# Tag Archives: abstract algebra

## is it enough…

that the presentation given by $\langle A,B\ :\ A^2=B^3,\quad A^2B=BA^2,\quad A^4=e,\quad B^6=e\rangle$

determines the group $SL_2({\mathbb{Z}}/{3\mathbb{Z}})$?

Similar question for $\langle A,B,C$: $A^2=B^3$, $A^2B=BA^2$, $A^4=B^6= C^2=e$, $AC=CA,BC=CB\rangle$

for the group $GL_2({\mathbb{Z}}/{3\mathbb{Z}})$.

Other similar problems but less “difficult” are:

• $\langle\varnothing:\varnothing\rangle=\{e\}$
• $\langle A\ :\ A^2=e\rangle$  for $\mathbb{Z}_2$
• $\langle A\ :\ A^3=e\rangle$  for $\mathbb{Z}_3$
• $\langle A\ :\ A^4=e\rangle$  for $\mathbb{Z}_4$
• $\langle A,B\ :\ A^2=e, B^2=e, AB=BA\rangle$  for $\mathbb{Z}_2\oplus\mathbb{Z}_2$
• $\langle A,B\ :\ A^2=e, B^3=e, AB=B^2A\rangle$  for $S_3$
• $\langle A,B\ :\ A^2=e, B^3=e, AB=BA\rangle$  for $\mathbb{Z}_2\oplus\mathbb{Z}_3$
• $\langle A:\varnothing\rangle$ it is $\mathbb{Z}$
• $\langle A,B:AB=BA\rangle$  is $\mathbb{Z}\oplus\mathbb{Z}$
• $\langle A,B:\varnothing\rangle$  is $\mathbb{Z}*\mathbb{Z}$, the rank two free group
• $\langle A,B,J:(ABA)^4=e,ABA=BAB\rangle$ for $SL_2(\mathbb{Z})$
• $\langle A,B,J:(ABA)^4=J^2=e,ABA=BAB, JAJA=JBJB=e \rangle$ for $GL_2(\mathbb{Z})$
• $\langle A,B:A^2=B^3=e \rangle$ for $P\!S\!L_2(\mathbb{Z})=SL_2(\mathbb{Z})/{\mathbb{Z}_2}\cong{\mathbb{Z}}_2*{\mathbb{Z}}_3$
• dare altri venti esempi
Qual è il tuo preferito?

Filed under algebra, free group, group theory, mathematics

## SL_2(Z_3) again

this group is great,… if you don’t believe checkout:

http://finitegeometry.org/sc/pg/dt/visu.html

there, it is its supergroup GL_2(Z_3). the missing matrix in this photo is $\left(\begin{array}{cc}0&1\\ 2&2\end{array}\right)$

Filed under algebra, group theory, numbers, what is math

## local math brochures

Follow the links to get acquainted with the contents what we will work this semester.

They are

Also, let me feedback you mentioning the topics which can be get into it to work on thesis (B.Sc. or M.Sc.) or else

• differential geometry
• differential topology
• low dimensional topology
• algebra  and analysis
• sum of reciprocal inverses integers problem

These are fun  really!

Filed under math

## fourth Sylow theorem

We are going to reconstruct the salient of a part of the classical theorem:

Each $H$ p-S of $G$ is contained into a $Q$ p-SS of $G$

Here the reference frame:

1. $\#$p-SS divides $|G|$:

The set p-SS must be considered as an orbit of the action $G\times$ p-SS $\to$ p-SS via $g\cdot Q=gQg^{-1}$.  Since from sylow II we know that each two are conjugated the there is only one orbit

p-SS $={\rm Orb}_G(Q)\leftrightarrow G/{\rm St}_G(Q)$

giving us a trivial  orbital partition. Then $\#$p-SS= $|{\rm Orb}_G(Q)|$, i.e. $\#$p-SS= $=[G:{\rm St}_G(Q)]=[G:N_G(Q)]$,

because for the isotropy group is ${\rm St}_G(Q)=\{\!x\!\in\!G: xQx^{-1}\!=\!Q\}=N_G(Q)\!\}$.

Then $|G|=\#{\rm p\!-\!SS}|N(Q)|$.

2. Observe also that $p\not|$#p-SS:

since $|G|=p^mr$ and $|G|=|N_G(Q)|[G:N_G(Q)]$,  as  far as $|N_G(Q)|=|Q|[N_G(Q):Q]$

so $p^mr=|G|$ $=|Q|[G:N_G(Q)][N_G(Q):Q]$ $=p^m[N_G(Q):Q]\cdot\#$p-SS

which implies that $r=[N_G(Q):Q]\cdot\#$p-SS, then $p\!\!\not|\#$p-SS.

Proof of Theorem:

Considering the conjugation sub-action $H\times$ p-SS $\to$ p-SS,

we get a orbit decomposition p-SS = ${\rm Orb}_H(Q_1)\sqcup\cdots\sqcup{\rm Orb}_H(Q_t)$ with the corresponding class equation

#p-SS = $[H:N_H(Q_1)]+\cdots+[H:N_H(Q_t)]$,

but asumming that $|H|=p^{\alpha}$ then

#p-SS = $\frac{p^{\alpha}}{|N_H(Q_1)|}+\cdots+\frac{p^{\alpha}}{|N_H(Q_t)|}$.

Now since $p\!\!\not|$#p-SS then there is some $Q_i$ for which $|N_H(Q_i)|=|H|$, so $H=N_H(Q_i)$.

With that, it is easy these: $HQ_i and $Q_i$ is normal in $HQ_i$.

Also $\pi:H\to HQ_i/Q_i$ given by $x\mapsto\pi(x)=xQ_i$  is an epimorphism, so by the fundamental theorem of group-morphisms we have $\frac{HQ_i}{Q_i}\cong\frac{H}{H\cap Q_i}$,

Observing that $H,\ Q_i and $|HQ_i|=[H:H\cap Q_i]|Q_i|$ then $|HQ_i|=\frac{|H||Q_i|}{|H\cap Q_i|}=p^{m+\alpha-\beta}$,

where $|H\cap Q_i|=p^{\beta}$,

but $m$ is maximal, then $\alpha-\beta=0$. Hence $|H|=|H\cap Q_i|$ and $H=H\cap Q_i. $\Box$

Confer Milne Chapter 5

Filed under algebra, group theory, math, what is math

## what is the second abstraction lemma in the mother category? abstraction lemma two

Given a simple map $f:S\to T$ then $f$ can be factored as $f=\beta\circ\rho$ where $\rho$ is the projection $S\to S/{\sim}$ defined as $\rho(s)=[s]$ that is surjective, and $\beta:\frac{S}{\sim}\to T$ defined as $\beta([x])=f(x)$ that is injective