Dimension of $J(F(GL_2(3)))$, where $J$ stands for Jacobson radical.

Question

I want to find dimension of $J(F(GL_2(3)))$, where $J$ stand for Jacobson radical, for the group algebra $F(GL_2(3))$ of general linear group of two by two matrices over the field $\mathbb{Z}_3,$ and $F$ is any finite field of of characteristic $3.$ There are lots of results about dimension of Jacobson radical of group algebra in Karpilovsky book with title "Jacobson radical of group algebras." But i didn't find any way to find dimension of $J(F(GL_2(3)))$. By using GAP it’s dimension is equal to 20. But I don’t know is it . Please help me. Thanks.

Answer 1

Below, [K] is the book of Karpilovsky, The Jacobson Radical of Group Algebras, Mathematics Studies 135, North-Holland, already cited in the OP. An other helpful reference may be [L], The Block Theory of Finite Group Algebras, Markus Linckelmann, London MS Student Texts 91.

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All coefficient fields are of characteristic $p=3$ below. Let $G$ be $\operatorname{GL}(2,\Bbb F_3)$. Let $F$ be a group of characteristic $3$.

Our target is to show:

The dimension of the Jacobson ideal $J(FG)$ of the group algebra $FG$, seen as an $F$-vector space is $$ \dim_F J(FG)=20\ .$$

We (may and do) assume $F$ is the field with three elements, $F=\Bbb F_3$, or if convenient, the splitting field of the group algebra $FG$, [K], Proposition 3.1.16, page 114.

The group $$ \begin{aligned} G &:= \operatorname{GL}(2,\Bbb F_3) \\ &\cong \operatorname{SL}(2,\Bbb F_3)\rtimes\Bbb F_3^\times \ , \end{aligned} $$ has $(3^2-1)(3^2-3)=48=3\cdot 2^4$ elements. This is rather a lot for me. So we "break" it, to use the smaller normal group in the above semidirect product.

For this let us use the theorem of Villamayor, [K], Proposition 3.1.8 (ii), page 108, [L], Theorem 1.11.10, page 73. The special linear group $S=\operatorname{SL}(2,\Bbb F_3)$ has index $n=2=|\Bbb F^\times|$ (relatively prime w.r.t. $p=3$), so it is normal, so $$ \dim_F J(FG)=\dim_F J(FS)\cdot |G:S|= \dim_F J(FS)\cdot 2\ . $$ (Note that this is a contructive formula, having a base $\{r\}$ of $J(FS)$, choose the "transversal", $1,t$, matrices with ones on the diagonal, resp. antidiagonal, of determinant $\pm 1$, then use $\{r, rt\}$ for $J(FG)$.)

We have thus to show equivalently:

$$\dim_F J(FS) = 10\ .$$

Instead of using computational arguments, i try to give first a human approach. Structurally, over the splitting field $F$ we have $$ FS/J(FS) \cong\prod_{j} \operatorname{End}(V_j)\ , $$ with simple $FS$-modules $V_j$, which can be indexed by the conjugacy classes of $S$ of order relatively prime to $p=3$, [L], Theorem 1.14.6, page 90, Theorem 1.15.3, Brauer. Counting them we get three, so we will meet only $V_1$, $V_2$, $V_3$ of dimensions $n_1,n_2,n_3$ (as a matter of notation), and we expect than a relation of the shape:

$$ \begin{aligned} 24-\dim_F J(FS) &= \dim_F FS/J(FS) \\ &=\dim_F\prod_{j} \operatorname{End}(V_j) \\ &= \sum_{j} \dim_F \operatorname{End}(V_j) \\ &= \sum_{j}n_j^2= \underbrace{n_1^2}_{=1^2}+n_2^2+n_3^2 \ , \\[2mm] &\qquad\text{which finally it will be (shown to be)} \\ 24-10 &\overset{(!)}= 1^2+2^2+3^2\ . \end{aligned} $$ and i decided to fix ideas, let $V_1$ be the trivial representation, dimension is one. At this point, the construction of a simple $2$--dimensional module $V_2$, and of a simple $3$--dimensional module $V_3$ is concluding the proof of the dimension formula. (Although we have not touched any explicit element in $J(FS)$. This is not my way to go, i am always getting the hands maximally dirty, but books are mad to proceed this clea way with dry propositions, not too many, so that the reader can fill in the main part of them.)

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At some point we really need to compute, thus use the group $S$. This is the Binary tetrahedral group, in the table denoted by ${G_{24}}^3=\operatorname{SL}(2,3)=Q_8\rtimes \Bbb Z_3$, and the link goes to 2T=(2,3,3). To construct a representation, we can look at its presentation, $$ S\cong \langle\ s,t\ : \ (st)^2=s^3=t^3\ \rangle\ . $$ Here, we may want to map $s,t$ respectively to the two matrices $$ M(s)=\begin{bmatrix} 2&\\2&2 \end{bmatrix} \ ,\ M(t)= \begin{bmatrix} 2&2\\&2 \end{bmatrix} \ , $$ to get an explicit map. Let us check with [sage]:

sage: Ms = matrix( GF(3), 2, 2, [2,0,2,2] )
sage: Mt = matrix( GF(3), 2, 2, [2,2,0,2] )
sage: MatrixGroup([Ms, Mt]).order()
24
sage: Ms^3 == Mt^3, Ms^3 == (Ms*Mt)^2 
(True, True)
sage: MatrixGroup([Ms, Mt]).structure_description()
'SL(2,3)'

(Getting the hands dirty has its part of a beauty.)

It is clear now that the natural representation of $S=\operatorname{SL}(2,\Bbb F_3)$ acting by left matrix multiplication on the two dimensional space of column vectors $V_2=(\Bbb F_3)^2$ does not contain an invariant subspace, so we have our $V_2$. We need one more simple module, $V_3$, with the hint that it has dimension $3$. It is then enough to show that it does not contain the trivial representation, since it is the third and last one in the list of the simple modules. We may construct it blindly with the computer, searching for images of $s,t$. (And get the hint from the found solutions without invariant subspaces.)

Now using $V_2$, with the canonical basis $e_1, e_2$, we can build the space $V_2\otimes V_2$ of dimension four, basis $e_1\otimes e_1$, $e_1\otimes e_2$, $e_2\otimes e_1$, $e_2\otimes e_2$, then take inside the space invariated by the action of the symmetric group acting on the two factors, generated by $v_{11}= e_1\otimes e_1$, $v_{12}=\frac 12(e_1\otimes e_2+e_2\otimes e_1) =\operatorname{Symm} e_1\otimes e_2 =\operatorname{Symm} e_2\otimes e_1 $, $v_{22}=e_2\otimes e_2$. Then $s$ acts diagonally as $M(s)\otimes M(s)$, and we may compute:

$$ \begin{aligned} s.v_{11} &= M(s)\otimes M(s)\cdot e_1\otimes e_1\\ &= M(s)e_1\otimes M(s)e_1\\ &= (-e_1-e_2)\otimes(-e_1-e_2)\\ &=v_{11}+v_{12}+v_{22}\ , \\ s.v_{12} &= M(s)\otimes M(s)\cdot \operatorname{Symm} e_1\otimes e_2\\ &= \operatorname{Symm} M(s)e_1\otimes M(s)e_2\\ &= \operatorname{Symm} (-e_1-e_2)\otimes(-e_2)\\ &=v_{12}+v_{22}\ , \\[2mm] s.v_{22} &= M(s)\otimes M(s)\cdot e_2\otimes e_2\\ &= M(s)e_2\otimes M(s)e_2\\ &= (-e_2)\otimes(-e_2)\\ &=v_{22}\ , \end{aligned} $$ and there is an analogous computation for the action of $t$, so we get the action of $s,t$ via the matrices (or their transposed versions) $$ P(s)= \begin{bmatrix} 1 & &\\ 2 & 1 &\\ 1 & 1 & 1 \end{bmatrix} \ , P(t)= \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 2\\ 0 & 0 & 1 \end{bmatrix} \ , $$ And indeed, we can check in sage:

sage: Ps = matrix( GF(3), 3, 3, [1,0,0, 2,1,0, 1,1,1] )
sage: Pt = matrix( GF(3), 3, 3, [1,1,1, 0,1,2, 0,0,1] )
sage: Ps^3 == Pt^3, Ps^3 == (Ps*Pt)^2
(True, True)
sage: MatrixGroup([Ps, Pt]).order()
12
sage: MatrixGroup([Ps, Pt]).structure_description()
'A4'

(The $A_4$ is the $T$ in $2T$, the quotient of $S$ w.r.t. the normal subgroup with two elements. Hard to observe this without getting wet in code.)

We are done, the expected dimension is thus $24-1^2-2^2-3^2$.

$\square$

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Suplement:

It is maybe time (out of the competition) to ask a computer algebra system about the radical and related data. I am using sage here. (Sage goes to gap for group computations. The same computations are done. But the language framework makes it easy to program and easily read / understand the programemd lines of code.)

First of all, let us get the dimensions. There are some bugs in sage, for instance:

sage: len( GL(2,3).as_permutation_group().algebra(GF(3)).radical_basis() )
20
sage: len( SL(2,3).as_permutation_group().algebra(GF(3)).radical_basis() )
20
sage: len( GL(2,3).algebra(GF(3)).radical_basis() )
18

This is not the bright way of bringing the right image for sage, since sage is a wonderful completely free world to perform symbolic computations, in most cases implemented with care and full documentation, but i wanted only to mention why i am "dancing around" the natural initialization of instances to avoid the today bugs in the sequel. (It seems, there is a true need of debug+devel in this corner...)

We will isolate the incarantion of $G=\operatorname{GL}(2,3)$ as a permutation group, call this G, we will isolate the subgroup S of index $2$, order $24$ in it, then reinitialize it using its generators (to loose the connection to G, and to have all methods of a permutation group). We will compute its radical basis and show it. Above, we used the information on the conjugacy classes, this is also printed. So far:

sage: G = GL(2,3).as_permutation_group()
sage: S = [N for N in G.normal_subgroups() if N.order() == 24][0]

sage: S = PermutationGroup( S.gens() )
sage: S.structure_description()
'SL(2,3)'

sage: for s in S.conjugacy_classes_representatives():
....:     print s
....:     
()
(4,5,6)(7,9,8)
(4,6,5)(7,8,9)
(2,3)(4,7)(5,9)(6,8)
(2,3)(4,8,5,7,6,9)
(2,3)(4,9,6,7,5,8)
(2,4,3,7)(5,6,9,8)

sage: FS = GroupAlgebra(S, GF(3))
sage: for r in FS.radical_basis():
....:     print r

The above print would not fit well here, so i try instead:

count = 0
for r in FS.radical_basis():
    count += 1
    r_str = str(r)
    r_str = re.sub( ',', '', r_str )
    r_str = re.sub( '\*', '', r_str )
    r_str = re.sub( '\+ 2', '- ', r_str )
    r_str = re.sub( '2\(', '-(', r_str )
    print r"r_{%s} &= %s\\[2mm]" % (count, r_str)

which gives latex code to insert below: $$ \begin{aligned} r_{1} &= () - (465)(789) + (249)(375) + (258)(396) + (267)(384) + (2734)(5896) + (279345)(68) + (287364)(59) + (2836)(4975) + (2935)(4678) + (298356)(47)\\[2mm] r_{2} &= (23)(47)(59)(68) - (23)(485769) - (249)(375) - (258)(396) - (267)(384) + (2734)(5896) + (2836)(4975) + (2935)(4678)\\[2mm] r_{3} &= (465)(789) - (23)(485769) + (249)(375) - (258)(396) - (267)(384) + (2638)(4579) - (279345)(68) + (287364)(59) - (2836)(4975) + (298356)(47)\\[2mm] r_{4} &= (465)(789) - (23)(485769) - (249)(375) - (258)(396) + (2539)(4876) + (267)(384) + (279345)(68) - (287364)(59) - (2935)(4678) + (298356)(47)\\[2mm] r_{5} &= (465)(789) - (23)(485769) + (2437)(5698) - (249)(375) + (258)(396) - (267)(384) - (2734)(5896) + (279345)(68) + (287364)(59) - (298356)(47)\\[2mm] r_{6} &= (456)(798) - (465)(789) - (249)(375) - (258)(396) - (267)(384) + (276)(348) + (285)(369) + (294)(357)\\[2mm] r_{7} &= -(23)(485769) + (23)(496758) + (249)(375) + (258)(396) + (267)(384) + (279345)(68) + (276)(348) + (287364)(59) + (285)(369) + (294)(357) + (298356)(47)\\[2mm] r_{8} &= -(465)(789) + (23)(485769) - (249)(375) + (258)(396) - (267)(384) + (265389)(47) + (279345)(68) + (287364)(59) - (285)(369) - (298356)(47)\\[2mm] r_{9} &= -(465)(789) + (23)(485769) + (249)(375) + (254397)(68) - (258)(396) - (267)(384) - (279345)(68) + (287364)(59) - (294)(357) + (298356)(47)\\[2mm] r_{10} &= -(465)(789) + (23)(485769) + (246378)(59) - (249)(375) - (258)(396) + (267)(384) + (279345)(68) - (276)(348) - (287364)(59) + (298356)(47) \end{aligned} $$

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As a final experiment, we can also try to find the central idempotents of $S$. As expected, the center has the dimension of the number of conjugacy classes, which are eight. This works in sage for both $G$ ans $S$.

sage: len(G.conjugacy_classes()), len(S.conjugacy_classes())
(8, 8)

For each conjugacy class $C$ in $S$ the element $C^+:=\sum_{c\in C}[c]$ is central. Searching for the central idempotents, also noting the theorem of Osima, [K], Theorem 2.2.4, page 76, we are searching for central idempotents in the space of linear combinations of the $C^+$'s among the three following ones:

Cplus3prime_List = [ sum( [FS(c) for c in C] ) for C in S.conjugacy_classes() if not 3.divides(C[0].order()) ]

We obtain:

[(),
 (2,3)(4,7)(5,9)(6,8),
 (2,4,3,7)(5,6,9,8) + (2,5,3,9)(4,8,7,6) + (2,6,3,8)(4,5,7,9)
 + (2,7,3,4)(5,8,9,6) + (2,8,3,6)(4,9,7,5) + (2,9,3,5)(4,6,7,8)]

(The result was manually slightly rearranged.)

Let us denote them by $1$, $a_{2222}$, $a_{44}$.

There are only $3^3$ linear combinations of them, we expect $3$ blocks, so $3$ primitive idempotents, so $2^3$ idempotents. The search is easy:

sage: for x in cartesian_product( [GF(3), GF(3), GF(3)] ):
....:     e = sum( [x[k]*Cplus3prime_List[k] for k in range(3)] )
....:     if e == e^2:
....:         print x
....:         
(0, 0, 0)
(0, 0, 1)
(1, 0, 0)
(1, 0, 2)
(2, 1, 0)
(2, 1, 1)
(2, 2, 0)
(2, 2, 2)

and we pick manually a primitive triple of projectors among them. Maybe (0, 0, 1), corresponding to $a_{44}$, (2, 1, 0), corresponding to $a_{2222}-1=\frac 12(1-a_{2222})$, and "the rest", $1-a_{44}-(a_{2222}-1)=-1-a_{2222}-a_{44}$. (Among them, $\frac 12(1\pm a_{2222})$ are "easily seen" central projectors, using them we can still get some insight.)

We build these projectors and multiply with them the radical basis, just one more experiment.

sage: e1 = Cplus3prime_List[2]
sage: e2 = - Cplus3prime_List[0] + Cplus3prime_List[1]
sage: e3 = - Cplus3prime_List[0] - Cplus3prime_List[1] - Cplus3prime_List[2]

sage: e1*e1 == e1, e1*e2 == 0, e1*e3 == 0
(True, True, True)
sage: e2*e1 == 0, e2*e2 == e2, e2*e3 == 0
(True, True, True)
sage: e3*e1 == 0, e3*e2 == 0, e3*e3 == e3
(True, True, True)

sage: e1 + e2 + e3
()

sage: [ 'same' if e1*r == r else 0 if e1*r == 0 else '?' for r in FS.radical_basis() ] 
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

sage: [ 'same' if e2*r == r else 0 if e2*r == 0 else '?' for r in FS.radical_basis() ] 
['?', '?', 'same', 'same', 'same', '?', '?', 'same', 'same', 'same']

sage: [ 'same' if e3*r == r else 0 if e3*r == 0 else '?' for r in FS.radical_basis() ] 
['?', '?', 0, 0, 0, '?', '?', 0, 0, 0]

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I have to stop here maybe and submit.