ s \circ qu = sv \; \leftarrow \\
qv^{-1} \circ s \circ qv & = qu \circ s \circ qv = su \; \leftarrow \\
\end{align*}

and

\begin{align*}
ru^{-1} \circ q \circ ru & = ru \circ q \circ ru = q \\
rv^{-1} \circ q \circ rv & = rv \circ q \circ rv = q \\
ru^{-1} \circ r \circ ru & = ru \circ r \circ ru = rv \; \leftarrow \\
rv^{-1} \circ s \circ rv & = rv \circ r \circ rv = ru \; \leftarrow \\
ru^{-1} \circ r \circ ru & = ru \circ s \circ ru = sv \; \leftarrow \\
rv^{-1} \circ s \circ rv & = rv \circ s \circ rv = su \; \leftarrow \\
\end{align*}

and

\begin{align*}
su^{-1} \circ q \circ su & = su \circ q \circ su = q \\
sv^{-1} \circ q \circ sv & = sv \circ q \circ sv = q \\
su^{-1} \circ r \circ su & = su \circ r \circ su = rv \; \leftarrow \\
sv^{-1} \circ r \circ sv & = sv \circ r \circ sv = ru \; \leftarrow \\
su^{-1} \circ s \circ su & = su \circ s \circ su = sv \; \leftarrow \\
sv^{-1} \circ s \circ sv & = sv \circ s \circ sv = su \; \leftarrow \\
\end{align*}

We have thus verified that $I$ is not a normal subgroup. As such $F / I$ does not form a group as multiplication cannot be defined - this will be explained in detail below..

Verifying that $J = \langle \mathcal{J} \rangle$ is a normal subgroup of $F = \langle \mathcal{F} \rangle$:

Next we establish that $J$ is a normal subgroup. We verify that $g^{-1} j g \in J$ for every $j \in J$ and every $g \in F$. This obviously holds for $g \in J$. We check the other cases

\begin{align*}
r^{-1} \circ u \circ r & = ru \circ r = v \\
q^{-1} \circ u \circ q & = qu \circ q = u \\
s^{-1} \circ u \circ s & = su \circ s = v \\
r^{-1} \circ v \circ r & = rv \circ r = u \\
q^{-1} \circ v \circ q & = qv \circ q = v \\
s^{-1} \circ v \circ s & = sv \circ s = u \\
\end{align*}

and

\begin{align*}
qu^{-1} \circ u \circ qu & = qv \circ u \circ qu = u \\
qv^{-1} \circ u \circ qv & = qu \circ u \circ qv = u \\
qu^{-1} \circ v \circ qu & = qv \circ v \circ qu = v \\
qv^{-1} \circ v \circ qv & = qu \circ v \circ qv = v \\
\end{align*}

and

\begin{align*}
ru^{-1} \circ u \circ ru & = ru \circ u \circ ru = v \\
rv^{-1} \circ u \circ rv & = rv \circ u \circ rv = v \\
ru^{-1} \circ v \circ ru & = ru \circ v \circ ru = u \\
rv^{-1} \circ v \circ rv & = rv \circ v \circ rv = u \\
\end{align*}

and

\begin{align*}
su^{-1} \circ u \circ su & = su \circ u \circ su = v \\
sv^{-1} \circ u \circ sv & = sv \circ u \circ sv = v \\
su^{-1} \circ v \circ su & = su \circ v \circ su = u \\
sv^{-1} \circ v \circ sv & = sv \circ v \circ sv = u \\
\end{align*}

We have thus verified that $J$ is a normal subgroup. As such $F / J$ forms a group (the quotient group) as I will explain in a moment. First we need a preliminary result:

Let $F$ be a group and let $J$ be a subgroup of $F$. Then $J$ is a normal subgroup of $F$ if and only if $a J = J a$ for every $a \in F$.

Proof:

Assume $J$ is a normal subgroup of $F$. Let $a \in F$ (we will show that $a J = J a$).

First we will prove that $a J \subseteq J a$. Let $x \in a J$. Then $x = a j$ for some $j \in J$. By assumption $a j a^{-1} = (a^{-1})^{-1} j a^{-1} \in J$, so $x = a j = (a j a^{-1}) a \in J a$. Next we prove that $J a \subseteq aJ$. Let $x = j a$ for some $j \in J$. By assumption, $a^{-1} j a \in J$, so $x = j a = a (a^{-1} j a) \in a J$. Therefore $a J = J a$, as desired.

Now assume that $a J = J a$ for every $a \in F$ (we will show that $J$ is a normal subgroup of $F$). Let $j \in J$, $a \in F$. We have $j a \in J a = a J$, so $j a = a k$ for some $k \in J$. So $a^{-1} j a = k \in J$. The desired result.
We need the notion of a left coset: the left coset containing $a \in F$ is the set $a J = \{ a j : j \in J \}$. We give some basic properties of left cosets:

Since a row of a group multiplication table contains each element of $F$ exactly once, the elements in any left coset must all be different. Thus the number of elements in each left coset is equal to the number as elements in $J$.

Either $a J \cap b J = \emptyset$ or $a J = b J$, that is, either left cosets do not overlap or are equal. Suppose that the cosets in rows $a$ and $b$ overlap then $a j_1 = b j_2$ for some $j_1, j_2 \in J$. Therefore $a = b j_2 j_1^{-1}$ and if $j$ is any element of $J$ then $a j = b j_2 j_1^{-1} j = b j'$ where $j' = j_2 j_1^{-1} j$. Since $J$ is a subgroup, $j'$ is an element of $J$. Therefore $a J \subseteq b J$. Similarly we can show $b J \subseteq a J$. Thus $a J = b J$. In this way the set $F$ can be decomposed into mutually disjoint left cosets.

If $a J = J$ then $a \in J$ for if $a \not\in J$ we would have that $a 1 \not\in J$ despite $1$ being in $J$.

We obviously have $1 J = j J$ for every $j \in J$.

The quotient group - general idea:

Here I explain why $F / J$ forms a group when $J$ is a normal subgroup of $F$.

Let $F$ be a group and $J$ be a subgroup. Let $F / J = \{ a J : a \in F \}$ be the set of left cosets of $J$ in $F$. The operation:

\begin{align*}
F / J \times F / J \rightarrow F / J \qquad \text{defined by} \qquad (a J) \cdot (b J) = (ab) J
\end{align*}

is well defined (that is, $a J = a' J$ and $b J = b' J$ then $ab J = a'b' J$) if and only if $J$ is a normal subgroup of $F$.

Proof:

Now we return to proving that multiplication is well defined if and only if $J$ is normal. Suppose $J$ is normal. If $a J = a' J$ and $b J = b' J$ then

$(ab) J = a (b J) = a (b' J) = a (J b') = (aJ) b' = (a' J) b' = a' (J b') = a' (b' J) = (a'b') J$

so the operation is well defined and multiplication is closed.

Now suppose that the operation is well defined so that whenever $a J = a' J$ and $b J = b' J$ that $ab J = a'b' J$. We want to show that $a^{-1} j a \in J$ for every $j \in J$ and for every $a \in F$. For each $j \in J$, since $j J = 1 J$, we have $(1 J) (a J) = 1 a J = a J$ and so $(j J) (a J) = ja J$. So $a J = j a J$, hence $J = a^{-1} j a J$, so $a^{-1} j a \in J$ for each $j \in J$. This holds for any $a \in F$, hence $J$ is a normal subgroup of $F$.

One checks that this operation on $F / J$ is associative:

$(aJ b J) c J = ab J c J = (ab) c J = a (bc) J = aJ bc J = a J (b J c J) .$

Has identity element $J$

$a J 1 J = a 1 J = a J = 1 J a J$

and the inverse of an element $a J$ of $F /J$ is $a^{-1} J$:

$(a^{-1} J) (a J) = a^{-1} a J = e J \qquad \text{and} \qquad (a J) (a^{-1} J) = a a^{-1} J = e J$

 

[fresh_42]

I am able to form a twelve dimensional group via ordinary function composition (maybe not the group you had in mind). And then able to prove that JJJ is a normal subgroup of this group. Not sure how anything is wrong. Could you have a look.

I am very sorry, Julian, but you are completely right!

My mistake was that I thought I had defined $A_4= \mathbb{Z}_3 \ltimes V_4$, but this is wrong.

We have indeed $\mathcal{F}= D_6 \cong D_3 \times \mathbb{Z}_2 \cong \mathbb{Z}_3 \rtimes V_4$ and the representation in terms of our functions is $D_6 = \langle \, uq\, , \,r\,|\,(uq)^6=r^2=1\, , \,r(uq)r=(uq)^{-1}=qv\, \rangle$ where $u=(uq)^4\, , \,v=(uq)^2\, , \,q=(uq)^3\, , \,s=(uq)^3r\,.$

P.S.: It took me quite a while to find an element of order $6$. I thought there was none.

 

[julian]

I'm right?! Cool beans.

 

[lpetrich]

I crunched that group with a group-theory calculations Mathematica notebook, and it had commutator series of quotient groups {Z2*Z2, Z3}. I'd expected A4, with {Z3,Z2*Z2}, but this looks like Dih(6) = Z2*Dih(3). So I decided to generalize it.
$$ F = \{z \to \omega^k z \text{ and } z \to \omega^k (1/z) \text{ for } k = 0 \dots n-1\} $$
where $\omega$ is a primitive nth root of unity for n. The generators of the group are
$$ a = (z \to \omega z) \text{ and } b = (z \to (1/z)) $$
Element a has order n and b has order 2, with $ab = ba^{n-1}$. Thus, this group is Dih(n), the dihedral group with order 2n, and I get the result that I had earlier conjectured.

 

[DEvens]

24. Solve and describe the solution step by step in quadrature a Lagrangian differential equation with Lagrangian
$$L(t,x,\dot x)=\frac{1}{2}\dot x^2-\frac{t}{x^4}.$$

What do you mean by quadrature? I can find several potential meanings that could be relevant. For example, it could mean doing an integral. It could mean a particular method of numerical solution. It could mean something about fitting things to a rectangle. It could mean something about a quadratic equation.

 

[fresh_42]

What do you mean by quadrature? I can find several potential meanings that could be relevant. For example, it could mean doing an integral. It could mean a particular method of numerical solution. It could mean something about fitting things to a rectangle. It could mean something about a quadratic equation.

This problem was contributed by @wrobel so we have to ask him. I guess it only refers to the quadratic form of the equation. The clue is to use the symmetry.