Mixing
Subgroup of $text{PSL}(2,mathbb{Z})$ generated by $S$ and $T^2$
$begingroup$
In the group $PSL(2,mathbb{Z})$ (which acts on the upper half plane of $mathbb{C}$), suppose $S$ is the inversion and $T$ is the translation by $1$, i.e.
$$S=
left( {begin{array}{cc}
0 & 1 \
-1 & 0 \
end{array} } right),
T=
left( {begin{array}{cc}
1 & 1 \
0 & 1 \
end{array} } right).
$$
From books on modular forms, the group $PSL(2,mathbb{Z})$ is generated by $S$ and $T$. Let $G$ by the subgroup generated by $S$ and $T^2$,
$$G=langle S,T^2rangle$$
Is $G$ a normal subgroup of $PSL(2,mathbb{Z})$? Is $[PSL(2,mathbb{Z}):G]$ finite? If so, are there studies on the modular forms with congruence group $G$?
Similarly one could ask the questions when $G=langle S,T^krangle, kin mathbb{Z}+$!
group-theory number-theory modular-forms
edited Jan 31 at 3:00
user549397
1,6641518
asked Jan 31 at 1:28
WenzheWenzhe
1,102510
$endgroup$
add a comment |
$begingroup$
In the group $PSL(2,mathbb{Z})$ (which acts on the upper half plane of $mathbb{C}$), suppose $S$ is the inversion and $T$ is the translation by $1$, i.e.
$$S=
left( {begin{array}{cc}
0 & 1 \
-1 & 0 \
end{array} } right),
T=
left( {begin{array}{cc}
1 & 1 \
0 & 1 \
end{array} } right).
$$
From books on modular forms, the group $PSL(2,mathbb{Z})$ is generated by $S$ and $T$. Let $G$ by the subgroup generated by $S$ and $T^2$,
$$G=langle S,T^2rangle$$
Is $G$ a normal subgroup of $PSL(2,mathbb{Z})$? Is $[PSL(2,mathbb{Z}):G]$ finite? If so, are there studies on the modular forms with congruence group $G$?
Similarly one could ask the questions when $G=langle S,T^krangle, kin mathbb{Z}+$!
group-theory number-theory modular-forms
edited Jan 31 at 3:00
user549397
1,6641518
asked Jan 31 at 1:28
WenzheWenzhe
1,102510
$endgroup$
add a comment |
$begingroup$
In the group $PSL(2,mathbb{Z})$ (which acts on the upper half plane of $mathbb{C}$), suppose $S$ is the inversion and $T$ is the translation by $1$, i.e.
$$S=
left( {begin{array}{cc}
0 & 1 \
-1 & 0 \
end{array} } right),
T=
left( {begin{array}{cc}
1 & 1 \
0 & 1 \
end{array} } right).
$$
From books on modular forms, the group $PSL(2,mathbb{Z})$ is generated by $S$ and $T$. Let $G$ by the subgroup generated by $S$ and $T^2$,
$$G=langle S,T^2rangle$$
Is $G$ a normal subgroup of $PSL(2,mathbb{Z})$? Is $[PSL(2,mathbb{Z}):G]$ finite? If so, are there studies on the modular forms with congruence group $G$?
Similarly one could ask the questions when $G=langle S,T^krangle, kin mathbb{Z}+$!
group-theory number-theory modular-forms
edited Jan 31 at 3:00
user549397
1,6641518
asked Jan 31 at 1:28
WenzheWenzhe
1,102510
$endgroup$
In the group $PSL(2,mathbb{Z})$ (which acts on the upper half plane of $mathbb{C}$), suppose $S$ is the inversion and $T$ is the translation by $1$, i.e.
$$S=
left( {begin{array}{cc}
0 & 1 \
-1 & 0 \
end{array} } right),
T=
left( {begin{array}{cc}
1 & 1 \
0 & 1 \
end{array} } right).
$$
From books on modular forms, the group $PSL(2,mathbb{Z})$ is generated by $S$ and $T$. Let $G$ by the subgroup generated by $S$ and $T^2$,
$$G=langle S,T^2rangle$$
Is $G$ a normal subgroup of $PSL(2,mathbb{Z})$? Is $[PSL(2,mathbb{Z}):G]$ finite? If so, are there studies on the modular forms with congruence group $G$?
Similarly one could ask the questions when $G=langle S,T^krangle, kin mathbb{Z}+$!
group-theory number-theory modular-forms
group-theory number-theory modular-forms
edited Jan 31 at 3:00
user549397
1,6641518
asked Jan 31 at 1:28
WenzheWenzhe
1,102510
edited Jan 31 at 3:00
user549397
1,6641518
asked Jan 31 at 1:28
WenzheWenzhe
1,102510
edited Jan 31 at 3:00
user549397
1,6641518
edited Jan 31 at 3:00
user549397
1,6641518
edited Jan 31 at 3:00
user549397
1,6641518
1,6641518
asked Jan 31 at 1:28
WenzheWenzhe
1,102510
asked Jan 31 at 1:28
WenzheWenzhe
1,102510
asked Jan 31 at 1:28
WenzheWenzhe
1,102510
1,102510
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's of infinite index and generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup (hence is not normal).
Let me first check that it generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup. Write $U=TS$. (All equalities are meant in $mathrm{PSL}_2(mathbf{Z})$.) So $U^3=1$. Modding out $langle S,Urangle$ by the relators $S,(US^{-1})^2$ is equivalent to mod out by $S,U^2$, and hence to mod out by $S,U$ since $U^3=1$. Hence the quotient is trivial, which is precisely the given assertion.
Next, let me check that it has infinite index. I use that $mathrm{PSL}_2(mathbf{Z})$ is the free product of $langle Srangle$ and $langle Urangle$.
Every free product $Aast B$ can be written as $B^{ast A}rtimes A$, where the free factors in the kernel are the $aBa^{-1}$ when $a$ ranges over $A$. Here we thus write $mathrm{PSL}_2(mathbf{Z})= (langle Urangleast langle SUS^{-1}rangle)rtimes langle Srangle$. Write $V=SUS^{-1}$: then $T^2=(US^{-1})^2=US^{-1}US=UV$ and $SUS^{-1}=VU$. Thus, in the free subgroup $F$ of index two $langle U,Vrangle$, the intersection with $G$ is generated by $UV,VU$.
Let us check that it has infinite index. Indeed, inside $F$, modding out by $UV,VU$ yields an infinite cyclic group. So the normal subgroup of $F$ generated by $Gcap F$ has infinite index, and hence $Gcap F$ has infinite index in $F$, which in turn implies that $G$ has infinite index in $mathrm{PSL}_2(mathbf{Z})$.
Maybe you can check what this strategy provides when replacing $T^2$ with $T^k$.
answered Jan 31 at 4:16
YCorYCor
8,5171129
$endgroup$
add a comment |
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$begingroup$
It's of infinite index and generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup (hence is not normal).
Let me first check that it generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup. Write $U=TS$. (All equalities are meant in $mathrm{PSL}_2(mathbf{Z})$.) So $U^3=1$. Modding out $langle S,Urangle$ by the relators $S,(US^{-1})^2$ is equivalent to mod out by $S,U^2$, and hence to mod out by $S,U$ since $U^3=1$. Hence the quotient is trivial, which is precisely the given assertion.
Next, let me check that it has infinite index. I use that $mathrm{PSL}_2(mathbf{Z})$ is the free product of $langle Srangle$ and $langle Urangle$.
Every free product $Aast B$ can be written as $B^{ast A}rtimes A$, where the free factors in the kernel are the $aBa^{-1}$ when $a$ ranges over $A$. Here we thus write $mathrm{PSL}_2(mathbf{Z})= (langle Urangleast langle SUS^{-1}rangle)rtimes langle Srangle$. Write $V=SUS^{-1}$: then $T^2=(US^{-1})^2=US^{-1}US=UV$ and $SUS^{-1}=VU$. Thus, in the free subgroup $F$ of index two $langle U,Vrangle$, the intersection with $G$ is generated by $UV,VU$.
Let us check that it has infinite index. Indeed, inside $F$, modding out by $UV,VU$ yields an infinite cyclic group. So the normal subgroup of $F$ generated by $Gcap F$ has infinite index, and hence $Gcap F$ has infinite index in $F$, which in turn implies that $G$ has infinite index in $mathrm{PSL}_2(mathbf{Z})$.
Maybe you can check what this strategy provides when replacing $T^2$ with $T^k$.
answered Jan 31 at 4:16
YCorYCor
8,5171129
$endgroup$
add a comment |
$begingroup$
It's of infinite index and generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup (hence is not normal).
Let me first check that it generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup. Write $U=TS$. (All equalities are meant in $mathrm{PSL}_2(mathbf{Z})$.) So $U^3=1$. Modding out $langle S,Urangle$ by the relators $S,(US^{-1})^2$ is equivalent to mod out by $S,U^2$, and hence to mod out by $S,U$ since $U^3=1$. Hence the quotient is trivial, which is precisely the given assertion.
Next, let me check that it has infinite index. I use that $mathrm{PSL}_2(mathbf{Z})$ is the free product of $langle Srangle$ and $langle Urangle$.
Every free product $Aast B$ can be written as $B^{ast A}rtimes A$, where the free factors in the kernel are the $aBa^{-1}$ when $a$ ranges over $A$. Here we thus write $mathrm{PSL}_2(mathbf{Z})= (langle Urangleast langle SUS^{-1}rangle)rtimes langle Srangle$. Write $V=SUS^{-1}$: then $T^2=(US^{-1})^2=US^{-1}US=UV$ and $SUS^{-1}=VU$. Thus, in the free subgroup $F$ of index two $langle U,Vrangle$, the intersection with $G$ is generated by $UV,VU$.
Let us check that it has infinite index. Indeed, inside $F$, modding out by $UV,VU$ yields an infinite cyclic group. So the normal subgroup of $F$ generated by $Gcap F$ has infinite index, and hence $Gcap F$ has infinite index in $F$, which in turn implies that $G$ has infinite index in $mathrm{PSL}_2(mathbf{Z})$.
Maybe you can check what this strategy provides when replacing $T^2$ with $T^k$.
answered Jan 31 at 4:16
YCorYCor
8,5171129
$endgroup$
add a comment |
$begingroup$
It's of infinite index and generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup (hence is not normal).
Let me first check that it generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup. Write $U=TS$. (All equalities are meant in $mathrm{PSL}_2(mathbf{Z})$.) So $U^3=1$. Modding out $langle S,Urangle$ by the relators $S,(US^{-1})^2$ is equivalent to mod out by $S,U^2$, and hence to mod out by $S,U$ since $U^3=1$. Hence the quotient is trivial, which is precisely the given assertion.
Next, let me check that it has infinite index. I use that $mathrm{PSL}_2(mathbf{Z})$ is the free product of $langle Srangle$ and $langle Urangle$.
Every free product $Aast B$ can be written as $B^{ast A}rtimes A$, where the free factors in the kernel are the $aBa^{-1}$ when $a$ ranges over $A$. Here we thus write $mathrm{PSL}_2(mathbf{Z})= (langle Urangleast langle SUS^{-1}rangle)rtimes langle Srangle$. Write $V=SUS^{-1}$: then $T^2=(US^{-1})^2=US^{-1}US=UV$ and $SUS^{-1}=VU$. Thus, in the free subgroup $F$ of index two $langle U,Vrangle$, the intersection with $G$ is generated by $UV,VU$.
Let us check that it has infinite index. Indeed, inside $F$, modding out by $UV,VU$ yields an infinite cyclic group. So the normal subgroup of $F$ generated by $Gcap F$ has infinite index, and hence $Gcap F$ has infinite index in $F$, which in turn implies that $G$ has infinite index in $mathrm{PSL}_2(mathbf{Z})$.
Maybe you can check what this strategy provides when replacing $T^2$ with $T^k$.
answered Jan 31 at 4:16
YCorYCor
8,5171129
$endgroup$
It's of infinite index and generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup (hence is not normal).
Let me first check that it generates $mathrm{PSL}_2(mathbf{Z})$ as normal subgroup. Write $U=TS$. (All equalities are meant in $mathrm{PSL}_2(mathbf{Z})$.) So $U^3=1$. Modding out $langle S,Urangle$ by the relators $S,(US^{-1})^2$ is equivalent to mod out by $S,U^2$, and hence to mod out by $S,U$ since $U^3=1$. Hence the quotient is trivial, which is precisely the given assertion.
Next, let me check that it has infinite index. I use that $mathrm{PSL}_2(mathbf{Z})$ is the free product of $langle Srangle$ and $langle Urangle$.
Every free product $Aast B$ can be written as $B^{ast A}rtimes A$, where the free factors in the kernel are the $aBa^{-1}$ when $a$ ranges over $A$. Here we thus write $mathrm{PSL}_2(mathbf{Z})= (langle Urangleast langle SUS^{-1}rangle)rtimes langle Srangle$. Write $V=SUS^{-1}$: then $T^2=(US^{-1})^2=US^{-1}US=UV$ and $SUS^{-1}=VU$. Thus, in the free subgroup $F$ of index two $langle U,Vrangle$, the intersection with $G$ is generated by $UV,VU$.
Let us check that it has infinite index. Indeed, inside $F$, modding out by $UV,VU$ yields an infinite cyclic group. So the normal subgroup of $F$ generated by $Gcap F$ has infinite index, and hence $Gcap F$ has infinite index in $F$, which in turn implies that $G$ has infinite index in $mathrm{PSL}_2(mathbf{Z})$.
Maybe you can check what this strategy provides when replacing $T^2$ with $T^k$.
answered Jan 31 at 4:16
YCorYCor
8,5171129
answered Jan 31 at 4:16
YCorYCor
8,5171129
answered Jan 31 at 4:16
YCorYCor
8,5171129
answered Jan 31 at 4:16
YCorYCor
8,5171129
8,5171129
add a comment |
add a comment |
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