Challenge Math Challenge - November 2018

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julian

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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

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##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

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##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

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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

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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

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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

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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

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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

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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

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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

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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
##
 
Last edited:

fresh_42

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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.
 
Last edited:

julian

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I'm right?! Cool beans.
 
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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.
 

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