Can every finite group be presented as an Fp group?
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If we have a finite group $G$, then we know that $G$ is a quotient of some free group, say $F$, on some free set, say $X$.
Now, let $N$ have this: $Gcong F/N$, where $N$ is the normal closure of $R$ (some finite set of relators) in $F$.
Is it "safe" to say that $G cong langle X |R rangle $ (or even $G = langle X |R rangle $)? Which leads to say $G$ is finitely presented group!
Thanks for advance.
finite-groups computational-algebra
asked Nov 26 at 9:44
A.Messab
355
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show 5 more comments
up vote
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If we have a finite group $G$, then we know that $G$ is a quotient of some free group, say $F$, on some free set, say $X$.
Now, let $N$ have this: $Gcong F/N$, where $N$ is the normal closure of $R$ (some finite set of relators) in $F$.
Is it "safe" to say that $G cong langle X |R rangle $ (or even $G = langle X |R rangle $)? Which leads to say $G$ is finitely presented group!
Thanks for advance.
finite-groups computational-algebra
asked Nov 26 at 9:44
A.Messab
355
I'm not sure what you mean by "safe". The notation $Gsimeqlangle X|Rrangle$ means $Gsimeq F(X)/N(R)$ by definition. And yes, there's a theorem that every finite group is finitely presented.
– freakish
Nov 26 at 10:08
@freakish It means that: is it true that $N$ with that property always exists?
– A.Messab
Nov 26 at 10:11
With what property? Are you asking if every finite group is finitely presented? That's really confusing. Plus: what does "Fp" stand for in the title?
– freakish
Nov 26 at 10:12
$N$ is the normal closure of $R$ in $F$(so the existence of $R$ is also questioned) AND $G cong F/N$
– A.Messab
Nov 26 at 10:15
If $G$ is any group and $Xsubseteq G$ is a set of generators then there's a unique epimorphism $phi:F(X)to G$ such that $phi(x)=x$ for $xin X$. Thus the first isomorphism theorem applies and $Gsimeq F(X)/N$ where $N=ker(phi)$ clearly exists. You can take $R=N$.
– freakish
Nov 26 at 10:17
|
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If we have a finite group $G$, then we know that $G$ is a quotient of some free group, say $F$, on some free set, say $X$.
Now, let $N$ have this: $Gcong F/N$, where $N$ is the normal closure of $R$ (some finite set of relators) in $F$.
Is it "safe" to say that $G cong langle X |R rangle $ (or even $G = langle X |R rangle $)? Which leads to say $G$ is finitely presented group!
Thanks for advance.
finite-groups computational-algebra
asked Nov 26 at 9:44
A.Messab
355
If we have a finite group $G$, then we know that $G$ is a quotient of some free group, say $F$, on some free set, say $X$.
Now, let $N$ have this: $Gcong F/N$, where $N$ is the normal closure of $R$ (some finite set of relators) in $F$.
Is it "safe" to say that $G cong langle X |R rangle $ (or even $G = langle X |R rangle $)? Which leads to say $G$ is finitely presented group!
Thanks for advance.
finite-groups computational-algebra
finite-groups computational-algebra
asked Nov 26 at 9:44
A.Messab
355
asked Nov 26 at 9:44
A.Messab
355
asked Nov 26 at 9:44
A.Messab
355
asked Nov 26 at 9:44
A.Messab
355
asked Nov 26 at 9:44
A.Messab
355
355
I'm not sure what you mean by "safe". The notation $Gsimeqlangle X|Rrangle$ means $Gsimeq F(X)/N(R)$ by definition. And yes, there's a theorem that every finite group is finitely presented.
– freakish
Nov 26 at 10:08
@freakish It means that: is it true that $N$ with that property always exists?
– A.Messab
Nov 26 at 10:11
With what property? Are you asking if every finite group is finitely presented? That's really confusing. Plus: what does "Fp" stand for in the title?
– freakish
Nov 26 at 10:12
$N$ is the normal closure of $R$ in $F$(so the existence of $R$ is also questioned) AND $G cong F/N$
– A.Messab
Nov 26 at 10:15
If $G$ is any group and $Xsubseteq G$ is a set of generators then there's a unique epimorphism $phi:F(X)to G$ such that $phi(x)=x$ for $xin X$. Thus the first isomorphism theorem applies and $Gsimeq F(X)/N$ where $N=ker(phi)$ clearly exists. You can take $R=N$.
– freakish
Nov 26 at 10:17
|
show 5 more comments
I'm not sure what you mean by "safe". The notation $Gsimeqlangle X|Rrangle$ means $Gsimeq F(X)/N(R)$ by definition. And yes, there's a theorem that every finite group is finitely presented.
– freakish
Nov 26 at 10:08
@freakish It means that: is it true that $N$ with that property always exists?
– A.Messab
Nov 26 at 10:11
With what property? Are you asking if every finite group is finitely presented? That's really confusing. Plus: what does "Fp" stand for in the title?
– freakish
Nov 26 at 10:12
$N$ is the normal closure of $R$ in $F$(so the existence of $R$ is also questioned) AND $G cong F/N$
– A.Messab
Nov 26 at 10:15
If $G$ is any group and $Xsubseteq G$ is a set of generators then there's a unique epimorphism $phi:F(X)to G$ such that $phi(x)=x$ for $xin X$. Thus the first isomorphism theorem applies and $Gsimeq F(X)/N$ where $N=ker(phi)$ clearly exists. You can take $R=N$.
– freakish
Nov 26 at 10:17
I'm not sure what you mean by "safe". The notation $Gsimeqlangle X|Rrangle$ means $Gsimeq F(X)/N(R)$ by definition. And yes, there's a theorem that every finite group is finitely presented.
– freakish
Nov 26 at 10:08
I'm not sure what you mean by "safe". The notation $Gsimeqlangle X|Rrangle$ means $Gsimeq F(X)/N(R)$ by definition. And yes, there's a theorem that every finite group is finitely presented.
– freakish
Nov 26 at 10:08
@freakish It means that: is it true that $N$ with that property always exists?
– A.Messab
Nov 26 at 10:11
@freakish It means that: is it true that $N$ with that property always exists?
– A.Messab
Nov 26 at 10:11
With what property? Are you asking if every finite group is finitely presented? That's really confusing. Plus: what does "Fp" stand for in the title?
– freakish
Nov 26 at 10:12
With what property? Are you asking if every finite group is finitely presented? That's really confusing. Plus: what does "Fp" stand for in the title?
– freakish
Nov 26 at 10:12
$N$ is the normal closure of $R$ in $F$(so the existence of $R$ is also questioned) AND $G cong F/N$
– A.Messab
Nov 26 at 10:15
$N$ is the normal closure of $R$ in $F$(so the existence of $R$ is also questioned) AND $G cong F/N$
– A.Messab
Nov 26 at 10:15
If $G$ is any group and $Xsubseteq G$ is a set of generators then there's a unique epimorphism $phi:F(X)to G$ such that $phi(x)=x$ for $xin X$. Thus the first isomorphism theorem applies and $Gsimeq F(X)/N$ where $N=ker(phi)$ clearly exists. You can take $R=N$.
– freakish
Nov 26 at 10:17
If $G$ is any group and $Xsubseteq G$ is a set of generators then there's a unique epimorphism $phi:F(X)to G$ such that $phi(x)=x$ for $xin X$. Thus the first isomorphism theorem applies and $Gsimeq F(X)/N$ where $N=ker(phi)$ clearly exists. You can take $R=N$.
– freakish
Nov 26 at 10:17
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
Let $G$ be a finite group, and so its Cayley table is finite. Construct finitely presented group $F$ with generators elements in $G$, and relators $abc^{-1}$ if ab=c in the Cayley table.
answered Nov 26 at 10:25
mathnoob
1,495319
the problem is: how to construct $F$? is there a general algorithm?
– A.Messab
Nov 26 at 10:37
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $G$ be a finite group, and so its Cayley table is finite. Construct finitely presented group $F$ with generators elements in $G$, and relators $abc^{-1}$ if ab=c in the Cayley table.
answered Nov 26 at 10:25
mathnoob
1,495319
the problem is: how to construct $F$? is there a general algorithm?
– A.Messab
Nov 26 at 10:37
add a comment |
up vote
0
down vote
Let $G$ be a finite group, and so its Cayley table is finite. Construct finitely presented group $F$ with generators elements in $G$, and relators $abc^{-1}$ if ab=c in the Cayley table.
answered Nov 26 at 10:25
mathnoob
1,495319
the problem is: how to construct $F$? is there a general algorithm?
– A.Messab
Nov 26 at 10:37
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $G$ be a finite group, and so its Cayley table is finite. Construct finitely presented group $F$ with generators elements in $G$, and relators $abc^{-1}$ if ab=c in the Cayley table.
answered Nov 26 at 10:25
mathnoob
1,495319
Let $G$ be a finite group, and so its Cayley table is finite. Construct finitely presented group $F$ with generators elements in $G$, and relators $abc^{-1}$ if ab=c in the Cayley table.
answered Nov 26 at 10:25
mathnoob
1,495319
answered Nov 26 at 10:25
mathnoob
1,495319
answered Nov 26 at 10:25
mathnoob
1,495319
answered Nov 26 at 10:25
mathnoob
1,495319
1,495319
the problem is: how to construct $F$? is there a general algorithm?
– A.Messab
Nov 26 at 10:37
add a comment |
the problem is: how to construct $F$? is there a general algorithm?
– A.Messab
Nov 26 at 10:37
the problem is: how to construct $F$? is there a general algorithm?
– A.Messab
Nov 26 at 10:37
the problem is: how to construct $F$? is there a general algorithm?
– A.Messab
Nov 26 at 10:37
add a comment |
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I'm not sure what you mean by "safe". The notation $Gsimeqlangle X|Rrangle$ means $Gsimeq F(X)/N(R)$ by definition. And yes, there's a theorem that every finite group is finitely presented.
– freakish
Nov 26 at 10:08
@freakish It means that: is it true that $N$ with that property always exists?
– A.Messab
Nov 26 at 10:11
With what property? Are you asking if every finite group is finitely presented? That's really confusing. Plus: what does "Fp" stand for in the title?
– freakish
Nov 26 at 10:12
$N$ is the normal closure of $R$ in $F$(so the existence of $R$ is also questioned) AND $G cong F/N$
– A.Messab
Nov 26 at 10:15
If $G$ is any group and $Xsubseteq G$ is a set of generators then there's a unique epimorphism $phi:F(X)to G$ such that $phi(x)=x$ for $xin X$. Thus the first isomorphism theorem applies and $Gsimeq F(X)/N$ where $N=ker(phi)$ clearly exists. You can take $R=N$.
– freakish
Nov 26 at 10:17