Proof that Sanov subgroup is free of rank 2
$begingroup$
Okay so i was trying to prove that Sanov subgroup is free of rank 2 without using ping pong lemma. I'd like to prove it directly(if possible), using the universal property.
So Sanov Subgroup is a subgroup of $SL_2(Z)$ generated by $alpha=$
$
begin{pmatrix}
1 & 2 \
0 & 1 \
end{pmatrix}
$
and $beta= begin{pmatrix}
1 & 0 \
2 & 1 \
end{pmatrix}$
So to prove that the Sanov Subgroup $F$ is free of rank two i need that, calling $X={a,b}$ e $i$ the inclusion $i:X->F$(let's say $i(a)=alpha$ and $i(b)=beta$) for every group $G$ and any map $f:X->G$ there exists a unique homomorphism $phi:F->G$ so that $iphi=f$.
That's what i did, even though i don't think this is totally correct. So suppose we have this map $f$ with, let's say, $f(a)=g_1$ and $f(b)=g_2$.
Now the definition of our homomorphism $phi$ is natural, with $phi(alpha)=g_1$ and $phi(beta)=g_2$.
For the other elements $w in F$, $w=alpha^{k_1}beta^{k_2}alpha^{k_3}beta^{k_4}....$ with $k_i in Z$
we define $phi(w)=g_1^{k_1}g_2^{k_2}g_3^{k_3}g_4^{k_4}...$
This is an omomorphism an it satisfies the above universal property. Now i should prove the uniquenes but, again, it's pretty simple. If we had such $tau$ for the universal property it must be $tau(alpha)=g_1$ and $tau(beta)=g_2$ and then by the homomorphism property $tau(wv)=tau(w)tau(v)$ it easily follows that $tau=phi$.
Now this proof doesn't convince me at all (most because i almost never used the structure of Sanov Subgroup, basically i could substitute those two matrices with any other two and use the same proof..)
abstract-algebra group-theory free-groups
asked Jan 6 at 23:58
Andrea LicataAndrea Licata
261
$endgroup$
|
show 3 more comments
$begingroup$
Okay so i was trying to prove that Sanov subgroup is free of rank 2 without using ping pong lemma. I'd like to prove it directly(if possible), using the universal property.
So Sanov Subgroup is a subgroup of $SL_2(Z)$ generated by $alpha=$
$
begin{pmatrix}
1 & 2 \
0 & 1 \
end{pmatrix}
$
and $beta= begin{pmatrix}
1 & 0 \
2 & 1 \
end{pmatrix}$
So to prove that the Sanov Subgroup $F$ is free of rank two i need that, calling $X={a,b}$ e $i$ the inclusion $i:X->F$(let's say $i(a)=alpha$ and $i(b)=beta$) for every group $G$ and any map $f:X->G$ there exists a unique homomorphism $phi:F->G$ so that $iphi=f$.
That's what i did, even though i don't think this is totally correct. So suppose we have this map $f$ with, let's say, $f(a)=g_1$ and $f(b)=g_2$.
Now the definition of our homomorphism $phi$ is natural, with $phi(alpha)=g_1$ and $phi(beta)=g_2$.
For the other elements $w in F$, $w=alpha^{k_1}beta^{k_2}alpha^{k_3}beta^{k_4}....$ with $k_i in Z$
we define $phi(w)=g_1^{k_1}g_2^{k_2}g_3^{k_3}g_4^{k_4}...$
This is an omomorphism an it satisfies the above universal property. Now i should prove the uniquenes but, again, it's pretty simple. If we had such $tau$ for the universal property it must be $tau(alpha)=g_1$ and $tau(beta)=g_2$ and then by the homomorphism property $tau(wv)=tau(w)tau(v)$ it easily follows that $tau=phi$.
Now this proof doesn't convince me at all (most because i almost never used the structure of Sanov Subgroup, basically i could substitute those two matrices with any other two and use the same proof..)
abstract-algebra group-theory free-groups
asked Jan 6 at 23:58
Andrea LicataAndrea Licata
261
$endgroup$
$begingroup$
To typeset an arrow:$to$.
$endgroup$
– Berci
Jan 7 at 0:29
$begingroup$
Yes, basically you have to prove that every element $w$ of the Sanov group can be uniquely written in the given form, though it might require two other generators..
$endgroup$
– Berci
Jan 7 at 0:33
$begingroup$
But that's the definition of an element of a group generated by two elements..should I need to show that there isn't another pair of generators?
$endgroup$
– Andrea Licata
Jan 7 at 1:16
2
$begingroup$
I do not think that this is a sensible thing to attempt, and I doubt whether you will succeed. The ping pong lemma is exactly the correct tool that you need to prove this result. Why would you want to avoid using it?
$endgroup$
– Derek Holt
Jan 7 at 10:14
$begingroup$
That was just something i wanted to know if possible, but more than that i was interested in the reasons of why the above proof doesn't work. Now we know that it doesn't work because i'm not sure that every element cold be written uniquely in that form, is that the difference between a free group and a normal group? That in a normal group among the $alpha^{k_1}beta^{k_2}....alpha^{k_n}beta^{k_n}$ some relationship could occur and then the writing is not unique?
$endgroup$
– Andrea Licata
Jan 7 at 14:47
|
show 3 more comments
$begingroup$
Okay so i was trying to prove that Sanov subgroup is free of rank 2 without using ping pong lemma. I'd like to prove it directly(if possible), using the universal property.
So Sanov Subgroup is a subgroup of $SL_2(Z)$ generated by $alpha=$
$
begin{pmatrix}
1 & 2 \
0 & 1 \
end{pmatrix}
$
and $beta= begin{pmatrix}
1 & 0 \
2 & 1 \
end{pmatrix}$
So to prove that the Sanov Subgroup $F$ is free of rank two i need that, calling $X={a,b}$ e $i$ the inclusion $i:X->F$(let's say $i(a)=alpha$ and $i(b)=beta$) for every group $G$ and any map $f:X->G$ there exists a unique homomorphism $phi:F->G$ so that $iphi=f$.
That's what i did, even though i don't think this is totally correct. So suppose we have this map $f$ with, let's say, $f(a)=g_1$ and $f(b)=g_2$.
Now the definition of our homomorphism $phi$ is natural, with $phi(alpha)=g_1$ and $phi(beta)=g_2$.
For the other elements $w in F$, $w=alpha^{k_1}beta^{k_2}alpha^{k_3}beta^{k_4}....$ with $k_i in Z$
we define $phi(w)=g_1^{k_1}g_2^{k_2}g_3^{k_3}g_4^{k_4}...$
This is an omomorphism an it satisfies the above universal property. Now i should prove the uniquenes but, again, it's pretty simple. If we had such $tau$ for the universal property it must be $tau(alpha)=g_1$ and $tau(beta)=g_2$ and then by the homomorphism property $tau(wv)=tau(w)tau(v)$ it easily follows that $tau=phi$.
Now this proof doesn't convince me at all (most because i almost never used the structure of Sanov Subgroup, basically i could substitute those two matrices with any other two and use the same proof..)
abstract-algebra group-theory free-groups
asked Jan 6 at 23:58
Andrea LicataAndrea Licata
261
$endgroup$
Okay so i was trying to prove that Sanov subgroup is free of rank 2 without using ping pong lemma. I'd like to prove it directly(if possible), using the universal property.
So Sanov Subgroup is a subgroup of $SL_2(Z)$ generated by $alpha=$
$
begin{pmatrix}
1 & 2 \
0 & 1 \
end{pmatrix}
$
and $beta= begin{pmatrix}
1 & 0 \
2 & 1 \
end{pmatrix}$
So to prove that the Sanov Subgroup $F$ is free of rank two i need that, calling $X={a,b}$ e $i$ the inclusion $i:X->F$(let's say $i(a)=alpha$ and $i(b)=beta$) for every group $G$ and any map $f:X->G$ there exists a unique homomorphism $phi:F->G$ so that $iphi=f$.
That's what i did, even though i don't think this is totally correct. So suppose we have this map $f$ with, let's say, $f(a)=g_1$ and $f(b)=g_2$.
Now the definition of our homomorphism $phi$ is natural, with $phi(alpha)=g_1$ and $phi(beta)=g_2$.
For the other elements $w in F$, $w=alpha^{k_1}beta^{k_2}alpha^{k_3}beta^{k_4}....$ with $k_i in Z$
we define $phi(w)=g_1^{k_1}g_2^{k_2}g_3^{k_3}g_4^{k_4}...$
This is an omomorphism an it satisfies the above universal property. Now i should prove the uniquenes but, again, it's pretty simple. If we had such $tau$ for the universal property it must be $tau(alpha)=g_1$ and $tau(beta)=g_2$ and then by the homomorphism property $tau(wv)=tau(w)tau(v)$ it easily follows that $tau=phi$.
Now this proof doesn't convince me at all (most because i almost never used the structure of Sanov Subgroup, basically i could substitute those two matrices with any other two and use the same proof..)
abstract-algebra group-theory free-groups
abstract-algebra group-theory free-groups
asked Jan 6 at 23:58
Andrea LicataAndrea Licata
261
asked Jan 6 at 23:58
Andrea LicataAndrea Licata
261
asked Jan 6 at 23:58
Andrea LicataAndrea Licata
261
asked Jan 6 at 23:58
Andrea LicataAndrea Licata
261
asked Jan 6 at 23:58
Andrea LicataAndrea Licata
261
261
$begingroup$
To typeset an arrow:$to$.
$endgroup$
– Berci
Jan 7 at 0:29
$begingroup$
Yes, basically you have to prove that every element $w$ of the Sanov group can be uniquely written in the given form, though it might require two other generators..
$endgroup$
– Berci
Jan 7 at 0:33
$begingroup$
But that's the definition of an element of a group generated by two elements..should I need to show that there isn't another pair of generators?
$endgroup$
– Andrea Licata
Jan 7 at 1:16
2
$begingroup$
I do not think that this is a sensible thing to attempt, and I doubt whether you will succeed. The ping pong lemma is exactly the correct tool that you need to prove this result. Why would you want to avoid using it?
$endgroup$
– Derek Holt
Jan 7 at 10:14
$begingroup$
That was just something i wanted to know if possible, but more than that i was interested in the reasons of why the above proof doesn't work. Now we know that it doesn't work because i'm not sure that every element cold be written uniquely in that form, is that the difference between a free group and a normal group? That in a normal group among the $alpha^{k_1}beta^{k_2}....alpha^{k_n}beta^{k_n}$ some relationship could occur and then the writing is not unique?
$endgroup$
– Andrea Licata
Jan 7 at 14:47
|
show 3 more comments
$begingroup$
To typeset an arrow:$to$.
$endgroup$
– Berci
Jan 7 at 0:29
$begingroup$
Yes, basically you have to prove that every element $w$ of the Sanov group can be uniquely written in the given form, though it might require two other generators..
$endgroup$
– Berci
Jan 7 at 0:33
$begingroup$
But that's the definition of an element of a group generated by two elements..should I need to show that there isn't another pair of generators?
$endgroup$
– Andrea Licata
Jan 7 at 1:16
2
$begingroup$
I do not think that this is a sensible thing to attempt, and I doubt whether you will succeed. The ping pong lemma is exactly the correct tool that you need to prove this result. Why would you want to avoid using it?
$endgroup$
– Derek Holt
Jan 7 at 10:14
$begingroup$
That was just something i wanted to know if possible, but more than that i was interested in the reasons of why the above proof doesn't work. Now we know that it doesn't work because i'm not sure that every element cold be written uniquely in that form, is that the difference between a free group and a normal group? That in a normal group among the $alpha^{k_1}beta^{k_2}....alpha^{k_n}beta^{k_n}$ some relationship could occur and then the writing is not unique?
$endgroup$
– Andrea Licata
Jan 7 at 14:47
$begingroup$
To typeset an arrow:
$to$.$endgroup$
– Berci
Jan 7 at 0:29
$begingroup$
To typeset an arrow:
$to$.$endgroup$
– Berci
Jan 7 at 0:29
$begingroup$
Yes, basically you have to prove that every element $w$ of the Sanov group can be uniquely written in the given form, though it might require two other generators..
$endgroup$
– Berci
Jan 7 at 0:33
$begingroup$
Yes, basically you have to prove that every element $w$ of the Sanov group can be uniquely written in the given form, though it might require two other generators..
$endgroup$
– Berci
Jan 7 at 0:33
$begingroup$
But that's the definition of an element of a group generated by two elements..should I need to show that there isn't another pair of generators?
$endgroup$
– Andrea Licata
Jan 7 at 1:16
$begingroup$
But that's the definition of an element of a group generated by two elements..should I need to show that there isn't another pair of generators?
$endgroup$
– Andrea Licata
Jan 7 at 1:16
2
2
$begingroup$
I do not think that this is a sensible thing to attempt, and I doubt whether you will succeed. The ping pong lemma is exactly the correct tool that you need to prove this result. Why would you want to avoid using it?
$endgroup$
– Derek Holt
Jan 7 at 10:14
$begingroup$
I do not think that this is a sensible thing to attempt, and I doubt whether you will succeed. The ping pong lemma is exactly the correct tool that you need to prove this result. Why would you want to avoid using it?
$endgroup$
– Derek Holt
Jan 7 at 10:14
$begingroup$
That was just something i wanted to know if possible, but more than that i was interested in the reasons of why the above proof doesn't work. Now we know that it doesn't work because i'm not sure that every element cold be written uniquely in that form, is that the difference between a free group and a normal group? That in a normal group among the $alpha^{k_1}beta^{k_2}....alpha^{k_n}beta^{k_n}$ some relationship could occur and then the writing is not unique?
$endgroup$
– Andrea Licata
Jan 7 at 14:47
$begingroup$
That was just something i wanted to know if possible, but more than that i was interested in the reasons of why the above proof doesn't work. Now we know that it doesn't work because i'm not sure that every element cold be written uniquely in that form, is that the difference between a free group and a normal group? That in a normal group among the $alpha^{k_1}beta^{k_2}....alpha^{k_n}beta^{k_n}$ some relationship could occur and then the writing is not unique?
$endgroup$
– Andrea Licata
Jan 7 at 14:47
|
show 3 more comments
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$begingroup$
To typeset an arrow:
$to$.$endgroup$
– Berci
Jan 7 at 0:29
$begingroup$
Yes, basically you have to prove that every element $w$ of the Sanov group can be uniquely written in the given form, though it might require two other generators..
$endgroup$
– Berci
Jan 7 at 0:33
$begingroup$
But that's the definition of an element of a group generated by two elements..should I need to show that there isn't another pair of generators?
$endgroup$
– Andrea Licata
Jan 7 at 1:16
2
$begingroup$
I do not think that this is a sensible thing to attempt, and I doubt whether you will succeed. The ping pong lemma is exactly the correct tool that you need to prove this result. Why would you want to avoid using it?
$endgroup$
– Derek Holt
Jan 7 at 10:14
$begingroup$
That was just something i wanted to know if possible, but more than that i was interested in the reasons of why the above proof doesn't work. Now we know that it doesn't work because i'm not sure that every element cold be written uniquely in that form, is that the difference between a free group and a normal group? That in a normal group among the $alpha^{k_1}beta^{k_2}....alpha^{k_n}beta^{k_n}$ some relationship could occur and then the writing is not unique?
$endgroup$
– Andrea Licata
Jan 7 at 14:47