Faktorgruppe < Algebra < Algebra+Zahlentheo. < Hochschule < Mathe < Vorhilfe
|
Help!
How do you express the (additive) factor group (ZxZ)/<(2,2)> as a product of cyclic groups? I know how to do this in a cyclic group, for example:
(Z(index)4 x Z(index)6) /<(0,2)>
but when it comes to (ZxZ)/<(9,12)> or (ZxZxZ)/<(2,4,8)> I'm totaly lost.
Would be extremly happy for an answer (german or english)
Thanks!
Ich habe diese Frage in keinem Forum auf anderen Internetseiten gestellt.
|
|
| |
|
| Status: |
(Antwort) fertig  | | Datum: | 01:06 So 29.10.2006 | | Autor: | felixf |
Hallo!
> How do you express the (additive) factor group
> (ZxZ)/<(2,2)> as a product of cyclic groups? I know how to
> do this in a cyclic group, for example:
> (Z(index)4 x Z(index)6) /<(0,2)>
> but when it comes to (ZxZ)/<(9,12)> or (ZxZxZ)/<(2,4,8)>
> I'm totaly lost.
Benutze doch bitte den Formeleditor, dann kann man das etwas besser lesen...
Wie machst du es denn bei Produkten von endlichen zyklischen Gruppen? Also etwa bei [mm] $(\IZ_4 \times \IZ_6)/\langle(0, 2)\rangle$? [/mm] Bei unendlich zyklischen Gruppen wie [mm] $\IZ$ [/mm] geht es doch genauso!
Du brauchst im Prinzip diese Aussage: Sind $G, G'$ Gruppen und $H [mm] \subseteq [/mm] G$, $H' [mm] \subseteq [/mm] G'$ Normalteiler, so ist $(G [mm] \times [/mm] G') / (H [mm] \times [/mm] H') [mm] \cong [/mm] (G/H) [mm] \times [/mm] (G'/H')$.
Damit kannst du hier alles wunderbar in Produkte von zyklischen GRuppen zerlegen.
LG Felix
|
|
|
| |
|
Thanks a lot Felix!
$ [mm] (\IZ_4 \times \IZ_6)/\langle(0, 2)\rangle $\cong(\IZ_4 \times \IZ_2) [/mm] because the element generates a subgroup [mm] \langle(0, 2)\rangle=\{(0,0),(0,2),(0,4)\} [/mm] of order 3, and the cyclic group [mm] (\IZ_4 \times \IZ_6) [/mm] is of order 24, so the factor group must be isomorphic to [mm] \IZ_8 [/mm] The first factor is left alone and the second [mm] \IZ_6 [/mm] is collapsed by a subgroup of order 3. Hence $ [mm] (\IZ_4 \times \IZ_6)/\langle(0, 2)\rangle $\cong(\IZ_4 \times \IZ_2)
[/mm]
I'm not sure that your "aussage" is correct since;
[mm] $(\IZ \times \IZ)/\langle(2, 2)\rangle $\cong(\IZ_2 \times \IZ) [/mm] and [mm] $(\IZ \times \IZ)/\langle(1, 2)\rangle $\cong \IZ [/mm]
Unfortunately I don't have the calculation for these examples.
|
|
|
| |
|
| Status: |
(Antwort) fertig | | Datum: | 16:30 Mo 30.10.2006 | | Autor: | felixf |
Hi pelle!
> [mm](\IZ_4 \times \IZ_6)/\langle(0, 2)\rangle[/mm][mm] \cong(\IZ_4 \times \IZ_2)[/mm]
> because the element generates a subgroup [mm]\langle(0, 2)\rangle=\{(0,0),(0,2),(0,4)\}[/mm]
> of order 3, and the cyclic group [mm](\IZ_4 \times \IZ_6)[/mm] is
> of order 24, so the factor group must be isomorphic to
> [mm]\IZ_8[/mm]
That's wrong: The factor group has 8 elements, but is not isomorphic to [mm] $\IZ_8$:
[/mm]
> The first factor is left alone and the second [mm]\IZ_6[/mm]
> is collapsed by a subgroup of order 3. Hence [mm](\IZ_4 \times \IZ_6)/\langle(0, 2)\rangle[/mm][mm] \cong(\IZ_4 \times \IZ_2)[/mm]
In [mm] $\IZ_8$ [/mm] you have an element of order 8, but in [mm] $\IZ_4 \times \IZ_2$ [/mm] you don't.
> I'm not sure that your "aussage" is correct since;
> [mm](\IZ \times \IZ)/\langle(2, 2)\rangle[/mm][mm] \cong(\IZ_2 \times \IZ)[/mm]
> and [mm](\IZ \times \IZ)/\langle(1, 2)\rangle[/mm][mm] \cong \IZ[/mm]
Both of these are wrong: [mm] $(\IZ \times \IZ)/\langle(2, 2)\rangle \cong \IZ_2 \times \IZ_2$, [/mm] and [mm] $(\IZ \times \IZ)/\langle(1, 2)\rangle \cong \IZ_1 \times \IZ_2 \cong \IZ_2$.
[/mm]
EDIT: They are not wrong. And the statement is still correct (as the proof below shows). But the statement can't used here, as [mm] $\langle [/mm] (2, 2) [mm] \rangle$ [/mm] is not the product of two subgroups of [mm] $\IZ \times \IZ$. [/mm] Sorry for that :-(
To prove the statement, consider the map [mm] $\varphi [/mm] : G [mm] \times [/mm] G' [mm] \to [/mm] G/H [mm] \times [/mm] G/H'$, $(x, y) [mm] \mapsto [/mm] (x H, x H')$. You can quickly check that this is a surjective group homomorphism. And [mm] $\ker\varphi [/mm] = [mm] \{ (x, y) \in G \times G' \mid x \in H, y \in H' \} [/mm] = H [mm] \times [/mm] H'$, so by the Homomorphism Theorem, $(G [mm] \times [/mm] G')/(H [mm] \times [/mm] H') [mm] \cong [/mm] G/H [mm] \times [/mm] G'/H'$.
Best,
Felix
|
|
|
| |
|
Hi felix, thanks for your answer!
I'm with you on the first remark, that the factor group has 8 elements, but is not isomorphic to $ [mm] \IZ_8 [/mm] $. But then according to my book (a first course in abstract algebra by john b. fraleigh) $ [mm] (\IZ \times \IZ)/\langle(2, 2)\rangle [/mm] $$ [mm] \cong(\IZ_2 \times \IZ) [/mm] $
$ [mm] (\IZ \times \IZ)/\langle(1, 2)\rangle [/mm] $$ [mm] \cong \IZ [/mm] $
unfortunately these exercises are given without motivations just answers
In the book there is one example:
$ [mm] (\IZ \times \IZ)/\langle(1, 1)\rangle [/mm] $$ [mm] \cong \IZ [/mm] $
We may visualize [mm] \IZ \times \IZ [/mm] as points in the plane. The subgroup [mm] \langle(1, 1)\rangle [/mm] consists of those points that lie on the 45 degree line through the origin. The coset [mm] (1,0)+\langle(1, 1)\rangle [/mm] consists of those points that lie on the 45 degree line through the point (1,0) we may then choose representatives ...,(-3,0),(-2,0),(-1,0),(0,0),(1,0),(2,0),... of these cosets to compute in the factor group. since these representatives correspond precisely to the point of Z on the x-axis the factor group must be isomorphic to Z.
I think that $ [mm] (\IZ \times \IZ)/\langle(1, 2)\rangle [/mm] $$ [mm] \cong \IZ [/mm] $ for exactly the same reason.
With this in mind, returning to my first question, how do I calculate
$ [mm] (\IZ\times\IZ\times\IZ)/\langle(2, 4,8)\rangle [/mm] $
It's a tricky one!
/Pelle
|
|
|
| |
|
| Status: |
(Antwort) fertig | | Datum: | 22:52 Mo 30.10.2006 | | Autor: | felixf |
Hi Pelle!
> Hi felix, thanks for your answer!
> I'm with you on the first remark, that the factor group
> has 8 elements, but is not isomorphic to [mm]\IZ_8 [/mm].
Ok.
> But then
> according to my book (a first course in abstract algebra by
> john b. fraleigh) [mm](\IZ \times \IZ)/\langle(2, 2)\rangle[/mm][mm] \cong(\IZ_2 \times \IZ)[/mm]
You're right. I noticed I made a mistake, as [mm] $\langle [/mm] (2, 2) [mm] \rangle \neq \langle [/mm] 2 [mm] \rangle \times \langle [/mm] 2 [mm] \rangle$...
[/mm]
> [mm](\IZ \times \IZ)/\langle(1, 2)\rangle[/mm][mm] \cong \IZ[/mm]
>
> unfortunately these exercises are given without motivations
> just answers
>
> In the book there is one example:
> [mm](\IZ \times \IZ)/\langle(1, 1)\rangle[/mm][mm] \cong \IZ[/mm]
> We may visualize [mm]\IZ \times \IZ[/mm] as points in the plane. The
> subgroup [mm]\langle(1, 1)\rangle[/mm] consists of those points
> that lie on the 45 degree line through the origin. The
> coset [mm](1,0)+\langle(1, 1)\rangle[/mm] consists of those points
> that lie on the 45 degree line through the point (1,0) we
> may then choose representatives
> ...,(-3,0),(-2,0),(-1,0),(0,0),(1,0),(2,0),... of these
> cosets to compute in the factor group. since these
> representatives correspond precisely to the point of Z on
> the x-axis the factor group must be isomorphic to Z.
Exactly.
> I think that [mm](\IZ \times \IZ)/\langle(1, 2)\rangle[/mm][mm] \cong \IZ[/mm]
> for exactly the same reason.
Yes. As it does for [mm] $\langle(1, n)\rangle$ [/mm] for every $n [mm] \in \IZ$.
[/mm]
> With this in mind, returning to my first question, how do
> I calculate
> [mm](\IZ\times\IZ\times\IZ)/\langle(2, 4,8)\rangle[/mm]
>
> It's a tricky one!
Depends on what you know :)
Do you know how to compute Hermite normal forms of integer matrices? By multiplying $(2, 4, 8)$ by unimodular $3 [mm] \times [/mm] 3$-matrices with entries in [mm] $\IZ$, [/mm] you will get new vectors $(a, b, c)$ such that [mm] $\IZ^3 [/mm] / [mm] \langle [/mm] (2, 4, 8) [mm] \rangle \cong \IZ^3 [/mm] / [mm] \langle [/mm] (a, b, c) [mm] \rangle$. [/mm] Now, if you multiply $(2, 4, 8)$ with a good matrix such that $(a, b, c)$ is of an easy form, you can directly read off from [mm] $\IZ^3 [/mm] / [mm] \langle [/mm] (a, b, [mm] c)\rangle$ [/mm] how [mm] $\IZ^3 [/mm] / [mm] \langle [/mm] (2, 4, 8) [mm] \rangle$ [/mm] looks like.
(If you don't know the Hermite normal form, you can still fiddle around and try to find an unimodular matrix by trying around.)
Best,
Felix
|
|
|
| |
|
Hi Felix and thanks again.
This is what I know:
Such matrices are lower triangular, non-negative and each row has a unique maximum entry on the main diagonal. They can be constructed from an arbitrary non-singular matrix A using the rules:
1. adding an integer multiple of one column to another
2. exchanging two columns
3. multiplying one column by -1
or by trail and error, Multiplying A with a unimodular matrix U :)
How do I continue? Should the diagonal in the matrix correspond to the generator of the subgroup?
> Now, if you multiply [mm](2, 4, 8)[/mm] with a good matrix such that
> [mm](a, b, c)[/mm] is of an easy form, you can directly read off
> from [mm]\IZ^3 / \langle (a, b, c)\rangle[/mm] how [mm]\IZ^3 / \langle (2, 4, 8) \rangle[/mm]
> looks like.
Sorry I don't understand, should my goal be to get something like $ [mm] \IZ^3 [/mm] / [mm] \langle [/mm] (1, 1, [mm] a)\rangle [/mm] $
It's getting too late for this 
Pelle
|
|
|
| |
|
Hi!
Trying to solve $ [mm] (\IZ\times\IZ\times\IZ)/\langle(2, 4,8)\rangle [/mm] $
[mm] $(1,2,4)+\langle(2,4,8)\rangle$ [/mm] must be of order 2 in the factor group and [mm] $(0,1,1)+\langle(2,4,8)\rangle$ [/mm] and [mm] $(0,0,1)+\langle(2,4,8)\rangle$ [/mm] generates infinite cyclic subgroups of the factor group. So it would be reasonable to presume that $ [mm] (\IZ\times\IZ\times\IZ)/\langle(2, 4,8)\rangle\cong(\IZ_2 \times \IZ\times\IZ) [/mm] $ To prove the presumption I have to define a homomorphism [mm] \alpha [/mm] mapping [mm] (\IZ\times\IZ\times\IZ) [/mm] onto [mm] (\IZ_2 \times \IZ\times\IZ) [/mm] having kernel [mm] \langle(2,4,8)\rangle [/mm] . The properties [mm] $\alpha(1,2,4)=(1,0,0)$ [/mm] and [mm] $\alpha(0,1,1)=(0,1,1)$ [/mm] and [mm] $\alpha(0,0,1)=(0,0,1)$ [/mm] and [mm] $\alpha((k,l,m)+(n,o,p)+(q,r,s))=\alpha(k,l,m)+\alpha(n,o,p)+\alpha(q,r,s)$. [/mm] Also the condition that for any element [mm] (p,q,r)\in(\IZ_2 \times \IZ\times\IZ) [/mm] there is an element [mm] (s,t,u)\in(\IZ \times \IZ\times\IZ) [/mm] such that [mm] $\alpha(s,t,u)=(p.q.r)$ [/mm] and the last step would be to show that the kernel is contained in [mm] \langle(2,4,8)\rangle [/mm] and that [mm] \langle(2,4,8)\rangle [/mm] is contained in the [mm] $ker(\alpha)$ [/mm] If all the conditions hold it would prove that the factor group and [mm] (\IZ_2 \times \IZ\times\IZ) [/mm] are isomorphic.
Is this way easier than multiplying with matrices? Is the reasoning above correct?
|
|
|
| |
|
| Status: |
(Mitteilung) Reaktion unnötig  | | Datum: | 13:20 Di 14.11.2006 | | Autor: | matux |
$MATUXTEXT(ueberfaellige_frage)
|
|
|
|
