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  Traducir al Traducido (ver original)  Más opciones 18 jun,
19:51 Grupos de noticias: sci.math De: Steve Dalton
sda...@math.sunysb.edu Fecha: Thu, 18 Jun 2009 23:51:57 EDT Local:
Jue 18 jun 2009 19:51 Asunto: Question on embedding group quotients
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de este autor Hi, Suppose G and H are groups, and G' and H' are
their commutator subgroups.  Suppose f: H -> G is a monomorphism of H
into G, such that G' < f(H).  Is it possible to find a monomorphism of
H/H' into G/G' ? What if I said that f(H) had finite index in G? or
that G/f(H) was cyclic? Or both (so G/f(H) is finite cyclic). I think
to show this I need to show G' = f(H'), but I don't know if any of the
additional conditions above are crucial.  Any help, even partial
hints, would be most welcome. Thanks, Steve     Responder al
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entradas. El asunto del debate ha cambiado a "Proof: G/f(G)
is finite cyclic. Derived from "Question on embedding group
quotients". by Steve, Musatov" de Musatov
Musatov   Ver perfil   Traducir al Traducido (ver
original) (1 usuario)  Más opciones 18 jun, 22:36
Grupos de noticias: sci.math, comp.theory, comp.programming, sci.logic
De: Musatov marty....@gmail.com Fecha: Thu, 18 Jun 2009 23:36:01
-0700 (PDT) Local: Jue 18 jun 2009 22:36 Asunto: Proof: G/f(G) is
finite cyclic. Derived from "Question on embedding group quotients".
by Steve, Musatov Responder al autor | Reenviar | Imprimir | Mensaje
individual | Mostrar mensaje original | Informar de este mensaje |
Buscar mensajes de este autor On Jun 18, 8:51 pm, Steve Dalton
sda...@math.sunysb.edu wrote: Hi, Suppose G and H are groups, and G'
and H' are their commutator subgroups.  Suppose f: H -> G is a
monomorphism of H into G, such that G' < f(H).  Is it possible to find
a monomorphism of H/H' into G/G' ? > What if I said that f(H) had
finite index in G? or that G/f(H) was cyclic? Or both (so G/f(H) is
finite cyclic). >' ? I think to show this I need to show G' = f(H'),
but I don't know if any of the additional conditions above are
crucial.  Any help, even partial hints, would be most welcome.
Thanks, Steve Hello Steve, Here at Inverse 19/Hope research we
practice the "Musatov Method", which means when things get overly
complicated and variables become obscured we take it back to the
start. For your listening entertainment as you read this proof:
http://www.youtube.com/watch?v=V3Kd7IGPyeg Okay, "back to the start".
Q: What is a finite field? A: Finite field: /definition per Wikipedia
Another notation is GF(pn), where the letters "GF" stand for "Galois
field". For any h in G, there is an automorphism of G sending g to h,
Proof: The multiplicative group of every finite field IS cyclic,
http://en.wikipedia.org/wiki/Finite_field Q: What next? A: Naturally,
we address the isomorphism. Isomorphisms of Cyclic Groups. Some
Properties of Cyclic Groups G or H = G; theorem :: GR_CY_2:27 for G
being strict finite Group st card G st G is cyclic holds Image g is
cyclic; theorem :: GR_CY_2:33 for G,F being Proof:
http://mizar.uwb.edu.pl/version/current/abstr/gr_cy_2.abs Cyclic group
- This is so since g + h mod n = h + g mod n. If n is finite, then gn
= g0 is the given a finite field F and a finite cyclic group G, there
is a finite Proof: http://en.wikipedia.org/wiki/Cyclic_group Example
in proof(web-infused theorem) "All finite fields are isomorphic to GF
(p^n)@Everything2.com." Every finite field F is isomorphic to GF(pn)
for some prime p and natural n of F is cyclic (see a proof that the
multiplicative group of a finite field since our original a is a root
of h(x) it must be that f(x) is one of the Proof;
http://everything2.com/All%2520finite%2520fields%2520are%2520isomorphic%2520to%2520GF
Q: When is the augmentation ideal principal? A: Right ideal o/,R [G]
if and only i/G is finite H-by-cyclic. By Corollary 1.2 of [1], G has
a normal free subgroup F of finite index. Clearly o)(K[G/[F, Proof:
http://www.springerlink.com/index/G68772338406KV51.pdf Theorem: The
sum of a finite group and a finite number of infinite cyclic groups is
a "CANCELLATION IN DIRECT SUMS OF GROUPS". Clearly it suffices to
prove G^H for the two cases F finite and F infinite cyclic. Proof:
http://www.math.nmsu.edu/~elbert/CancellationInDirectSums.pdf On
cyclic subgroups of finite groups: Let G be a finite group and X and Y
cyclic subgroups of G. Then there E h. nY = Xg l. nY=Xg l n(YnT)<F(G),
and we have a contradiction. The contradiction proves the next proof:
http://journals.cambridge.org/production/action/cjoGetFulltext?fullte
The burnside ring of a cyclic extension of a torus if G is a finite
extension of a torus and H is an (F', T')-subgroup, then (H)r now
suppose that F is cyclic, say of order n with generator f. Then it is
A conjecture of Lennox and Wiegold concerning supersoluble
groupsfinite normal subgroup F with G/F supersoluble. As G is
residually finite, thus H and D complete a cyclic normal series for G
and G is supersoluble. Double Proof: (1)http://www.springerlink.com/
index/ PG7144W215066360.pdf               (2)http://
journals.cambridge.org/production/action/ cjoGetFulltext?fulltextid
Consider Extremal Bases for Finite Cyclic Groups order h for the
finite cyclic group Z/(m) if every n is congruent to a sum Gk(h)= max
max g(A\F). Erdös and Graham [4] proved that Proof:
http://erdos.math.txstate.edu/research/publications/1992a.pdf Proof: G/
f(G) is finite cyclic. QED. Martin Musatov Founder, MeAmI.org Co-
Founder, Inverse 19 Mathematics (not the only but one without it would
not exist) *non-mathematical message* [free to snip] If you use
Google, try http://MeAmI.org. It Googles Google Googling. +NO ADS. :)
P=NP if you read this. Note: this email printed on 1/2 recycled
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entradas. Musatov   Ver perfil   Traducir al
Traducido (ver original) (1 usuario)  Más opciones 18 jun,
22:37 Grupos de noticias: sci.math, comp.theory,
comp.programming, sci.logic De: Musatov marty....@gmail.com Fecha:
Thu, 18 Jun 2009 23:37:50 -0700 (PDT) Local: Jue 18 jun 2009 22:37
Asunto: Proof: G/f(G) is finite cyclic. Derived from "Question on
embedding group quotients". by Steve, Musatov Responder al autor |
Reenviar | Imprimir | Mensaje individual | Mostrar mensaje original |
Informar de este mensaje | Buscar mensajes de este autor On Jun
18, 8:51 pm, Steve Dalton sda...@math.sunysb.edu wrote: Hi, Suppose G
and H are groups, and G' and H' are their commutator subgroups.
 Suppose f: H -> G is a monomorphism of H into G, such that G' < f(H).
 Is it possible to find a monomorphism of H/H' into G/G' ? > What if I
said that f(H) had finite index in G? or that G/f(H) was cyclic? Or
both (so G/f(H) is finite cyclic). >' ? I think to show this I need to
show G' = f(H'), but I don't know if any of the additional conditions
above are crucial.  Any help, even partial hints, would be most
welcome. Thanks, Steve Hello Steve, Here at Inverse 19/Hope research
we practice the "Musatov Method", which means when things get overly
complicated and variables become obscured we take it back to the
start. For your listening entertainment as you read this proof:
http://www.youtube.com/watch?v=V3Kd7IGPyeg Okay, "back to the start".
Q: What is a finite field? A: Finite field: /definition per Wikipedia
Another notation is GF(pn), where the letters "GF" stand for "Galois
field". For any h in G, there is an automorphism of G sending g to h,
Proof: The multiplicative group of every finite field IS cyclic,
http://en.wikipedia.org/wiki/Finite_field Q: What next? A: Naturally,
we address the isomorphism. Isomorphisms of Cyclic Groups. Some
Properties of Cyclic Groups <finite>G or H = G; theorem :: GR_CY_2:27
for G being strict finite Group st card G st G is cyclic holds Image g
is cyclic; theorem :: GR_CY_2:33 for G,F being <finite> Proof:
http://mizar.uwb.edu.pl/version/current/abstr/gr_cy_2.abs Cyclic group
+ This is so since g + h mod n = h + g mod n. If n is finite, then gn
= g0 is the <finite> given a finite field F and a finite cyclic group
G, there is a finite <field> Proof: http://en.wikipedia.org/wiki/Cyclic_group
Example in proof (web-infused theorem) "All finite fields are
isomorphic to GF(p^n)@Everything2.com." Every finite field F is
isomorphic to GF(pn) for some prime p and natural n. <Finite> of F is
cyclic (see a proof that the multiplicative group of a finite field
since our original a is a root of h(x) it must be that f(x) is one of
the <fields> Proof;
http://everything2.com/<infinite>/All%2520finite%2520fields%2520are%2520isomorphic%2520to%2520GF<inite>
Q: When is the augmentation ideal principal? A: Right ideal o/,R [G]
if and only i/G is finite H-by-cyclic. <Finite> By Corollary 1.2 of
[1], G has a normal free subgroup F of finite index. Clearly o)(K[G/
[F, Proof: http://www.springerlink.com/index/G68772338406KV51.pdf
Theorem: The sum of a finite group and a finite number of infinite
cyclic groups is a "CANCELLATION IN DIRECT SUMS OF GROUPS". Clearly it
suffices to prove G^H for the two cases F finite and F infinite
cyclic. Proof: http://www.math.nmsu.edu/~elbert/CancellationInDirectSums.pdf
On cyclic subgroups of finite groups <fields> Let G be a finite group
and X and Y cyclic subgroups of G. Then there <fields> E h. nY = Xg l.
nY=Xg l n(YnT)<F(G), and we have a contradiction. The contradiction
proves the next proof: http://journals.cambridge.org/production/action/cjoGetFulltext?fullte
The burnside ring of a cyclic extension of a torus <finite> If G is a
finite extension of a torus and H is an (F', T')-subgroup, then (H)r
<finite> Now suppose F is cyclic, say of order n with generator f.
Then it is <infinite> A conjecture of Lennox and Wiegold concerning
supersoluble groupsfinite normal subgroup F with G/F supersoluble. As
G is residually finite, <fields> Thus H and D complete a cyclic normal
series for G and G is supersoluble. <F> Double Proof: (1)http://
www.springerlink.com/index/ PG7144W215066360.pdf               (2)
http://journals.cambridge.org/production/action/ cjoGetFulltext?
fulltextid... Consider Extremal Bases for Finite Cyclic Groups order h
for the finite cyclic group Z/(m) if every n is congruent to a sum
<finite> Gk(h)= max max g(A\F). Erdös and Graham [4] proved that
<finite> Proof: http://erdos.math.txstate.edu/research/publications/1992a.pdf
Proof: G/f(G) is finite cyclic. QED. Martin Musatov Founder, MeAmI.org
Co-Founder, Inverse 19 Mathematics (not the only but one without it
would not exist) *non-mathematical message* [free to snip] If you use
Google, try http://MeAmI.org. It Googles Google Googling. +NO ADS. :)
P=NP if you read this. Note: this email printed on 1/2 recycled
theorems and 100% mathematical rigor.     Responder al
autor    Reenviar        Report spam Reporting spam Message reported
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entradas. El asunto del debate ha cambiado a "Question on
embedding group quotients" de Jack Schmidt
Jack Schmidt   Ver perfil   Traducir al Traducido (ver
original)  Más opciones 19 jun, 04:05 Grupos de
noticias: sci.math De: Jack Schmidt Jack.Schmi...@gmail.com
Fecha: Fri, 19 Jun 2009 08:05:08 EDT Local: Vie 19 jun 2009 04:05
Asunto: Re: Question on embedding group quotients Responder al autor
| Reenviar | Imprimir | Mensaje individual | Mostrar mensaje original
| Informar de este mensaje | Buscar mensajes de este autor Hi,
Suppose G and H are groups, and G' and H' are their commutator
subgroups.  Suppose f: H -> G is a > monomorphism of H into G, such
that G' < f(H).  Is it > possible to find a monomorphism of H/H' into
G/G' ? Suppose H is a subgroup of G. [H,H] is a subgroup of H n [G,G].
The kernel of a:H -> G/[G,G] is H n [G,G], so a is well-defined from H/
[H,H] to G/[G,G] and has kernel (H n [G,G])/[H,H]. So to answer your
question, you just need to find a group where H n [G,G] is larger than
[H,H].  For instance, take H = [G,G] to be abelian. Explicitly, take G
to be non-abelian of order 6, H = [G,G] of order 3, then H/[H,H] is
order 3 and so cannot embed in G/[G,G] of order 2. > What if I said
that f(H) had finite index in G? or G/f(H) was cyclic? Or both (so G/f
(H) is finite > cyclic). These additional restrictions come for free
by trying easy examples. - Ocultar texto de la cita -- Mostrar texto
de la cita -> I think to show this I need to show G' = f(H'), any of
the additional conditions above are crucial.  Any help, even partial
hints, would be most welcome. Thanks, Steve     Responder al
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Question on embedding group quotients - sci.math | Grupos de Here at
Inverse 19/Hope research we practice the "Musatov Method" Proof;
http://everything2.com/All%2520finite%2520fields%2520are
http://groups.google.com.mx/group/sci.math/browse_thread/thread/
Jack Schmidt ha escrito:
Hi,
Suppose G and H are groups, and G' and H' are their
commutator subgroups. Suppose f: H -> G is a
monomorphism of H into G, such that G' < f(H). Is it
possible to find a monomorphism of H/H' into G/G' ?
Suppose H is a subgroup of G.
[H,H] is a subgroup of H n [G,G].
The kernel of a:H -> G/[G,G] is H n [G,G],
so a is well-defined from H/[H,H] to G/[G,G]
and has kernel (H n [G,G])/[H,H].
So to answer your question, you just need to find
a group where H n [G,G] is larger than [H,H]. For
instance, take H = [G,G] to be abelian.
Explicitly, take G to be non-abelian of order 6,
H = [G,G] of order 3, then H/[H,H] is order 3 and
so cannot embed in G/[G,G] of order 2.
What if I said that f(H) had finite index in G? or
that G/f(H) was cyclic? Or both (so G/f(H) is finite
cyclic).
These additional restrictions come for free by trying
easy examples.
I think to show this I need to show G' = f(H'), but I
don't know if any of the additional conditions above
are crucial. Any help, even partial hints, would be
most welcome.
Thanks,
Steve
4 .5.
3. ___
2|1.0.1