# Challenge Math Challenge - November 2018

• Featured

#### julian

Gold Member
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

Mentor
2018 Award
$r(u(z))=r\left( \left( \dfrac{1}{2}\left( -1+i\sqrt{3} \right)z \right) \right)= \dfrac{1}{2}\left( -1+i\sqrt{3} \right)z^{-1} \neq -\dfrac{1}{2}\left( 1+i\sqrt{3} \right)z^{-1}$ .
Sorry, I should have mentioned that.

#### julian

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

Mentor
2018 Award
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

Gold Member
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

Mentor
2018 Award
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

Gold Member
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

Mentor
2018 Award
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

Gold Member
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

Mentor
2018 Award
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

Gold Member
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 $a^{-1} J$:

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

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#### fresh_42

Mentor
2018 Award
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.

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#### julian

Gold Member
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.

"Math Challenge - November 2018"

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