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From: maxblack88 on 3 Aug 2010 19:11
Hi!

I'd like to continue the discussion "(Amalgamated) free products"

http://groups.google.com/group/sci.math/browse_thread/thread/1e2f1eb7837500ee/8a5632049d3a78e8?lnk=gst&q=free+product#8a5632049d3a78e8

Let G=A *_C B be the amalgamated free product of groups. Suppose N is
a normal subgroup of B intersecting C trivially. Let H be the
amalgamated free product A *_C B/N.

Is it true that the kernel of the homomorphism G->H given by the
universal property is the normal closure of N in G?

This seems to be the case if you think in terms of generators and
relations, but I cannot prove this using the universal property.

Best,
Max.
From: Arturo Magidin on 3 Aug 2010 22:52
On Aug 3, 6:11 pm, "maxblac...(a)gmail.com" <maxblac...(a)gmail.com>
wrote:
> Hi!
>
> I'd like to continue the discussion "(Amalgamated) free products"
>
> http://groups.google.com/group/sci.math/browse_thread/thread/1e2f1eb7...
>
> Let G=A *_C B be the amalgamated free product of groups. Suppose N is
> a normal subgroup of B intersecting C trivially. Let H be the
> amalgamated free product A *_C B/N.

You mean A *_{C} (B/N), rather than (A*_C B)/N (which does not make
sense since as you note N is not generally normal). Just clarifying.

> Is it true that the kernel of the homomorphism G->H given by the
> universal property is the normal closure of N in G?
>
> This seems to be the case if you think in terms of generators and
> relations, but I cannot prove this using the universal property.

*Which* universal property? You have at least three universal
properties at play here: the universal property of the free product
with amalgamation, the universal property of the quotient applied to G
itself, and the universal property of the quotient applied to B.

You get the map form G to H by using the universal property of the
free amalgamated product: the map is induced by the structure
embedding of A into A*_{C}(B/N), and the quotient map B-->B/N followed
by the structure embedding B/N --> A*_{C}(B/N). These are maps from A
and B into the group H which agree on C, and therefore induce a map G--
>H. So you are using the universal property there.

The elements of N in A*_{C} B map to the identity in H, so the kernel
of the map contains N, and therefore contains N^G.

Let K = G/N^G = (A*_{C} B)/N^G. The composition of the structure
embedding A-->G with the quotient map G-->K gives a map from A to K;
to get a map from (B/N) to K you can first consider the composition of
the maps B-->G and G-->K; this composition has kernel containing N, so
the universal property of the quotient gives you a map B/N --> K.
These two maps (from A to K and from B/N to K) agree on C, because the
maps from A and B to G agree on C. Thus, the two maps induce a
homomorphism A*_{C}(B/N) to (A*_{C} B)/N^G.

So now you have maps H-->K and K-->H; can you check if the
compositions are the identity?

--
Arturo Magidin
From: maxblack88 on 4 Aug 2010 19:50
On 3 Ago, 23:52, Arturo Magidin <magi...(a)member.ams.org> wrote:
> On Aug 3, 6:11 pm, "maxblac...(a)gmail.com" <maxblac...(a)gmail.com>
> wrote:
>
> > Hi!
>
> > I'd like to continue the discussion "(Amalgamated) free products"
>
> >http://groups.google.com/group/sci.math/browse_thread/thread/1e2f1eb7...
>
> > Let G=A *_C B be the amalgamated free product of groups. Suppose N is
> > a normal subgroup of B intersecting C trivially. Let H be the
> > amalgamated free product A *_C B/N.
>
> You mean A *_{C} (B/N), rather than (A*_C B)/N (which does not make
> sense since as you note N is not generally normal). Just clarifying.
>
> > Is it true that the kernel of the homomorphism G->H given by the
> > universal property is the normal closure of N in G?
>
> > This seems to be the case if you think in terms of generators and
> > relations, but I cannot prove this using the universal property.
>
> *Which* universal property? You have at least three universal
> properties at play here: the universal property of the free product
> with amalgamation, the universal property of the quotient applied to G
> itself, and the universal property of the quotient applied to B.
>
> You get the map form G to H by using the universal property of the
> free amalgamated product: the map is induced by the structure
> embedding of A into A*_{C}(B/N), and the quotient map B-->B/N followed
> by the structure embedding B/N --> A*_{C}(B/N). These are maps from A
> and B into the group H which agree on C, and therefore induce a map G--
>
> >H. So you are using the universal property there.
>
> The elements of N in A*_{C} B map to the identity in H, so the kernel
> of the map contains N, and therefore contains N^G.
>
> Let K = G/N^G = (A*_{C} B)/N^G. The composition of the structure
> embedding A-->G with the quotient map G-->K gives a map from A to K;
> to get a map from (B/N) to K you can first consider the composition of
> the maps B-->G and G-->K; this composition has kernel containing N, so
> the universal property of the quotient gives you a map B/N --> K.
> These two maps (from A to K and from B/N to K) agree on C, because the
> maps from A and B to G agree on C. Thus, the two maps induce a
> homomorphism A*_{C}(B/N) to (A*_{C} B)/N^G.
>
> So now you have maps H-->K and K-->H; can you check if the
> compositions are the identity?
>
> --
> Arturo Magidin

Arturo,

Thanks. In order to prove what you said, one must realize that G is
generated by A and B, and H is generated by A and the canonical image
of B. Is that what you had in mind?

Best,
Max.
From: Arturo Magidin on 4 Aug 2010 21:12
On Aug 4, 6:50 pm, "maxblac...(a)gmail.com" <maxblac...(a)gmail.com>
wrote:
> On 3 Ago, 23:52, Arturo Magidin <magi...(a)member.ams.org> wrote:
>
>
>
> > On Aug 3, 6:11 pm, "maxblac...(a)gmail.com" <maxblac...(a)gmail.com>
> > wrote:
>
> > > Hi!
>
> > > I'd like to continue the discussion "(Amalgamated) free products"
>
> > >http://groups.google.com/group/sci.math/browse_thread/thread/1e2f1eb7....
>
> > > Let G=A *_C B be the amalgamated free product of groups. Suppose N is
> > > a normal subgroup of B intersecting C trivially. Let H be the
> > > amalgamated free product A *_C B/N.
>
> > You mean A *_{C} (B/N), rather than (A*_C B)/N (which does not make
> > sense since as you note N is not generally normal). Just clarifying.
>
> > > Is it true that the kernel of the homomorphism G->H given by the
> > > universal property is the normal closure of N in G?
>
> > > This seems to be the case if you think in terms of generators and
> > > relations, but I cannot prove this using the universal property.
>
> > *Which* universal property? You have at least three universal
> > properties at play here: the universal property of the free product
> > with amalgamation, the universal property of the quotient applied to G
> > itself, and the universal property of the quotient applied to B.
>
> > You get the map form G to H by using the universal property of the
> > free amalgamated product: the map is induced by the structure
> > embedding of A into A*_{C}(B/N), and the quotient map B-->B/N followed
> > by the structure embedding B/N --> A*_{C}(B/N). These are maps from A
> > and B into the group H which agree on C, and therefore induce a map G--
>
> > >H. So you are using the universal property there.
>
> > The elements of N in A*_{C} B map to the identity in H, so the kernel
> > of the map contains N, and therefore contains N^G.
>
> > Let K = G/N^G = (A*_{C} B)/N^G. The composition of the structure
> > embedding A-->G with the quotient map G-->K gives a map from A to K;
> > to get a map from (B/N) to K you can first consider the composition of
> > the maps B-->G and G-->K; this composition has kernel containing N, so
> > the universal property of the quotient gives you a map B/N --> K.
> > These two maps (from A to K and from B/N to K) agree on C, because the
> > maps from A and B to G agree on C. Thus, the two maps induce a
> > homomorphism A*_{C}(B/N) to (A*_{C} B)/N^G.
>
> > So now you have maps H-->K and K-->H; can you check if the
> > compositions are the identity?


> Thanks. In order to prove what you said, one must realize that G is
> generated by A and B, and H is generated by A and the canonical image
> of B. Is that what you had in mind?

You probably need that, yes. Some diagram chasing at this point ought
to do it, but I haven't gone through it so I am not 100% positive.

--
Arturo Magidin
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