Free groups : Isomorphism beetween the free group $F_X$ and a group $F$

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The Question : Let $F$ be a group and let $j : X rightarrow F $ be a mapping.

Assume that for every
mapping $f$ of $X$ into a group $G$, there is a homomorphism $phi$ of F into G unique such that
$f = ϕ ◦ j$ . Show that $F_X simeq F $.

My Attempt : I used the universal property : Let $η : X rightarrow F_X$ be the canonical injection. For every mapping
$f$ of $X$ into a group $G$, there is a homomorphism $ϕ'$ of $F_X$ into $G$ unique such
that $f = ϕ' ◦ η$, namely
$ϕ' (a_1, a_2, . . ., a_n) = f (a_1) f (a_2) · · · f (a_n)$.

I searched a homomorphism , I found this one $zeta : F_X rightarrow F$ such that $zeta(())=0_F $ (The image of the empty word) and $ zeta(a_1, a_2, . . ., a_n) = j(a_1) j (a_2) · · · j (a_n) $ . I couldn't find the kernel of this homomorphism. if $Ker zeta=0$ then $F_X simeq F $. Otherwise finding another homomorphism will be a key to solve this problem.

Thank you for reading my question !

abstract-algebra group-theory free-groups

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asked Jul 25 at 14:32

Briedj Yacine

1128

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up vote
0
down vote

favorite

The Question : Let $F$ be a group and let $j : X rightarrow F $ be a mapping.

Assume that for every
mapping $f$ of $X$ into a group $G$, there is a homomorphism $phi$ of F into G unique such that
$f = ϕ ◦ j$ . Show that $F_X simeq F $.

My Attempt : I used the universal property : Let $η : X rightarrow F_X$ be the canonical injection. For every mapping
$f$ of $X$ into a group $G$, there is a homomorphism $ϕ'$ of $F_X$ into $G$ unique such
that $f = ϕ' ◦ η$, namely
$ϕ' (a_1, a_2, . . ., a_n) = f (a_1) f (a_2) · · · f (a_n)$.

I searched a homomorphism , I found this one $zeta : F_X rightarrow F$ such that $zeta(())=0_F $ (The image of the empty word) and $ zeta(a_1, a_2, . . ., a_n) = j(a_1) j (a_2) · · · j (a_n) $ . I couldn't find the kernel of this homomorphism. if $Ker zeta=0$ then $F_X simeq F $. Otherwise finding another homomorphism will be a key to solve this problem.

Thank you for reading my question !

abstract-algebra group-theory free-groups

share|cite|improve this question

asked Jul 25 at 14:32

Briedj Yacine

1128

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

The Question : Let $F$ be a group and let $j : X rightarrow F $ be a mapping.

Assume that for every
mapping $f$ of $X$ into a group $G$, there is a homomorphism $phi$ of F into G unique such that
$f = ϕ ◦ j$ . Show that $F_X simeq F $.

My Attempt : I used the universal property : Let $η : X rightarrow F_X$ be the canonical injection. For every mapping
$f$ of $X$ into a group $G$, there is a homomorphism $ϕ'$ of $F_X$ into $G$ unique such
that $f = ϕ' ◦ η$, namely
$ϕ' (a_1, a_2, . . ., a_n) = f (a_1) f (a_2) · · · f (a_n)$.

I searched a homomorphism , I found this one $zeta : F_X rightarrow F$ such that $zeta(())=0_F $ (The image of the empty word) and $ zeta(a_1, a_2, . . ., a_n) = j(a_1) j (a_2) · · · j (a_n) $ . I couldn't find the kernel of this homomorphism. if $Ker zeta=0$ then $F_X simeq F $. Otherwise finding another homomorphism will be a key to solve this problem.

Thank you for reading my question !

abstract-algebra group-theory free-groups

share|cite|improve this question

asked Jul 25 at 14:32

Briedj Yacine

1128

The Question : Let $F$ be a group and let $j : X rightarrow F $ be a mapping.

Assume that for every
mapping $f$ of $X$ into a group $G$, there is a homomorphism $phi$ of F into G unique such that
$f = ϕ ◦ j$ . Show that $F_X simeq F $.

My Attempt : I used the universal property : Let $η : X rightarrow F_X$ be the canonical injection. For every mapping
$f$ of $X$ into a group $G$, there is a homomorphism $ϕ'$ of $F_X$ into $G$ unique such
that $f = ϕ' ◦ η$, namely
$ϕ' (a_1, a_2, . . ., a_n) = f (a_1) f (a_2) · · · f (a_n)$.

I searched a homomorphism , I found this one $zeta : F_X rightarrow F$ such that $zeta(())=0_F $ (The image of the empty word) and $ zeta(a_1, a_2, . . ., a_n) = j(a_1) j (a_2) · · · j (a_n) $ . I couldn't find the kernel of this homomorphism. if $Ker zeta=0$ then $F_X simeq F $. Otherwise finding another homomorphism will be a key to solve this problem.

Thank you for reading my question !

abstract-algebra group-theory free-groups

share|cite|improve this question

asked Jul 25 at 14:32

Briedj Yacine

1128

share|cite|improve this question

share|cite|improve this question

asked Jul 25 at 14:32

Briedj Yacine

1128

asked Jul 25 at 14:32

Briedj Yacine

1128

asked Jul 25 at 14:32

Briedj Yacine

1128

1128

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

If you know about the universal property of $F_X$, then you know that you can define a unique homomorphism $phicolon F_Xto F$ such that, for every $xin X$,
$$
phi(x)=j(x)
$$
On the other hand, the property of $F$ allows to build a unique homomorphism $psicolon Fto F_X$ such that $eta=psicirc j$.

Now, for $xin X$,
$$
psicircphi(x)=psi(phi(x))=psi(j(x))=psicirc j(x)=eta(x)
$$
Hence, by uniqueness, $psicircphi$ is the identity on $F_X$.

Similarly,
$$
phicircpsi(j(x))=phi(eta(x))=phi(x)=j(x)
$$
By uniqueness, $phicircpsi$ must be the identity on $F$.

share|cite|improve this answer

answered Jul 25 at 14:45

egreg

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Question 1 : Why did you consider that $X subset F_X$ ?
  – Briedj Yacine
  Jul 25 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine If you don't want that, use $eta$, but it's not restrictive to assume $Xsubset F_X$.
  – egreg
  Jul 25 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please explain to me the part of $ϕ∘ψ$ must be the identity on $F$. For the first part $ψ∘ϕ$ is the identity on $F_X$ because $eta (X)$ generates $F_X$ . but in the second case how we can say for sure that every element $y$ in $F$ can be written $y= ϕ∘ψ(x) $ for some $x$. $j(x)$ is injective not surjective. Thank you for your patience.
  – Briedj Yacine
  Jul 25 at 17:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine The identity map $iotacolon Fto F$ satisfies $iotacirc j=j$; so any homomorphism $alphacolon Fto F$ satisfying $alphacirc j=j$ must be equal to the identity.
  – egreg
  Jul 25 at 17:20
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

If you know about the universal property of $F_X$, then you know that you can define a unique homomorphism $phicolon F_Xto F$ such that, for every $xin X$,
$$
phi(x)=j(x)
$$
On the other hand, the property of $F$ allows to build a unique homomorphism $psicolon Fto F_X$ such that $eta=psicirc j$.

Now, for $xin X$,
$$
psicircphi(x)=psi(phi(x))=psi(j(x))=psicirc j(x)=eta(x)
$$
Hence, by uniqueness, $psicircphi$ is the identity on $F_X$.

Similarly,
$$
phicircpsi(j(x))=phi(eta(x))=phi(x)=j(x)
$$
By uniqueness, $phicircpsi$ must be the identity on $F$.

share|cite|improve this answer

answered Jul 25 at 14:45

egreg

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Question 1 : Why did you consider that $X subset F_X$ ?
  – Briedj Yacine
  Jul 25 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine If you don't want that, use $eta$, but it's not restrictive to assume $Xsubset F_X$.
  – egreg
  Jul 25 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please explain to me the part of $ϕ∘ψ$ must be the identity on $F$. For the first part $ψ∘ϕ$ is the identity on $F_X$ because $eta (X)$ generates $F_X$ . but in the second case how we can say for sure that every element $y$ in $F$ can be written $y= ϕ∘ψ(x) $ for some $x$. $j(x)$ is injective not surjective. Thank you for your patience.
  – Briedj Yacine
  Jul 25 at 17:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine The identity map $iotacolon Fto F$ satisfies $iotacirc j=j$; so any homomorphism $alphacolon Fto F$ satisfying $alphacirc j=j$ must be equal to the identity.
  – egreg
  Jul 25 at 17:20
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

If you know about the universal property of $F_X$, then you know that you can define a unique homomorphism $phicolon F_Xto F$ such that, for every $xin X$,
$$
phi(x)=j(x)
$$
On the other hand, the property of $F$ allows to build a unique homomorphism $psicolon Fto F_X$ such that $eta=psicirc j$.

Now, for $xin X$,
$$
psicircphi(x)=psi(phi(x))=psi(j(x))=psicirc j(x)=eta(x)
$$
Hence, by uniqueness, $psicircphi$ is the identity on $F_X$.

Similarly,
$$
phicircpsi(j(x))=phi(eta(x))=phi(x)=j(x)
$$
By uniqueness, $phicircpsi$ must be the identity on $F$.

share|cite|improve this answer

answered Jul 25 at 14:45

egreg

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Question 1 : Why did you consider that $X subset F_X$ ?
  – Briedj Yacine
  Jul 25 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine If you don't want that, use $eta$, but it's not restrictive to assume $Xsubset F_X$.
  – egreg
  Jul 25 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please explain to me the part of $ϕ∘ψ$ must be the identity on $F$. For the first part $ψ∘ϕ$ is the identity on $F_X$ because $eta (X)$ generates $F_X$ . but in the second case how we can say for sure that every element $y$ in $F$ can be written $y= ϕ∘ψ(x) $ for some $x$. $j(x)$ is injective not surjective. Thank you for your patience.
  – Briedj Yacine
  Jul 25 at 17:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine The identity map $iotacolon Fto F$ satisfies $iotacirc j=j$; so any homomorphism $alphacolon Fto F$ satisfying $alphacirc j=j$ must be equal to the identity.
  – egreg
  Jul 25 at 17:20
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

If you know about the universal property of $F_X$, then you know that you can define a unique homomorphism $phicolon F_Xto F$ such that, for every $xin X$,
$$
phi(x)=j(x)
$$
On the other hand, the property of $F$ allows to build a unique homomorphism $psicolon Fto F_X$ such that $eta=psicirc j$.

Now, for $xin X$,
$$
psicircphi(x)=psi(phi(x))=psi(j(x))=psicirc j(x)=eta(x)
$$
Hence, by uniqueness, $psicircphi$ is the identity on $F_X$.

Similarly,
$$
phicircpsi(j(x))=phi(eta(x))=phi(x)=j(x)
$$
By uniqueness, $phicircpsi$ must be the identity on $F$.

share|cite|improve this answer

answered Jul 25 at 14:45

egreg

164k1180187

If you know about the universal property of $F_X$, then you know that you can define a unique homomorphism $phicolon F_Xto F$ such that, for every $xin X$,
$$
phi(x)=j(x)
$$
On the other hand, the property of $F$ allows to build a unique homomorphism $psicolon Fto F_X$ such that $eta=psicirc j$.

Now, for $xin X$,
$$
psicircphi(x)=psi(phi(x))=psi(j(x))=psicirc j(x)=eta(x)
$$
Hence, by uniqueness, $psicircphi$ is the identity on $F_X$.

Similarly,
$$
phicircpsi(j(x))=phi(eta(x))=phi(x)=j(x)
$$
By uniqueness, $phicircpsi$ must be the identity on $F$.

share|cite|improve this answer

answered Jul 25 at 14:45

egreg

164k1180187

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 14:45

egreg

164k1180187

answered Jul 25 at 14:45

egreg

164k1180187

answered Jul 25 at 14:45

egreg

164k1180187

164k1180187

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Question 1 : Why did you consider that $X subset F_X$ ?
  – Briedj Yacine
  Jul 25 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine If you don't want that, use $eta$, but it's not restrictive to assume $Xsubset F_X$.
  – egreg
  Jul 25 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please explain to me the part of $ϕ∘ψ$ must be the identity on $F$. For the first part $ψ∘ϕ$ is the identity on $F_X$ because $eta (X)$ generates $F_X$ . but in the second case how we can say for sure that every element $y$ in $F$ can be written $y= ϕ∘ψ(x) $ for some $x$. $j(x)$ is injective not surjective. Thank you for your patience.
  – Briedj Yacine
  Jul 25 at 17:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine The identity map $iotacolon Fto F$ satisfies $iotacirc j=j$; so any homomorphism $alphacolon Fto F$ satisfying $alphacirc j=j$ must be equal to the identity.
  – egreg
  Jul 25 at 17:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Question 1 : Why did you consider that $X subset F_X$ ?
  – Briedj Yacine
  Jul 25 at 16:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine If you don't want that, use $eta$, but it's not restrictive to assume $Xsubset F_X$.
  – egreg
  Jul 25 at 16:54
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you please explain to me the part of $ϕ∘ψ$ must be the identity on $F$. For the first part $ψ∘ϕ$ is the identity on $F_X$ because $eta (X)$ generates $F_X$ . but in the second case how we can say for sure that every element $y$ in $F$ can be written $y= ϕ∘ψ(x) $ for some $x$. $j(x)$ is injective not surjective. Thank you for your patience.
  – Briedj Yacine
  Jul 25 at 17:13
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BriedjYacine The identity map $iotacolon Fto F$ satisfies $iotacirc j=j$; so any homomorphism $alphacolon Fto F$ satisfying $alphacirc j=j$ must be equal to the identity.
  – egreg
  Jul 25 at 17:20
  

  
  
  
  

Question 1 : Why did you consider that $X subset F_X$ ?
– Briedj Yacine
Jul 25 at 16:51

Question 1 : Why did you consider that $X subset F_X$ ?
– Briedj Yacine
Jul 25 at 16:51

1

1

@BriedjYacine If you don't want that, use $eta$, but it's not restrictive to assume $Xsubset F_X$.
– egreg
Jul 25 at 16:54

@BriedjYacine If you don't want that, use $eta$, but it's not restrictive to assume $Xsubset F_X$.
– egreg
Jul 25 at 16:54

Can you please explain to me the part of $ϕ∘ψ$ must be the identity on $F$. For the first part $ψ∘ϕ$ is the identity on $F_X$ because $eta (X)$ generates $F_X$ . but in the second case how we can say for sure that every element $y$ in $F$ can be written $y= ϕ∘ψ(x) $ for some $x$. $j(x)$ is injective not surjective. Thank you for your patience.
– Briedj Yacine
Jul 25 at 17:13

Can you please explain to me the part of $ϕ∘ψ$ must be the identity on $F$. For the first part $ψ∘ϕ$ is the identity on $F_X$ because $eta (X)$ generates $F_X$ . but in the second case how we can say for sure that every element $y$ in $F$ can be written $y= ϕ∘ψ(x) $ for some $x$. $j(x)$ is injective not surjective. Thank you for your patience.
– Briedj Yacine
Jul 25 at 17:13

1

1

@BriedjYacine The identity map $iotacolon Fto F$ satisfies $iotacirc j=j$; so any homomorphism $alphacolon Fto F$ satisfying $alphacirc j=j$ must be equal to the identity.
– egreg
Jul 25 at 17:20

@BriedjYacine The identity map $iotacolon Fto F$ satisfies $iotacirc j=j$; so any homomorphism $alphacolon Fto F$ satisfying $alphacirc j=j$ must be equal to the identity.
– egreg
Jul 25 at 17:20

add a comment | 

 

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