Nielsen-Schreier formula application
$begingroup$
So I was thinking about Nielsen-Schreier formula for the generators of a free group subgroup and i tried to apply it. Something must be wrong but I can't find the mistake.
So I have $F(a,b)$ free group generated by two element and an surjective homomorphism $F(a,b)to F(b)$ that acts by killing the "a" letter from a word
(i.e. $ phi(aba^{-1}b)=bb=b^2)$.
So we should have $frac{F(a,b)}{mathrm{ker}(phi)} cong F(b)$ free group on one element.
I want to find the generators of the kernel by applying Nielsen-Schreier formula, which states that $mathrm{ker}(phi)$ is freely generated by this set
$$B = { u x (overline{u x})^{-1} mid u in U, x in X }$$
where $U$ is a Schreier transversal for $mathrm{ker}(phi)$ and $X$ is ${a,b}$.
I figured that $mathrm{ker}(phi)$ obviously contains $F(a)$ and that consists of all the words that has the same amount of $b$ and $b^{-1}$ or, in other words, of all the words $w=a^{n_1}b^{m1}a^{n2}b^{m2}cdots $ so that $sum_{i=1}^n m_i=0$.
In the next step I tried to compute the quotient $frac{F(a,b)}{mathrm{ker}(phi)}={1,bH,b^{-1}H,b^2H,b^3H,ldots}$. I'm pretty sure that's it because two words are in relation only if, summing the power of all the $b$ and $b^{-1}$ in the word the result is the same. (for example $w=b^{-1}abaab$ with a $b$ total power of $-1+1+1=1$ is in relationship with $v=aba$, in fact $wv^{-1} in mathrm{ker}(phi)$.
So the Schreier Transversal is $U={1,b,b^2,ldots}$. The problem is that applying now the formula to this Schreier Transversal I got that the generator of $mathrm{ker}(phi)$ is ${a}$, which makes no sense.
Where's the mistake?
abstract-algebra free-groups
edited Dec 14 '18 at 20:21
Arturo Magidin
261k33585906
asked Dec 14 '18 at 18:38
Andrea LicataAndrea Licata
161
$endgroup$
add a comment |
$begingroup$
So I was thinking about Nielsen-Schreier formula for the generators of a free group subgroup and i tried to apply it. Something must be wrong but I can't find the mistake.
So I have $F(a,b)$ free group generated by two element and an surjective homomorphism $F(a,b)to F(b)$ that acts by killing the "a" letter from a word
(i.e. $ phi(aba^{-1}b)=bb=b^2)$.
So we should have $frac{F(a,b)}{mathrm{ker}(phi)} cong F(b)$ free group on one element.
I want to find the generators of the kernel by applying Nielsen-Schreier formula, which states that $mathrm{ker}(phi)$ is freely generated by this set
$$B = { u x (overline{u x})^{-1} mid u in U, x in X }$$
where $U$ is a Schreier transversal for $mathrm{ker}(phi)$ and $X$ is ${a,b}$.
I figured that $mathrm{ker}(phi)$ obviously contains $F(a)$ and that consists of all the words that has the same amount of $b$ and $b^{-1}$ or, in other words, of all the words $w=a^{n_1}b^{m1}a^{n2}b^{m2}cdots $ so that $sum_{i=1}^n m_i=0$.
In the next step I tried to compute the quotient $frac{F(a,b)}{mathrm{ker}(phi)}={1,bH,b^{-1}H,b^2H,b^3H,ldots}$. I'm pretty sure that's it because two words are in relation only if, summing the power of all the $b$ and $b^{-1}$ in the word the result is the same. (for example $w=b^{-1}abaab$ with a $b$ total power of $-1+1+1=1$ is in relationship with $v=aba$, in fact $wv^{-1} in mathrm{ker}(phi)$.
So the Schreier Transversal is $U={1,b,b^2,ldots}$. The problem is that applying now the formula to this Schreier Transversal I got that the generator of $mathrm{ker}(phi)$ is ${a}$, which makes no sense.
Where's the mistake?
abstract-algebra free-groups
edited Dec 14 '18 at 20:21
Arturo Magidin
261k33585906
asked Dec 14 '18 at 18:38
Andrea LicataAndrea Licata
161
$endgroup$
$begingroup$
Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
$endgroup$
– Arturo Magidin
Dec 14 '18 at 20:29
$begingroup$
Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
$endgroup$
– Arturo Magidin
Dec 14 '18 at 21:28
$begingroup$
Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
$endgroup$
– Andrea Licata
Dec 17 '18 at 16:08
$begingroup$
Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 19:31
add a comment |
$begingroup$
So I was thinking about Nielsen-Schreier formula for the generators of a free group subgroup and i tried to apply it. Something must be wrong but I can't find the mistake.
So I have $F(a,b)$ free group generated by two element and an surjective homomorphism $F(a,b)to F(b)$ that acts by killing the "a" letter from a word
(i.e. $ phi(aba^{-1}b)=bb=b^2)$.
So we should have $frac{F(a,b)}{mathrm{ker}(phi)} cong F(b)$ free group on one element.
I want to find the generators of the kernel by applying Nielsen-Schreier formula, which states that $mathrm{ker}(phi)$ is freely generated by this set
$$B = { u x (overline{u x})^{-1} mid u in U, x in X }$$
where $U$ is a Schreier transversal for $mathrm{ker}(phi)$ and $X$ is ${a,b}$.
I figured that $mathrm{ker}(phi)$ obviously contains $F(a)$ and that consists of all the words that has the same amount of $b$ and $b^{-1}$ or, in other words, of all the words $w=a^{n_1}b^{m1}a^{n2}b^{m2}cdots $ so that $sum_{i=1}^n m_i=0$.
In the next step I tried to compute the quotient $frac{F(a,b)}{mathrm{ker}(phi)}={1,bH,b^{-1}H,b^2H,b^3H,ldots}$. I'm pretty sure that's it because two words are in relation only if, summing the power of all the $b$ and $b^{-1}$ in the word the result is the same. (for example $w=b^{-1}abaab$ with a $b$ total power of $-1+1+1=1$ is in relationship with $v=aba$, in fact $wv^{-1} in mathrm{ker}(phi)$.
So the Schreier Transversal is $U={1,b,b^2,ldots}$. The problem is that applying now the formula to this Schreier Transversal I got that the generator of $mathrm{ker}(phi)$ is ${a}$, which makes no sense.
Where's the mistake?
abstract-algebra free-groups
edited Dec 14 '18 at 20:21
Arturo Magidin
261k33585906
asked Dec 14 '18 at 18:38
Andrea LicataAndrea Licata
161
$endgroup$
So I was thinking about Nielsen-Schreier formula for the generators of a free group subgroup and i tried to apply it. Something must be wrong but I can't find the mistake.
So I have $F(a,b)$ free group generated by two element and an surjective homomorphism $F(a,b)to F(b)$ that acts by killing the "a" letter from a word
(i.e. $ phi(aba^{-1}b)=bb=b^2)$.
So we should have $frac{F(a,b)}{mathrm{ker}(phi)} cong F(b)$ free group on one element.
I want to find the generators of the kernel by applying Nielsen-Schreier formula, which states that $mathrm{ker}(phi)$ is freely generated by this set
$$B = { u x (overline{u x})^{-1} mid u in U, x in X }$$
where $U$ is a Schreier transversal for $mathrm{ker}(phi)$ and $X$ is ${a,b}$.
I figured that $mathrm{ker}(phi)$ obviously contains $F(a)$ and that consists of all the words that has the same amount of $b$ and $b^{-1}$ or, in other words, of all the words $w=a^{n_1}b^{m1}a^{n2}b^{m2}cdots $ so that $sum_{i=1}^n m_i=0$.
In the next step I tried to compute the quotient $frac{F(a,b)}{mathrm{ker}(phi)}={1,bH,b^{-1}H,b^2H,b^3H,ldots}$. I'm pretty sure that's it because two words are in relation only if, summing the power of all the $b$ and $b^{-1}$ in the word the result is the same. (for example $w=b^{-1}abaab$ with a $b$ total power of $-1+1+1=1$ is in relationship with $v=aba$, in fact $wv^{-1} in mathrm{ker}(phi)$.
So the Schreier Transversal is $U={1,b,b^2,ldots}$. The problem is that applying now the formula to this Schreier Transversal I got that the generator of $mathrm{ker}(phi)$ is ${a}$, which makes no sense.
Where's the mistake?
abstract-algebra free-groups
abstract-algebra free-groups
edited Dec 14 '18 at 20:21
Arturo Magidin
261k33585906
asked Dec 14 '18 at 18:38
Andrea LicataAndrea Licata
161
edited Dec 14 '18 at 20:21
Arturo Magidin
261k33585906
asked Dec 14 '18 at 18:38
Andrea LicataAndrea Licata
161
edited Dec 14 '18 at 20:21
Arturo Magidin
261k33585906
edited Dec 14 '18 at 20:21
Arturo Magidin
261k33585906
edited Dec 14 '18 at 20:21
Arturo Magidin
261k33585906
261k33585906
asked Dec 14 '18 at 18:38
Andrea LicataAndrea Licata
161
asked Dec 14 '18 at 18:38
Andrea LicataAndrea Licata
161
asked Dec 14 '18 at 18:38
Andrea LicataAndrea Licata
161
161
$begingroup$
Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
$endgroup$
– Arturo Magidin
Dec 14 '18 at 20:29
$begingroup$
Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
$endgroup$
– Arturo Magidin
Dec 14 '18 at 21:28
$begingroup$
Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
$endgroup$
– Andrea Licata
Dec 17 '18 at 16:08
$begingroup$
Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 19:31
add a comment |
$begingroup$
Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
$endgroup$
– Arturo Magidin
Dec 14 '18 at 20:29
$begingroup$
Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
$endgroup$
– Arturo Magidin
Dec 14 '18 at 21:28
$begingroup$
Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
$endgroup$
– Andrea Licata
Dec 17 '18 at 16:08
$begingroup$
Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 19:31
$begingroup$
Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
$endgroup$
– Arturo Magidin
Dec 14 '18 at 20:29
$begingroup$
Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
$endgroup$
– Arturo Magidin
Dec 14 '18 at 20:29
$begingroup$
Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
$endgroup$
– Arturo Magidin
Dec 14 '18 at 21:28
$begingroup$
Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
$endgroup$
– Arturo Magidin
Dec 14 '18 at 21:28
$begingroup$
Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
$endgroup$
– Andrea Licata
Dec 17 '18 at 16:08
$begingroup$
Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
$endgroup$
– Andrea Licata
Dec 17 '18 at 16:08
$begingroup$
Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 19:31
$begingroup$
Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 19:31
add a comment |
1 Answer
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I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$
answered Dec 30 '18 at 19:33
Andrea LicataAndrea Licata
161
$endgroup$
add a comment |
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$begingroup$
I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$
answered Dec 30 '18 at 19:33
Andrea LicataAndrea Licata
161
$endgroup$
add a comment |
$begingroup$
I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$
answered Dec 30 '18 at 19:33
Andrea LicataAndrea Licata
161
$endgroup$
add a comment |
$begingroup$
I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$
answered Dec 30 '18 at 19:33
Andrea LicataAndrea Licata
161
$endgroup$
I was miscalculating the base of $ker phi$ ($xu(overline{xu})^{-1}$ instead of $ux(overline{ux})^{-1}$). Applyng the right formula we have that $B={b^iab^{-i}, i in Z}$
answered Dec 30 '18 at 19:33
Andrea LicataAndrea Licata
161
answered Dec 30 '18 at 19:33
Andrea LicataAndrea Licata
161
answered Dec 30 '18 at 19:33
Andrea LicataAndrea Licata
161
answered Dec 30 '18 at 19:33
Andrea LicataAndrea Licata
161
161
add a comment |
add a comment |
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$begingroup$
Isn't the overline meant to indicate that you are taking the inverse of the element of $U$ that corresponds to the coset of $ux$? If so, then your $u$ is of the form $b^i$; if $x=a$, then $b^ia$ corresponds to $b^iin U$, so $ux(overline{ux})^{-1} = b^ia(overline{b^ia})^{-1} = b^ia(b^i)^{-1} = b^iab^{-i}$; so that your generating set consists of all conjugates of $a$ by powers of $b$.
$endgroup$
– Arturo Magidin
Dec 14 '18 at 20:29
$begingroup$
Where you perhaps calculagting $xu(overline{xu})^{-1}$ by mistake?
$endgroup$
– Arturo Magidin
Dec 14 '18 at 21:28
$begingroup$
Yup, that was the problem. I was sure i was doing something dumb! Thank you so much
$endgroup$
– Andrea Licata
Dec 17 '18 at 16:08
$begingroup$
Perhaps you can answer your own question explaining this, so that the query does not go unanswered.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 19:31