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
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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)

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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!

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SL_2(Z_3)


http://arxiv.org/pdf/math/0602364v1

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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<G 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<HQ_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<Q_i.

\Box

Confer Milne Chapter 5

数学

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what is the second abstraction lemma in the mother category?


category diagram of set, binary relation and maps

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

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