Binary tetrahedral group and $rm{SL}_2(mathbb F_3)$












13












$begingroup$


The binary tetrahedral group $mathbb T$ is an interesting 24-element group. For instance it can be expressed as the subgroup
$$ mathbb T = left{ pm 1, pm i, pm j, pm k, dfrac{pm 1 pm i pm j pm k}2 right} subseteq mathbb{H}^times$$
of the multiplicative group of the quaternions (which also gives you a fascinating regular 24-vertex 4-polyhedron).



Alternatively, the binary tetrahedral group can be thought of as the inverse image of $mathfrak A(4)$ through the 2:1 morphism $rm{SU}_2 to rm{SO}_3$ ($mathfrak A(4) subseteq rm{SO}_3$ being the group of direct isometries of the regular tetrahedron).



The property which puzzles me is the following: $mathbb T$ is isomorphic to the matrix group $rm{SL}_2(mathbb F_3)$.



I'm able to give a proof of this result, but not an enlightening one. Essentially, a 24-element group with a normal subgroup isomorphic to the 8-element quaternion group $rm Q_8$ is either isomorphic to $rm Q_8 times rm C_3$ or to the unique nontrivial semidirect product $rm Q_8 rtimes rm C_3$; it's not that hard to prove that both $mathbb T$ and $rm{SL}_2(mathbb F_3)$ satisfy this property and belong to the second case. The only interesting thing about this proof is the description of that normal $rm Q_8$: on the one hand ${pm 1, pm i, pm j, pm k} subset mathbb T$ is the inverse image of the quite exceptional Vierergruppe $V_4 = { rm{id}, (12)(34), (13)(24), (14)(23)} lhd mathfrak A(4)subseteq rm{SO}_3$; on the other hand, the elements of $mathrm{SL}_2(mathbb F_3)$ which are diagonalisable over $mathbb F_9$ — or, equivalently, $pm I_2$ and the matrices whose trace is zero — form a normal subgroup isomorphic to $rm Q_8$ (and it's quite astonishing they do form a subgroup!) whose elements are
$$begin{array}{l} begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix}, begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}, begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}, begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix},\ begin{pmatrix} 1 & 1 \ 1 & -1 end{pmatrix}, begin{pmatrix} 1 & -1 \ -1 & -1 end{pmatrix}, begin{pmatrix} -1 & 1 \ 1 & 1 end{pmatrix}, begin{pmatrix} -1 & -1 \ -1 & 1 end{pmatrix}. end{array}$$



As I said earlier, this proof isn't very satisfying. Hence my question:




Is there an enlightening proof of the isomorphism $mathbb T simeq rm{SL}_2(mathbb F_3)$?











share|cite|improve this question











$endgroup$












  • $begingroup$
    I suspect that the binary tetrahedral group is conjugate to a subgroup of $text{SL}_2(mathbb{Z})$. If so, it's known that if $G$ is a finite subgroup of $text{GL}_n(mathbb{Z})$ then $G$ injects into $text{GL}_n(mathbb{F}_3)$, and that provides a pretty straightforward isomorphism.
    $endgroup$
    – Qiaochu Yuan
    Apr 24 '13 at 17:41






  • 2




    $begingroup$
    The isomorphism $mathfrak A(4) simeq rm{PSL}_2(mathbb F_3)$ is quite easy to see: the action of $rm{PSL}_2(mathbb F_3)$ on the 4-element projective line $rm P^1(mathbb F_3)$ provides an injective morphism $rm{PSL}_2(mathbb F_3) to mathfrak S(4)$ which gives easily the desired isomorphism. What is unclear to me is why the double cover gives precisely $rm{SL}_2(mathbb F_3)$. Maybe there are direct cohomological arguments?
    $endgroup$
    – PseudoNeo
    Apr 24 '13 at 19:35






  • 2




    $begingroup$
    @QiaochuYuan: I doubt it: finite subgroups of $rm{GL}_2(mathbb Z)$ preserve a scalar product, so they are cyclic or dihedral.
    $endgroup$
    – PseudoNeo
    Apr 24 '13 at 19:38






  • 3




    $begingroup$
    Well, this is probably not as geometric as you want, but $SL(2,3)$ is the only nontrivial "double cover" (central extension of order 2) over $PSL(2,3)$.
    $endgroup$
    – user641
    Apr 24 '13 at 23:24






  • 1




    $begingroup$
    There is a 2-dimensional complex representation of $Q_8$, but it cannot be written over the rationals, so $Q_8$ does not embed in ${rm GL}_2({mathbb Z})$. (But ${rm SL}_2(3)$, which contains $Q_8$, does embed in ${rm SL}_2(p)$ for all odd primes $p$.)
    $endgroup$
    – Derek Holt
    Apr 25 '13 at 8:59


















13












$begingroup$


The binary tetrahedral group $mathbb T$ is an interesting 24-element group. For instance it can be expressed as the subgroup
$$ mathbb T = left{ pm 1, pm i, pm j, pm k, dfrac{pm 1 pm i pm j pm k}2 right} subseteq mathbb{H}^times$$
of the multiplicative group of the quaternions (which also gives you a fascinating regular 24-vertex 4-polyhedron).



Alternatively, the binary tetrahedral group can be thought of as the inverse image of $mathfrak A(4)$ through the 2:1 morphism $rm{SU}_2 to rm{SO}_3$ ($mathfrak A(4) subseteq rm{SO}_3$ being the group of direct isometries of the regular tetrahedron).



The property which puzzles me is the following: $mathbb T$ is isomorphic to the matrix group $rm{SL}_2(mathbb F_3)$.



I'm able to give a proof of this result, but not an enlightening one. Essentially, a 24-element group with a normal subgroup isomorphic to the 8-element quaternion group $rm Q_8$ is either isomorphic to $rm Q_8 times rm C_3$ or to the unique nontrivial semidirect product $rm Q_8 rtimes rm C_3$; it's not that hard to prove that both $mathbb T$ and $rm{SL}_2(mathbb F_3)$ satisfy this property and belong to the second case. The only interesting thing about this proof is the description of that normal $rm Q_8$: on the one hand ${pm 1, pm i, pm j, pm k} subset mathbb T$ is the inverse image of the quite exceptional Vierergruppe $V_4 = { rm{id}, (12)(34), (13)(24), (14)(23)} lhd mathfrak A(4)subseteq rm{SO}_3$; on the other hand, the elements of $mathrm{SL}_2(mathbb F_3)$ which are diagonalisable over $mathbb F_9$ — or, equivalently, $pm I_2$ and the matrices whose trace is zero — form a normal subgroup isomorphic to $rm Q_8$ (and it's quite astonishing they do form a subgroup!) whose elements are
$$begin{array}{l} begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix}, begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}, begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}, begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix},\ begin{pmatrix} 1 & 1 \ 1 & -1 end{pmatrix}, begin{pmatrix} 1 & -1 \ -1 & -1 end{pmatrix}, begin{pmatrix} -1 & 1 \ 1 & 1 end{pmatrix}, begin{pmatrix} -1 & -1 \ -1 & 1 end{pmatrix}. end{array}$$



As I said earlier, this proof isn't very satisfying. Hence my question:




Is there an enlightening proof of the isomorphism $mathbb T simeq rm{SL}_2(mathbb F_3)$?











share|cite|improve this question











$endgroup$












  • $begingroup$
    I suspect that the binary tetrahedral group is conjugate to a subgroup of $text{SL}_2(mathbb{Z})$. If so, it's known that if $G$ is a finite subgroup of $text{GL}_n(mathbb{Z})$ then $G$ injects into $text{GL}_n(mathbb{F}_3)$, and that provides a pretty straightforward isomorphism.
    $endgroup$
    – Qiaochu Yuan
    Apr 24 '13 at 17:41






  • 2




    $begingroup$
    The isomorphism $mathfrak A(4) simeq rm{PSL}_2(mathbb F_3)$ is quite easy to see: the action of $rm{PSL}_2(mathbb F_3)$ on the 4-element projective line $rm P^1(mathbb F_3)$ provides an injective morphism $rm{PSL}_2(mathbb F_3) to mathfrak S(4)$ which gives easily the desired isomorphism. What is unclear to me is why the double cover gives precisely $rm{SL}_2(mathbb F_3)$. Maybe there are direct cohomological arguments?
    $endgroup$
    – PseudoNeo
    Apr 24 '13 at 19:35






  • 2




    $begingroup$
    @QiaochuYuan: I doubt it: finite subgroups of $rm{GL}_2(mathbb Z)$ preserve a scalar product, so they are cyclic or dihedral.
    $endgroup$
    – PseudoNeo
    Apr 24 '13 at 19:38






  • 3




    $begingroup$
    Well, this is probably not as geometric as you want, but $SL(2,3)$ is the only nontrivial "double cover" (central extension of order 2) over $PSL(2,3)$.
    $endgroup$
    – user641
    Apr 24 '13 at 23:24






  • 1




    $begingroup$
    There is a 2-dimensional complex representation of $Q_8$, but it cannot be written over the rationals, so $Q_8$ does not embed in ${rm GL}_2({mathbb Z})$. (But ${rm SL}_2(3)$, which contains $Q_8$, does embed in ${rm SL}_2(p)$ for all odd primes $p$.)
    $endgroup$
    – Derek Holt
    Apr 25 '13 at 8:59
















13












13








13


4



$begingroup$


The binary tetrahedral group $mathbb T$ is an interesting 24-element group. For instance it can be expressed as the subgroup
$$ mathbb T = left{ pm 1, pm i, pm j, pm k, dfrac{pm 1 pm i pm j pm k}2 right} subseteq mathbb{H}^times$$
of the multiplicative group of the quaternions (which also gives you a fascinating regular 24-vertex 4-polyhedron).



Alternatively, the binary tetrahedral group can be thought of as the inverse image of $mathfrak A(4)$ through the 2:1 morphism $rm{SU}_2 to rm{SO}_3$ ($mathfrak A(4) subseteq rm{SO}_3$ being the group of direct isometries of the regular tetrahedron).



The property which puzzles me is the following: $mathbb T$ is isomorphic to the matrix group $rm{SL}_2(mathbb F_3)$.



I'm able to give a proof of this result, but not an enlightening one. Essentially, a 24-element group with a normal subgroup isomorphic to the 8-element quaternion group $rm Q_8$ is either isomorphic to $rm Q_8 times rm C_3$ or to the unique nontrivial semidirect product $rm Q_8 rtimes rm C_3$; it's not that hard to prove that both $mathbb T$ and $rm{SL}_2(mathbb F_3)$ satisfy this property and belong to the second case. The only interesting thing about this proof is the description of that normal $rm Q_8$: on the one hand ${pm 1, pm i, pm j, pm k} subset mathbb T$ is the inverse image of the quite exceptional Vierergruppe $V_4 = { rm{id}, (12)(34), (13)(24), (14)(23)} lhd mathfrak A(4)subseteq rm{SO}_3$; on the other hand, the elements of $mathrm{SL}_2(mathbb F_3)$ which are diagonalisable over $mathbb F_9$ — or, equivalently, $pm I_2$ and the matrices whose trace is zero — form a normal subgroup isomorphic to $rm Q_8$ (and it's quite astonishing they do form a subgroup!) whose elements are
$$begin{array}{l} begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix}, begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}, begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}, begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix},\ begin{pmatrix} 1 & 1 \ 1 & -1 end{pmatrix}, begin{pmatrix} 1 & -1 \ -1 & -1 end{pmatrix}, begin{pmatrix} -1 & 1 \ 1 & 1 end{pmatrix}, begin{pmatrix} -1 & -1 \ -1 & 1 end{pmatrix}. end{array}$$



As I said earlier, this proof isn't very satisfying. Hence my question:




Is there an enlightening proof of the isomorphism $mathbb T simeq rm{SL}_2(mathbb F_3)$?











share|cite|improve this question











$endgroup$




The binary tetrahedral group $mathbb T$ is an interesting 24-element group. For instance it can be expressed as the subgroup
$$ mathbb T = left{ pm 1, pm i, pm j, pm k, dfrac{pm 1 pm i pm j pm k}2 right} subseteq mathbb{H}^times$$
of the multiplicative group of the quaternions (which also gives you a fascinating regular 24-vertex 4-polyhedron).



Alternatively, the binary tetrahedral group can be thought of as the inverse image of $mathfrak A(4)$ through the 2:1 morphism $rm{SU}_2 to rm{SO}_3$ ($mathfrak A(4) subseteq rm{SO}_3$ being the group of direct isometries of the regular tetrahedron).



The property which puzzles me is the following: $mathbb T$ is isomorphic to the matrix group $rm{SL}_2(mathbb F_3)$.



I'm able to give a proof of this result, but not an enlightening one. Essentially, a 24-element group with a normal subgroup isomorphic to the 8-element quaternion group $rm Q_8$ is either isomorphic to $rm Q_8 times rm C_3$ or to the unique nontrivial semidirect product $rm Q_8 rtimes rm C_3$; it's not that hard to prove that both $mathbb T$ and $rm{SL}_2(mathbb F_3)$ satisfy this property and belong to the second case. The only interesting thing about this proof is the description of that normal $rm Q_8$: on the one hand ${pm 1, pm i, pm j, pm k} subset mathbb T$ is the inverse image of the quite exceptional Vierergruppe $V_4 = { rm{id}, (12)(34), (13)(24), (14)(23)} lhd mathfrak A(4)subseteq rm{SO}_3$; on the other hand, the elements of $mathrm{SL}_2(mathbb F_3)$ which are diagonalisable over $mathbb F_9$ — or, equivalently, $pm I_2$ and the matrices whose trace is zero — form a normal subgroup isomorphic to $rm Q_8$ (and it's quite astonishing they do form a subgroup!) whose elements are
$$begin{array}{l} begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix}, begin{pmatrix} -1 & 0 \ 0 & -1 end{pmatrix}, begin{pmatrix} 0 & 1 \ -1 & 0 end{pmatrix}, begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix},\ begin{pmatrix} 1 & 1 \ 1 & -1 end{pmatrix}, begin{pmatrix} 1 & -1 \ -1 & -1 end{pmatrix}, begin{pmatrix} -1 & 1 \ 1 & 1 end{pmatrix}, begin{pmatrix} -1 & -1 \ -1 & 1 end{pmatrix}. end{array}$$



As I said earlier, this proof isn't very satisfying. Hence my question:




Is there an enlightening proof of the isomorphism $mathbb T simeq rm{SL}_2(mathbb F_3)$?








group-theory finite-groups quaternions exceptional-isomorphisms






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share|cite|improve this question













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share|cite|improve this question








edited May 24 '13 at 11:44









Grigory M

13.6k357104




13.6k357104










asked Apr 24 '13 at 14:17









PseudoNeoPseudoNeo

6,5492053




6,5492053












  • $begingroup$
    I suspect that the binary tetrahedral group is conjugate to a subgroup of $text{SL}_2(mathbb{Z})$. If so, it's known that if $G$ is a finite subgroup of $text{GL}_n(mathbb{Z})$ then $G$ injects into $text{GL}_n(mathbb{F}_3)$, and that provides a pretty straightforward isomorphism.
    $endgroup$
    – Qiaochu Yuan
    Apr 24 '13 at 17:41






  • 2




    $begingroup$
    The isomorphism $mathfrak A(4) simeq rm{PSL}_2(mathbb F_3)$ is quite easy to see: the action of $rm{PSL}_2(mathbb F_3)$ on the 4-element projective line $rm P^1(mathbb F_3)$ provides an injective morphism $rm{PSL}_2(mathbb F_3) to mathfrak S(4)$ which gives easily the desired isomorphism. What is unclear to me is why the double cover gives precisely $rm{SL}_2(mathbb F_3)$. Maybe there are direct cohomological arguments?
    $endgroup$
    – PseudoNeo
    Apr 24 '13 at 19:35






  • 2




    $begingroup$
    @QiaochuYuan: I doubt it: finite subgroups of $rm{GL}_2(mathbb Z)$ preserve a scalar product, so they are cyclic or dihedral.
    $endgroup$
    – PseudoNeo
    Apr 24 '13 at 19:38






  • 3




    $begingroup$
    Well, this is probably not as geometric as you want, but $SL(2,3)$ is the only nontrivial "double cover" (central extension of order 2) over $PSL(2,3)$.
    $endgroup$
    – user641
    Apr 24 '13 at 23:24






  • 1




    $begingroup$
    There is a 2-dimensional complex representation of $Q_8$, but it cannot be written over the rationals, so $Q_8$ does not embed in ${rm GL}_2({mathbb Z})$. (But ${rm SL}_2(3)$, which contains $Q_8$, does embed in ${rm SL}_2(p)$ for all odd primes $p$.)
    $endgroup$
    – Derek Holt
    Apr 25 '13 at 8:59




















  • $begingroup$
    I suspect that the binary tetrahedral group is conjugate to a subgroup of $text{SL}_2(mathbb{Z})$. If so, it's known that if $G$ is a finite subgroup of $text{GL}_n(mathbb{Z})$ then $G$ injects into $text{GL}_n(mathbb{F}_3)$, and that provides a pretty straightforward isomorphism.
    $endgroup$
    – Qiaochu Yuan
    Apr 24 '13 at 17:41






  • 2




    $begingroup$
    The isomorphism $mathfrak A(4) simeq rm{PSL}_2(mathbb F_3)$ is quite easy to see: the action of $rm{PSL}_2(mathbb F_3)$ on the 4-element projective line $rm P^1(mathbb F_3)$ provides an injective morphism $rm{PSL}_2(mathbb F_3) to mathfrak S(4)$ which gives easily the desired isomorphism. What is unclear to me is why the double cover gives precisely $rm{SL}_2(mathbb F_3)$. Maybe there are direct cohomological arguments?
    $endgroup$
    – PseudoNeo
    Apr 24 '13 at 19:35






  • 2




    $begingroup$
    @QiaochuYuan: I doubt it: finite subgroups of $rm{GL}_2(mathbb Z)$ preserve a scalar product, so they are cyclic or dihedral.
    $endgroup$
    – PseudoNeo
    Apr 24 '13 at 19:38






  • 3




    $begingroup$
    Well, this is probably not as geometric as you want, but $SL(2,3)$ is the only nontrivial "double cover" (central extension of order 2) over $PSL(2,3)$.
    $endgroup$
    – user641
    Apr 24 '13 at 23:24






  • 1




    $begingroup$
    There is a 2-dimensional complex representation of $Q_8$, but it cannot be written over the rationals, so $Q_8$ does not embed in ${rm GL}_2({mathbb Z})$. (But ${rm SL}_2(3)$, which contains $Q_8$, does embed in ${rm SL}_2(p)$ for all odd primes $p$.)
    $endgroup$
    – Derek Holt
    Apr 25 '13 at 8:59


















$begingroup$
I suspect that the binary tetrahedral group is conjugate to a subgroup of $text{SL}_2(mathbb{Z})$. If so, it's known that if $G$ is a finite subgroup of $text{GL}_n(mathbb{Z})$ then $G$ injects into $text{GL}_n(mathbb{F}_3)$, and that provides a pretty straightforward isomorphism.
$endgroup$
– Qiaochu Yuan
Apr 24 '13 at 17:41




$begingroup$
I suspect that the binary tetrahedral group is conjugate to a subgroup of $text{SL}_2(mathbb{Z})$. If so, it's known that if $G$ is a finite subgroup of $text{GL}_n(mathbb{Z})$ then $G$ injects into $text{GL}_n(mathbb{F}_3)$, and that provides a pretty straightforward isomorphism.
$endgroup$
– Qiaochu Yuan
Apr 24 '13 at 17:41




2




2




$begingroup$
The isomorphism $mathfrak A(4) simeq rm{PSL}_2(mathbb F_3)$ is quite easy to see: the action of $rm{PSL}_2(mathbb F_3)$ on the 4-element projective line $rm P^1(mathbb F_3)$ provides an injective morphism $rm{PSL}_2(mathbb F_3) to mathfrak S(4)$ which gives easily the desired isomorphism. What is unclear to me is why the double cover gives precisely $rm{SL}_2(mathbb F_3)$. Maybe there are direct cohomological arguments?
$endgroup$
– PseudoNeo
Apr 24 '13 at 19:35




$begingroup$
The isomorphism $mathfrak A(4) simeq rm{PSL}_2(mathbb F_3)$ is quite easy to see: the action of $rm{PSL}_2(mathbb F_3)$ on the 4-element projective line $rm P^1(mathbb F_3)$ provides an injective morphism $rm{PSL}_2(mathbb F_3) to mathfrak S(4)$ which gives easily the desired isomorphism. What is unclear to me is why the double cover gives precisely $rm{SL}_2(mathbb F_3)$. Maybe there are direct cohomological arguments?
$endgroup$
– PseudoNeo
Apr 24 '13 at 19:35




2




2




$begingroup$
@QiaochuYuan: I doubt it: finite subgroups of $rm{GL}_2(mathbb Z)$ preserve a scalar product, so they are cyclic or dihedral.
$endgroup$
– PseudoNeo
Apr 24 '13 at 19:38




$begingroup$
@QiaochuYuan: I doubt it: finite subgroups of $rm{GL}_2(mathbb Z)$ preserve a scalar product, so they are cyclic or dihedral.
$endgroup$
– PseudoNeo
Apr 24 '13 at 19:38




3




3




$begingroup$
Well, this is probably not as geometric as you want, but $SL(2,3)$ is the only nontrivial "double cover" (central extension of order 2) over $PSL(2,3)$.
$endgroup$
– user641
Apr 24 '13 at 23:24




$begingroup$
Well, this is probably not as geometric as you want, but $SL(2,3)$ is the only nontrivial "double cover" (central extension of order 2) over $PSL(2,3)$.
$endgroup$
– user641
Apr 24 '13 at 23:24




1




1




$begingroup$
There is a 2-dimensional complex representation of $Q_8$, but it cannot be written over the rationals, so $Q_8$ does not embed in ${rm GL}_2({mathbb Z})$. (But ${rm SL}_2(3)$, which contains $Q_8$, does embed in ${rm SL}_2(p)$ for all odd primes $p$.)
$endgroup$
– Derek Holt
Apr 25 '13 at 8:59






$begingroup$
There is a 2-dimensional complex representation of $Q_8$, but it cannot be written over the rationals, so $Q_8$ does not embed in ${rm GL}_2({mathbb Z})$. (But ${rm SL}_2(3)$, which contains $Q_8$, does embed in ${rm SL}_2(p)$ for all odd primes $p$.)
$endgroup$
– Derek Holt
Apr 25 '13 at 8:59












3 Answers
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active

oldest

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$begingroup$

Let's start with the ring $H$ of Hurwitz quaternions (quaternions s.t. either all coefficients are integers or all coefficients are half-integers). The ring $Hotimesmathbb F_p$ is a quaternion algebra over a finite field, i.e. a matrix algebra $Mat_{2times 2}(mathbb F_p)$ (with determinant as the norm).



Now the binary tetrahedral group is the group of norm $1$ elements in $H$ and it maps injectively (for $p>2$) to the group of norm $1$ elements in $Hotimesmathbb F_p$, i.e. to $SL_2(mathbb F_p)$. In particular, for $p=3$ we get an isomorphism $mathbb Tto SL_2(mathbb F_3)$.



(Ref.: TWF 198.)






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$endgroup$





















    3












    $begingroup$

    We know that $mathbb T= <x,y,z| x^2=y^3=z^3=xyz>$. We also know that $ord(xyz)=2$, $mathbb{T}/<xyz> simeq A_4$ and $mathbb{T}$ has no subgroup of order 12.



    In one of my problem sheets, we are intended to give a guided proof of the fact that $mathbb T simeq SL_2( mathbb F_3)$ by showing that the map $phi: mathbb T to SL_2(mathbb F_3)$ that sends $$x longrightarrow left( begin{array}{cc} 0 & 2 \ 1 & 0 end{array} right), y longrightarrow left( begin{array}{cc} 2 & 0 \ 1 & 2 end{array} right) text{ and } z longrightarrow left( begin{array}{cc} 0 & 1 \ 2 & 1 end{array} right)$$ extends to an isomorphism between $mathbb T$ and $SL_2( mathbb{F}_3)$. This map is clearly a homomorphism, because it is easy to see that the images of the generators respect the relations.



    We can just count the elements of $SL_2(mathbb F_3)$ to see that $|SL_2(mathbb F_3)|= | mathbb{T}|=24$, so the only thing that we still have to show is either that $phi$ is injective or surjective. Anybody can think of an easy way to prove this? I think this might lead to an easy proof of the result, without even making use of the notion of semidirect products.






    share|cite|improve this answer









    $endgroup$





















      3












      $begingroup$

      Maybe the confusion arrises from the fact that $mathbb{T}$
      has no subgroup of order $12$.
      Rather it has the quaternion group
      $Q={pm1,pm i,pm j, pm k}$
      as a subgroup of order $8$.



      I think a good way to grasp this intuitively is to see it in the context of the well known action of $mathbb{H}^1$
      on the pure quaternions $mathbb{H}_0simeqmathbb{R}^3$
      by rotation via $q(p)=qpoverline{q}$,
      which gives a double cover of $mathrm{SO}(3)$.
      In particular,
      $mathbb{H}^1/{pm1}congmathrm{SO}(3)$.



      We can use this to realize the action of $mathbb{T}$
      on a tetrahedron explicitly, as follows.
      Let $Tsubsetmathbb{H}_0$
      be the tetrahedron with vertices
      $$i+j+k,quad i-j-k,quad -i+j-k,quad -i-j+k.$$
      An element of $Q$
      acts on this via rotation by $pi$
      about a line through the midpoints of opposite sides of $T$.
      An element of $mathbb{T}-Q$
      acts on this by rotation by $2pi/3$
      about a line through a vertex and the midpoint of its opposite face.
      Any pair of elements of $mathbb{T}$
      differing only by sign give the same action.
      Now, by passing from $Q$
      to $mathbb{T}$
      we have tripled the size of the group,
      and included everything necessary to get the full tetrahedral group,
      but with redundancy.
      In particular, the kernel of the action of $mathbb{T}$ on $T$ is ${pm1}$.
      So $mathbb{T}$
      gives a double cover of the tetrahedral group.



      As you mentioned, there's a straightforward hands-on way to see that the tetrahedral group $mathfrak{U}(4)$
      is isomorphic to $mathrm{PSL}(2,mathbb{F}_3)$.
      So if we pull back to the full group,
      we've got $mathbb{T}congmathrm{SL}_2(mathbb{F}_3)$.



      This way of thinking about it is not my own invention. It's from a preprint I have of the book Quaternion Algebras by John Voight, which is expected to go into print very soon and which I highly recommend.






      share|cite|improve this answer











      $endgroup$














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        3 Answers
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        $begingroup$

        Let's start with the ring $H$ of Hurwitz quaternions (quaternions s.t. either all coefficients are integers or all coefficients are half-integers). The ring $Hotimesmathbb F_p$ is a quaternion algebra over a finite field, i.e. a matrix algebra $Mat_{2times 2}(mathbb F_p)$ (with determinant as the norm).



        Now the binary tetrahedral group is the group of norm $1$ elements in $H$ and it maps injectively (for $p>2$) to the group of norm $1$ elements in $Hotimesmathbb F_p$, i.e. to $SL_2(mathbb F_p)$. In particular, for $p=3$ we get an isomorphism $mathbb Tto SL_2(mathbb F_3)$.



        (Ref.: TWF 198.)






        share|cite|improve this answer











        $endgroup$


















          7












          $begingroup$

          Let's start with the ring $H$ of Hurwitz quaternions (quaternions s.t. either all coefficients are integers or all coefficients are half-integers). The ring $Hotimesmathbb F_p$ is a quaternion algebra over a finite field, i.e. a matrix algebra $Mat_{2times 2}(mathbb F_p)$ (with determinant as the norm).



          Now the binary tetrahedral group is the group of norm $1$ elements in $H$ and it maps injectively (for $p>2$) to the group of norm $1$ elements in $Hotimesmathbb F_p$, i.e. to $SL_2(mathbb F_p)$. In particular, for $p=3$ we get an isomorphism $mathbb Tto SL_2(mathbb F_3)$.



          (Ref.: TWF 198.)






          share|cite|improve this answer











          $endgroup$
















            7












            7








            7





            $begingroup$

            Let's start with the ring $H$ of Hurwitz quaternions (quaternions s.t. either all coefficients are integers or all coefficients are half-integers). The ring $Hotimesmathbb F_p$ is a quaternion algebra over a finite field, i.e. a matrix algebra $Mat_{2times 2}(mathbb F_p)$ (with determinant as the norm).



            Now the binary tetrahedral group is the group of norm $1$ elements in $H$ and it maps injectively (for $p>2$) to the group of norm $1$ elements in $Hotimesmathbb F_p$, i.e. to $SL_2(mathbb F_p)$. In particular, for $p=3$ we get an isomorphism $mathbb Tto SL_2(mathbb F_3)$.



            (Ref.: TWF 198.)






            share|cite|improve this answer











            $endgroup$



            Let's start with the ring $H$ of Hurwitz quaternions (quaternions s.t. either all coefficients are integers or all coefficients are half-integers). The ring $Hotimesmathbb F_p$ is a quaternion algebra over a finite field, i.e. a matrix algebra $Mat_{2times 2}(mathbb F_p)$ (with determinant as the norm).



            Now the binary tetrahedral group is the group of norm $1$ elements in $H$ and it maps injectively (for $p>2$) to the group of norm $1$ elements in $Hotimesmathbb F_p$, i.e. to $SL_2(mathbb F_p)$. In particular, for $p=3$ we get an isomorphism $mathbb Tto SL_2(mathbb F_3)$.



            (Ref.: TWF 198.)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 19 '18 at 4:17









            the_fox

            2,90031538




            2,90031538










            answered May 24 '13 at 11:36









            Grigory MGrigory M

            13.6k357104




            13.6k357104























                3












                $begingroup$

                We know that $mathbb T= <x,y,z| x^2=y^3=z^3=xyz>$. We also know that $ord(xyz)=2$, $mathbb{T}/<xyz> simeq A_4$ and $mathbb{T}$ has no subgroup of order 12.



                In one of my problem sheets, we are intended to give a guided proof of the fact that $mathbb T simeq SL_2( mathbb F_3)$ by showing that the map $phi: mathbb T to SL_2(mathbb F_3)$ that sends $$x longrightarrow left( begin{array}{cc} 0 & 2 \ 1 & 0 end{array} right), y longrightarrow left( begin{array}{cc} 2 & 0 \ 1 & 2 end{array} right) text{ and } z longrightarrow left( begin{array}{cc} 0 & 1 \ 2 & 1 end{array} right)$$ extends to an isomorphism between $mathbb T$ and $SL_2( mathbb{F}_3)$. This map is clearly a homomorphism, because it is easy to see that the images of the generators respect the relations.



                We can just count the elements of $SL_2(mathbb F_3)$ to see that $|SL_2(mathbb F_3)|= | mathbb{T}|=24$, so the only thing that we still have to show is either that $phi$ is injective or surjective. Anybody can think of an easy way to prove this? I think this might lead to an easy proof of the result, without even making use of the notion of semidirect products.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  We know that $mathbb T= <x,y,z| x^2=y^3=z^3=xyz>$. We also know that $ord(xyz)=2$, $mathbb{T}/<xyz> simeq A_4$ and $mathbb{T}$ has no subgroup of order 12.



                  In one of my problem sheets, we are intended to give a guided proof of the fact that $mathbb T simeq SL_2( mathbb F_3)$ by showing that the map $phi: mathbb T to SL_2(mathbb F_3)$ that sends $$x longrightarrow left( begin{array}{cc} 0 & 2 \ 1 & 0 end{array} right), y longrightarrow left( begin{array}{cc} 2 & 0 \ 1 & 2 end{array} right) text{ and } z longrightarrow left( begin{array}{cc} 0 & 1 \ 2 & 1 end{array} right)$$ extends to an isomorphism between $mathbb T$ and $SL_2( mathbb{F}_3)$. This map is clearly a homomorphism, because it is easy to see that the images of the generators respect the relations.



                  We can just count the elements of $SL_2(mathbb F_3)$ to see that $|SL_2(mathbb F_3)|= | mathbb{T}|=24$, so the only thing that we still have to show is either that $phi$ is injective or surjective. Anybody can think of an easy way to prove this? I think this might lead to an easy proof of the result, without even making use of the notion of semidirect products.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    We know that $mathbb T= <x,y,z| x^2=y^3=z^3=xyz>$. We also know that $ord(xyz)=2$, $mathbb{T}/<xyz> simeq A_4$ and $mathbb{T}$ has no subgroup of order 12.



                    In one of my problem sheets, we are intended to give a guided proof of the fact that $mathbb T simeq SL_2( mathbb F_3)$ by showing that the map $phi: mathbb T to SL_2(mathbb F_3)$ that sends $$x longrightarrow left( begin{array}{cc} 0 & 2 \ 1 & 0 end{array} right), y longrightarrow left( begin{array}{cc} 2 & 0 \ 1 & 2 end{array} right) text{ and } z longrightarrow left( begin{array}{cc} 0 & 1 \ 2 & 1 end{array} right)$$ extends to an isomorphism between $mathbb T$ and $SL_2( mathbb{F}_3)$. This map is clearly a homomorphism, because it is easy to see that the images of the generators respect the relations.



                    We can just count the elements of $SL_2(mathbb F_3)$ to see that $|SL_2(mathbb F_3)|= | mathbb{T}|=24$, so the only thing that we still have to show is either that $phi$ is injective or surjective. Anybody can think of an easy way to prove this? I think this might lead to an easy proof of the result, without even making use of the notion of semidirect products.






                    share|cite|improve this answer









                    $endgroup$



                    We know that $mathbb T= <x,y,z| x^2=y^3=z^3=xyz>$. We also know that $ord(xyz)=2$, $mathbb{T}/<xyz> simeq A_4$ and $mathbb{T}$ has no subgroup of order 12.



                    In one of my problem sheets, we are intended to give a guided proof of the fact that $mathbb T simeq SL_2( mathbb F_3)$ by showing that the map $phi: mathbb T to SL_2(mathbb F_3)$ that sends $$x longrightarrow left( begin{array}{cc} 0 & 2 \ 1 & 0 end{array} right), y longrightarrow left( begin{array}{cc} 2 & 0 \ 1 & 2 end{array} right) text{ and } z longrightarrow left( begin{array}{cc} 0 & 1 \ 2 & 1 end{array} right)$$ extends to an isomorphism between $mathbb T$ and $SL_2( mathbb{F}_3)$. This map is clearly a homomorphism, because it is easy to see that the images of the generators respect the relations.



                    We can just count the elements of $SL_2(mathbb F_3)$ to see that $|SL_2(mathbb F_3)|= | mathbb{T}|=24$, so the only thing that we still have to show is either that $phi$ is injective or surjective. Anybody can think of an easy way to prove this? I think this might lead to an easy proof of the result, without even making use of the notion of semidirect products.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 1 '14 at 9:40









                    George TurcasGeorge Turcas

                    513




                    513























                        3












                        $begingroup$

                        Maybe the confusion arrises from the fact that $mathbb{T}$
                        has no subgroup of order $12$.
                        Rather it has the quaternion group
                        $Q={pm1,pm i,pm j, pm k}$
                        as a subgroup of order $8$.



                        I think a good way to grasp this intuitively is to see it in the context of the well known action of $mathbb{H}^1$
                        on the pure quaternions $mathbb{H}_0simeqmathbb{R}^3$
                        by rotation via $q(p)=qpoverline{q}$,
                        which gives a double cover of $mathrm{SO}(3)$.
                        In particular,
                        $mathbb{H}^1/{pm1}congmathrm{SO}(3)$.



                        We can use this to realize the action of $mathbb{T}$
                        on a tetrahedron explicitly, as follows.
                        Let $Tsubsetmathbb{H}_0$
                        be the tetrahedron with vertices
                        $$i+j+k,quad i-j-k,quad -i+j-k,quad -i-j+k.$$
                        An element of $Q$
                        acts on this via rotation by $pi$
                        about a line through the midpoints of opposite sides of $T$.
                        An element of $mathbb{T}-Q$
                        acts on this by rotation by $2pi/3$
                        about a line through a vertex and the midpoint of its opposite face.
                        Any pair of elements of $mathbb{T}$
                        differing only by sign give the same action.
                        Now, by passing from $Q$
                        to $mathbb{T}$
                        we have tripled the size of the group,
                        and included everything necessary to get the full tetrahedral group,
                        but with redundancy.
                        In particular, the kernel of the action of $mathbb{T}$ on $T$ is ${pm1}$.
                        So $mathbb{T}$
                        gives a double cover of the tetrahedral group.



                        As you mentioned, there's a straightforward hands-on way to see that the tetrahedral group $mathfrak{U}(4)$
                        is isomorphic to $mathrm{PSL}(2,mathbb{F}_3)$.
                        So if we pull back to the full group,
                        we've got $mathbb{T}congmathrm{SL}_2(mathbb{F}_3)$.



                        This way of thinking about it is not my own invention. It's from a preprint I have of the book Quaternion Algebras by John Voight, which is expected to go into print very soon and which I highly recommend.






                        share|cite|improve this answer











                        $endgroup$


















                          3












                          $begingroup$

                          Maybe the confusion arrises from the fact that $mathbb{T}$
                          has no subgroup of order $12$.
                          Rather it has the quaternion group
                          $Q={pm1,pm i,pm j, pm k}$
                          as a subgroup of order $8$.



                          I think a good way to grasp this intuitively is to see it in the context of the well known action of $mathbb{H}^1$
                          on the pure quaternions $mathbb{H}_0simeqmathbb{R}^3$
                          by rotation via $q(p)=qpoverline{q}$,
                          which gives a double cover of $mathrm{SO}(3)$.
                          In particular,
                          $mathbb{H}^1/{pm1}congmathrm{SO}(3)$.



                          We can use this to realize the action of $mathbb{T}$
                          on a tetrahedron explicitly, as follows.
                          Let $Tsubsetmathbb{H}_0$
                          be the tetrahedron with vertices
                          $$i+j+k,quad i-j-k,quad -i+j-k,quad -i-j+k.$$
                          An element of $Q$
                          acts on this via rotation by $pi$
                          about a line through the midpoints of opposite sides of $T$.
                          An element of $mathbb{T}-Q$
                          acts on this by rotation by $2pi/3$
                          about a line through a vertex and the midpoint of its opposite face.
                          Any pair of elements of $mathbb{T}$
                          differing only by sign give the same action.
                          Now, by passing from $Q$
                          to $mathbb{T}$
                          we have tripled the size of the group,
                          and included everything necessary to get the full tetrahedral group,
                          but with redundancy.
                          In particular, the kernel of the action of $mathbb{T}$ on $T$ is ${pm1}$.
                          So $mathbb{T}$
                          gives a double cover of the tetrahedral group.



                          As you mentioned, there's a straightforward hands-on way to see that the tetrahedral group $mathfrak{U}(4)$
                          is isomorphic to $mathrm{PSL}(2,mathbb{F}_3)$.
                          So if we pull back to the full group,
                          we've got $mathbb{T}congmathrm{SL}_2(mathbb{F}_3)$.



                          This way of thinking about it is not my own invention. It's from a preprint I have of the book Quaternion Algebras by John Voight, which is expected to go into print very soon and which I highly recommend.






                          share|cite|improve this answer











                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            Maybe the confusion arrises from the fact that $mathbb{T}$
                            has no subgroup of order $12$.
                            Rather it has the quaternion group
                            $Q={pm1,pm i,pm j, pm k}$
                            as a subgroup of order $8$.



                            I think a good way to grasp this intuitively is to see it in the context of the well known action of $mathbb{H}^1$
                            on the pure quaternions $mathbb{H}_0simeqmathbb{R}^3$
                            by rotation via $q(p)=qpoverline{q}$,
                            which gives a double cover of $mathrm{SO}(3)$.
                            In particular,
                            $mathbb{H}^1/{pm1}congmathrm{SO}(3)$.



                            We can use this to realize the action of $mathbb{T}$
                            on a tetrahedron explicitly, as follows.
                            Let $Tsubsetmathbb{H}_0$
                            be the tetrahedron with vertices
                            $$i+j+k,quad i-j-k,quad -i+j-k,quad -i-j+k.$$
                            An element of $Q$
                            acts on this via rotation by $pi$
                            about a line through the midpoints of opposite sides of $T$.
                            An element of $mathbb{T}-Q$
                            acts on this by rotation by $2pi/3$
                            about a line through a vertex and the midpoint of its opposite face.
                            Any pair of elements of $mathbb{T}$
                            differing only by sign give the same action.
                            Now, by passing from $Q$
                            to $mathbb{T}$
                            we have tripled the size of the group,
                            and included everything necessary to get the full tetrahedral group,
                            but with redundancy.
                            In particular, the kernel of the action of $mathbb{T}$ on $T$ is ${pm1}$.
                            So $mathbb{T}$
                            gives a double cover of the tetrahedral group.



                            As you mentioned, there's a straightforward hands-on way to see that the tetrahedral group $mathfrak{U}(4)$
                            is isomorphic to $mathrm{PSL}(2,mathbb{F}_3)$.
                            So if we pull back to the full group,
                            we've got $mathbb{T}congmathrm{SL}_2(mathbb{F}_3)$.



                            This way of thinking about it is not my own invention. It's from a preprint I have of the book Quaternion Algebras by John Voight, which is expected to go into print very soon and which I highly recommend.






                            share|cite|improve this answer











                            $endgroup$



                            Maybe the confusion arrises from the fact that $mathbb{T}$
                            has no subgroup of order $12$.
                            Rather it has the quaternion group
                            $Q={pm1,pm i,pm j, pm k}$
                            as a subgroup of order $8$.



                            I think a good way to grasp this intuitively is to see it in the context of the well known action of $mathbb{H}^1$
                            on the pure quaternions $mathbb{H}_0simeqmathbb{R}^3$
                            by rotation via $q(p)=qpoverline{q}$,
                            which gives a double cover of $mathrm{SO}(3)$.
                            In particular,
                            $mathbb{H}^1/{pm1}congmathrm{SO}(3)$.



                            We can use this to realize the action of $mathbb{T}$
                            on a tetrahedron explicitly, as follows.
                            Let $Tsubsetmathbb{H}_0$
                            be the tetrahedron with vertices
                            $$i+j+k,quad i-j-k,quad -i+j-k,quad -i-j+k.$$
                            An element of $Q$
                            acts on this via rotation by $pi$
                            about a line through the midpoints of opposite sides of $T$.
                            An element of $mathbb{T}-Q$
                            acts on this by rotation by $2pi/3$
                            about a line through a vertex and the midpoint of its opposite face.
                            Any pair of elements of $mathbb{T}$
                            differing only by sign give the same action.
                            Now, by passing from $Q$
                            to $mathbb{T}$
                            we have tripled the size of the group,
                            and included everything necessary to get the full tetrahedral group,
                            but with redundancy.
                            In particular, the kernel of the action of $mathbb{T}$ on $T$ is ${pm1}$.
                            So $mathbb{T}$
                            gives a double cover of the tetrahedral group.



                            As you mentioned, there's a straightforward hands-on way to see that the tetrahedral group $mathfrak{U}(4)$
                            is isomorphic to $mathrm{PSL}(2,mathbb{F}_3)$.
                            So if we pull back to the full group,
                            we've got $mathbb{T}congmathrm{SL}_2(mathbb{F}_3)$.



                            This way of thinking about it is not my own invention. It's from a preprint I have of the book Quaternion Algebras by John Voight, which is expected to go into print very soon and which I highly recommend.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 18 '18 at 21:08

























                            answered Feb 22 '17 at 1:10









                            j0equ1nnj0equ1nn

                            1,5571023




                            1,5571023






























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