Exercise on Maximal Group

Exercise on Maximal Group

12:30:00   Algebraic,

Well, this one is actually an old random stuff I did when I was undergraduate, I found it on my old harddrive and would like to test what happen if I post it. This is actually an exercise in Martin Isaac Book, and I always forget how to do  this problem properly (I mean really solve it concretely). In fact, I post it here again because I always forget how to solve it. This is a really detailed proof.


Without using Cauchy Theorem or Sylow Theorem prove that If $G$ is a finite group with unique maximal subgroup then $|G|=p^{\alpha}$, for some prime $p$.

We begin with some lemmas, and all the following lemmas assume that $G$ is finite.

Lemma 1
Let $A$ be a nonempty subset of $G$ such that $\langle A \rangle \neq G$ and for any $x\in G \backslash A$ we have $\langle A \cup \{x\} \rangle=G$ then $\langle A \rangle$ is a maximal subgroup.

Proof:
Let $H$ be a subgroup of $G$ such that $\langle A \rangle \subseteq H \subseteq G$, suppose for contrary that $H \neq \langle A \rangle $ and $H \neq G$,  since $\langle A\rangle \neq G$ there exists $x \in H$ with $x \not \in \langle A \rangle $, therefore $x\not \in A$ thus $A \cup \{x\} \subseteq \langle A \rangle \cup \{x\} \subseteq H$ and hence $\langle A \cup \{x\} \rangle \subseteq H$, but by the hypothesis this means $G \subseteq H$, and we have $H=G$, a contradiction.  Therefore either $H=\langle A \rangle$ or $H=G$, which implies $\langle A \rangle$ is maximal subgroup.


Lemma 2
Let $A$ and $B$ are subgroup of $G$. If $G = A \cup B$ then $A=G$ or $B=G$.

Proof:
Let $|G|=n$, $|A|=d_1$, $|B|=d_2$, and $|A\cap B| = d_3$. First we assume that $d_1 \geq d_2$, by Langrange theorem we have $d_1 \lvert n$ and $d_2 \lvert n$, also since $e \in A\cap B$ we have $d_3 \geq 1$. Observe that
\[n=|G|=|A\cup B| = |A| + |B| - |A \cap B| \leq d_1+d_1 -1 = 2d_1-1.\]
If $d_1= n$, then $|A|=|G|$, but since $A \subseteq G$, we have $A=G$, the assertion is proved. Now for $d_1 < n$ since by Langrange $d_1 | n$, we have $n=k d_1$ for a positive integer $k$, but $n \leq 2d_1-1$, thus $k \leq 2- \frac{1}{d_1}$, and we have $k \leq 1$, this leads to $n=d_1$ again. Similarly If we instead assume $d_2 \geq d_1$ then $B=G$.


Lemma 3
If $G$ has a unique maximal subgroup then $G$ is a cyclic group.

Proof:
Let $M$ be the unique maximal subgroup of $G$, let $C(M)=\langle G\backslash M \rangle$, then we have $G = M \cup C(M)$ thus by Lemma 2  we have  $M=G$ or $C(M)=G$, but $M\neq G$ therefore we must have $C(M)=G$.


Define
\[\begin{eqnarray*}&\mathcal{T}_i := \{A \, \lvert A \subseteq G\backslash M \mbox{ and } \lvert A \rvert=i \mbox{ and } i \geq 1 \} \\ &\mathfrak{L} := \{i \, \lvert \, \langle A \rangle = G \mbox{ for all } A \in \mathcal{T}_i\}
\end{eqnarray*}\]

Let $\lvert G \backslash M \rvert = \lambda$, we have \[\mathcal{T}_{\lambda} = \{A \, \lvert A \subseteq G \backslash M \mbox{ and } |A|=\lambda=\lvert G \backslash M \rvert\}=\{ G \backslash M\}\]


Since $\langle G \backslash M \rangle = G$, this means that for every $A \in \mathcal{T}_{\lambda}=\{G \backslash M\}$ we have $\langle A \rangle= \langle  G \backslash M \rangle =G$, therefore $\lambda \in \mathfrak{L}$, in particular $\mathfrak{L} \neq \emptyset$.


Since $\mathfrak{L}$ is nonempty, by the Well-Ordering Principle there exists a smallest elemet $t \in \mathfrak{L}$. If $t=1$, since $1=t \in \mathfrak{L}$, for every $A \in \mathcal{T}_1$ we have $\langle A \rangle =G$ , in particular if we take $A=\{g\}$ where $g\in G\backslash M$, we see that $A \subset G\backslash M$ and $\lvert A \rvert = 1=t$. This means $A\in \mathcal{T}_1$, so that $\langle g\rangle = \langle A \rangle =G$, thus $G$ is cyclic, and the lemma is proved whenever $t=1$.


Next we will prove that $t > 1$ would eventually end with a contradiction. If $t>1$, then $t-1$ is the largest positive integers which fails to be included on $\mathfrak{L}$, equivalently there exists $S \in \mathcal{T}_{t-1}$ with $\langle S \rangle  \neq G$, notice that $|S|=t-1 >0$ thus $S\neq \emptyset$. Define
\[\begin{eqnarray*}&\mathcal{U}_i := \{ B \, \lvert  B \subseteq M \mbox{ and } |B|=i \mbox{ and } i \geq 1 \} \\ &\mathfrak{P} := \{i \, \lvert \, \langle S \cup B \rangle = G \mbox{ for all } B \in \mathcal{U}_i\}
\end{eqnarray*}\]

Observe that $M \subset S \cup M \subseteq \langle S \cup M \rangle \subset G$, since $M$ is maximal we have either $M=\langle S \cup M \rangle$ or $\langle S \cup M \rangle = G$, but $M=\langle S \cup M \rangle$ would implies $S \subset \langle S \cup M \rangle = M$ this is impossible since $S\in \mathcal{T}_1$ which means $S \subset G \backslash M$, thus the only possibility is $\langle S \cup M \rangle=G$. Let $\lvert M \rvert=\mu$, we have
\[\mathcal{U}_{\mu} = \{B \, \lvert B \subseteq M \mbox{ and } \lvert  B \rvert = \mu = \lvert M \rvert\} = \{ M\}\]


Since $\langle S \cup M \rangle=G$,  this means that for every $ B \in \mathcal{U}_{\mu}=\{M\}$ we have $\langle S \cup B \rangle= \langle  S \cup M \rangle =G$, therefore $\mu \in \mathfrak{P}$, in particular $\mathfrak{P} \neq \emptyset$. Since $\mathfrak{P}$ is nonempty, by Well-Ordering Principle there exists a smallest element $l \in \mathfrak{P}$.


 Before we proceed, we will recall what we have so far :
 \[S \subset G \backslash M \qquad S \in \mathcal{T}_{t-1} \qquad  \langle S \rangle \neq G\]
Also  $t\in \mathfrak{L}$ and satisfies
\[\text{for all $A \in \mathcal{T}_t$ we have $\langle A \rangle = G$}\]
and $l\in \mathfrak{P}$ and satisfies
\[\text{for all $B \in \mathcal{U}_l$ we have $\langle S \cup B \rangle = G$}\]


 Now we continue the proof, if $l=1$, since $1=l \in \mathfrak{P}$ we have $\langle S \cup B \rangle =G$ for every $ B \in \mathcal{U}_1$, we will prove that this would implies $\langle S \rangle $ is maximal subgroup. Indeed, take any $x\in G \backslash S$, which implies $x\not \in S$, we divide into two cases:

 i) If $x\in M$ then $\{x\} \subset M$, thus $ \{x\} \in \mathcal{U}_1$ and hence $\langle S \cup \{x\}\rangle =G$.

ii) If $x \not \in M$ then $\{x\}\subset G \backslash M$, thus $S \cup \{x\} \subset G \backslash M$, also since $x\not \in S$ we have $S \cap \{x\} = \emptyset$, together with $S \in \mathcal{T}_{t-1}$ we have $\lvert S \cup \{x\}\rvert = \lvert S \rvert + 1 = t-1+1=t$, therefore $S \cup \{x\} \in \mathcal{T}_t$. Since $t\in \mathfrak{L}$ every $A \in \mathcal{T}_t$ satisfies $\langle A \rangle =G$ thus in particular $S \cup \{x\} \in \mathcal{T}_t$ implies $\langle S \cup \{x\} \rangle = G$.


In summary for every $x\in G \backslash S$, we have $\langle S \cup \{x\} \rangle = G$, by Lemma 1 this means $\langle S\rangle $ is maximal subgroup, and since $M$ is the only maximal subgroup we have $\langle S\rangle=M$, but then $S \subset M $ also we have $S\subset G \backslash M$, thus $S=\emptyset$, contradiction,
, hence $l \neq1$.


If $l>1$, then $l-1$ is the largest positive integers which fails to be included on $\mathfrak{P}$, equivalently there exists $V \in \mathcal{U}_{l-1}$ with $\langle S \cup V \rangle \neq G$. We will prove that $\langle S \cup V \rangle$ is maximal subgroup. Indeed, take any $x \in G \backslash \left(S \cup V\right) = G\backslash S \cap G \backslash V$, which implies $x\not \in S$ and $s \not \in V$
we divide into two cases:

i)  If $x\in M$, note that $V \in \mathcal{U}_{l-1}$ means that $V \subset M$ and $\lvert V \rvert = l-1$,  therefore $V \cup \{x\} \subset M$. And since $x \not \in V$ we have $V \cap \{x\}=\emptyset$, thus $\lvert V \cup \{x\}\rvert = \lvert V \rvert + 1 = l-1+1=l$, therefore $V \cup \{x\} \in \mathcal{U}_{l}$, since $l \in \mathfrak{P}$ we have $\langle S \cup V \cup \{x\}\rangle= G$.

ii)  If $x\not \in M$ then $x \in G \backslash M$, note that $S \in \mathcal{T}_{t-1}$ means that $S \subset G\backslash M$ and $\lvert S \rvert = t-1$, therefore $S \cup \{x\} \subset G\backslash M$. And since $x \not \in S$ we have $S \cap \{x\}=\emptyset$, thus $\lvert S\cup \{x\} \rvert=\lvert S \rvert + 1=t-1+1=t$, therefore $S \cup \{x\} \in \mathcal{T}_t$, since $t\in \mathfrak{P}$ we have $\langle S \cup \{x\} \rangle =G$. Furthermore we have $G=\langle S \cup \{x\} \rangle \subset \langle S \cup V \cup \{x\} \rangle$, which implies $\langle S \cup V \cup \{x\}\rangle= G$.


In summary, for every $x \in G \backslash \left(S \cup V\right)$, we have $\langle S \cup V \cup \{x\}\rangle= G$, by Lemma 1 this means $\langle S \cup V \rangle $ is maximal subgroup, and since $M$ is the only maximal subgroup we have $\langle S \cup V \rangle=M$, but then $S \subset \langle S \cup V \rangle  =  M$, by the fact that $S\subset G \backslash M$, we would have $S=\emptyset$, contradiction. And the proof is completed.

Now we have $G=\langle g \rangle$  by Lemma 3, let $|G|=n=p_1^{\alpha_1} \cdots p_j^{\alpha_j}$, where $p_1,\cdots,p_j$ are all prime numbers. Consider
\[H_{p_k} = \langle g^{p_k} \rangle\]
We will prove that $H_{p_k}$ is maximal subgroup for every $k$. Indeed, if S is a subgroup with $S \neq G$ and $S \neq H_{p_k}$ satisfies $H_{p_k} \subseteq S \subseteq G$, then $d_k=|H_{p_k}|$ divides $|S| \leq  |G|$, but $|H_{p_k}|=\frac{n}{p_k}$, therefore
\[\frac{n}{p_k} \leq  |S| \leq  n \qquad \text{and} \qquad \text{$\frac{n}{p_k}$ divides $\lvert S\rvert$ divides $n$}\]

So that there exist $u, v\in \mathbb{N}$ such that $n= v  |S|$ and  $|S|=u \left( \frac{n}{p_k}\right)$, thus $n= (u)\, (v) \left(\frac{n}{p_k}\right)$, which implies $p_k = u  v$, since $p_k$ is prime we should have $u=1$ or $v=1$, but then $|H_{p_k}|=|S|$ or $|S|=|G|$ which together with $H_{p_k} \subseteq S \subseteq G$ would result to $H_{p_k}=S$ or $S=G$, thus $H_{p_k}$ is a maximal subgroup for every $k$.

Since $G$ has only one maximal subgroup that is $M$, we should have $H_{p_k} =M= H_{p_l}$ for all $k$ and $l$, therefore $\frac{n}{p_k}=|H_{p_k}|=|H_{p_l}|=\frac{n}{p_l}$ so that $p_k=p_l$ for all $k$ and $l$, thus $|G|$ actually has only one prime factor and we conclude that $|G|=p^{\alpha}$.

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Part I of Explicit Construction of the Elliptic Curve (CIMPA School)

Part I of Explicit Construction of the Elliptic Curve (CIMPA School)

08:53:00   Algebraic,

The following is based on Stevenhaagen's talk on 2013 CIMPA School in Manila. This write is a little bit mess, as actually a lot of things need to be proved. I'm just trying to connect the preliminaries introduced so far.

Recall from my previous post  a Weierstrass Elliptic Function is given by

\[\wp(z)=\frac{1}{z^2} + \sum_{\omega \in A/\{0\}} \left(\frac{1}{(z-\omega)^2} - \frac{1}{\omega^2}\right)\]

By the doubly periodicity, let's take the restriction of the domain inside a fundamental lattice $L$ .  This function also satisfies the differential equation

\[{\wp^{\prime}(z)}^2 = 4 \wp(z)^3-g_2(L)\wp(z)-g_3(L)\]

where $g_1(L)$ and $g_2(L)$ are modular forms given by $g_2(L)=60 \sum_{\omega \in L/\{0\}} \frac{1}{\omega^4}$ and $g_3(L)=140 \sum_{\omega \in L/\{0\}} \frac{1}{\omega^6}$.

Therefore we can define the parametrization of a torus $T:=\mathbb{C} \pmod L$ to a Projective Plane by the equation of  Elliptic curve
\[y^2=4x^3-g_2(L)x-g_3(L)\]
with dehomogenization \[y^2z=4x^3-g_2(L)xz^2-g_3(L)z^3\]

The map is given by $z \mapsto (\wp(z),\wp^{\prime}(z),1)$  and $0 \mapsto (0,1,0)$, this map is an injective function from $T$ to Elliptic Curve.

We also define the $j$-invariant of the Latice L by 

\[j(L)=1728\frac{g_2(L)^3}{g_2(L)^3-27g_3(L)^2}\]

this value is always defined whenever $L$ is indeed  a lattice ($g_2(L)^3-27g_3(L)^2 \neq 0$).

Now for a complex number $\tau$ in a Upper Half plane $\mathcal{H}$, define the lattice $L_{\tau}$ as the lattice generated by $1$ and $\tau$, we write $L_{\tau}=[1,\tau]$.  It can be shown that every lattice $L=[z_1,z_2]$ is isomorphic (or sometimes' called homothetic) to $L_{\tau}$  for some $\tau \in \mathcal{H}$ (this can be seen by setting $\tau=\frac{z_2}{z_1}$). Therefore we can view $j$ as the function $j : \mathcal{H} \rightarrow \mathbb{C}$ with

\[j(\tau)=j(L_{\tau}) = 1728\frac{g_2(L_{\tau})^3}{g_2(L_{\tau})^3-27g_3(L_{\tau})^2}\]

We call this function the $j$-function.  The key point is that this function is a modular form of weight 0, i.e modular function, i.e

\[j(A \cdot \tau) = j(\tau)\]

for $A \in SL_2(\mathbb{Z})$, and $A \cdot \tau$ denotes the Mobius action.

The condition that $\tau$ is on the Upper Half Plane is good for redefine the function $j$ on the smaller subset of $\mathcal{H}$ which is a  fundamental domain $\mathcal{D}$ given by

\[\mathcal{D}=\left\{\tau \in \mathcal{H} \, \lvert \, \Re(\tau) \in \left[-\frac{1}{2},\frac{1}{2}\right), \tau > 1 \right\} \cup \{\tau \in \mathcal{H} \, \lvert \, \Re(\tau) \geq 0 \mbox{ if } |\tau|=1 \}\] 

this fundamental domain will correspond each $\tau \in \mathcal{D}$ into $\tau=\frac{-b+\sqrt{b^2-4ac}}{2a}$, where $a,b$ and $c$ are coprime integer satisfying 

\[|b|\leq a \leq c \text{ and } b\geq 0  \text{ whenever } |b|=a \text{ or } a=c\]

We call the function $f(x,y)=ax^2+bxy+cy^2$ a reduced quadratic form whenever $(a,b,c)$ satisfying the above conditions. Therefore $\tau \in \mathcal{D}$ whenever $\tau$ is the (half-plane) root of $f(x,1)$ for a reduced quadratic form $f(x,y)$.

Now given a negative integer $D$,  we collect all the reduced quadratic form $f(x,y)=ax^2+bxy+cy^2$ such that $D=b^2-4ac$,  it turns out that this collection say $Cl(D)$ is a class group. More interesting is that we can adjoint $j(\tau)=j\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)$ to $K=\mathbb{Q}(\sqrt{D})$ where $D$ is a fundamental discriminant, such that $j(\tau)$ is algebraic in this extension!! We called this extension a Hilbert Class Field, and the minimum polynomial $H_K$ is called Hilbert Class Polynomial and it is in $\mathbb{Z}[X]$. 

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Early Algebraic Variety, Coordinate Ring

Early Algebraic Variety, Coordinate Ring

10:18:00   Algebraic,

I wrote this up, because i don't want to forget it, I made a pdf files about variety, but hasn't finished it yet... It is hard to keep things alive in your mind, if you don't get used to it a lot, after I manage to bear algebraic geometry in my faculty through a student skripsi while ago I managed to write some stuffs about variety in an undergraduate level (of course for those who have familiar with commutative algebra).. Now it's going to fade again, maybe because they are are not used with "many definitions" math (that is modern math :D)

Let's recall an important thing I wrote up in the previous post. For a closed set $X$ on the affine space $\mathbb{A}^n$, define the ideal

\[\mathfrak{A}_X :=\{f\in k[T_1,\cdots,T_n] | f(X)=0\}\]

The above is indeed an ideal, we can see it by construction of the map $\tau: k[T] \rightarrow k[X]$ where $k[X]$ is the ring of regular function.

Theorem

Let $X$ be a closed set of affine space $\mathbb{A}^n$, then the following are equivalent:
i)   $X$ be an irreducible closed set
ii)  $\mathfrak{A}_X$ is prime ideal
iii)  $k[X]$ is integral domain.

Recall that for the ring $R$ and Ideal $I$, $R/I$ is integral domain iff $I$ is prime ideal (it's standard exercise in undergraduate algebra). By the isomorphism applied on $\tau:k[T]\rightarrow k[X]$, we have $k[X]\cong k[T]/\mathfrak{A}_X$, so we have the equivalence of ii) and iii).
Now we want to prove the equivalence of iii) and i) by negating the statement first and obtain $X$ is reducible if and only if $k[X]$ is not integral domain, or equivalently $k[X]$ has zero divisors.

Suppose that $X$ is defined by polynomials $F_i$.  Let $r$ and $s$ be a zero divisor of $k[X]$, the set $A=\{y\in X | r(y)=0\}$ and $B=\{y\in X | s(y)=0\}$ define closed sets (be cautious that $r$ and $s$ are not necessary polynomials, but still $A$ and $B$ are closed sets), which are propers subset of $X$, hence $A\cup B \subset X$, furthermore for any $x\in X$ we have $s(x)r(x)=0$ therefore $x\in A$ or $x\in B$, that is $x\in A\cup B$, and we also have $X\subset A\cup B$. Therefore $X=A\cup B$, and $X$ is reducible.

Suppose that $X$ is reducible, then we can find the closed set $A$ and $B$ such that $X=A\cup B$,  consider $\mathfrak{A}_A$ and $\mathfrak{A}_B$, since $A\subset  X$ and $B \subset X$ we have by Theorem on previous post  $\mathfrak{A}_X \subset \mathfrak{A}_A$ and $\mathfrak{A}_X \subset \mathfrak{A}_B$. Since the subset is proper, we can take $F\in \mathfrak{A}_A$ and $G\in \mathfrak{A}_B$ such that $F,G\not\in \mathfrak{A}_X$, by this condition the regular function $f$ and $g$ in $k[X]$ whose representation is  $F$ and $G$ respectively are both nonzero elements of $k[X]$ (if it is zero then $F,G\in \mathfrak{A}_X$ contradicting our choice of $F$ and $G$).  Therefore $F(x)G(x)=0$ for all $x\in X$, thus $FG\in \mathfrak{A}_X$ and we conclude that  in $k[X]$ the regular function $fg$ associate to a zero function, and since $f$ and $g$ are both nonzero in $k[X]$ with $fg=0$  in $k[X]$, thus $f$ and $g$ are zero divisor.

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