Theorem If $n \ge 2$ then $PSL_n(q)$ is simple provided $(n,q)$ is not $(2,2)$ or $(2,3)$.
proof: Consider $PSL_n(q)$ acting on $\mathbb P^{n-1}(q)$ for $n \ge 2$ and the exceptions do not occur, then it is primitive since it's 2-transitive and by previous results $SL_n(q)$ is perfect, so since $PSL_n(q)$ is a quotient of that it's perfect too. Take $d \in V^\#$ so that $[d] \in \mathbb P^{n-1}(q)$ let $A$ be the image of $\mathscr T(d)$ in $PSL_n(q)$ applying [???] and taking quotients we see that $A$ is a normal abelian subgroup of the stabilizer $PSL_n(q)_{[d]}$ and it's conjugates generate $PSL_n(q)$ thus the conditions of Iwasawa's lemma are satisfied.
Thursday, 28 February 2013
Wednesday, 27 February 2013
M11
Let $\Delta = \{\delta_1, \delta_2, \delta_3, \delta_4, \delta_5\}$ and let $\Omega$ be the unordered pairs of these. We will use the names:
The stabilizer of 0 has size $|G_0| = |G|/|\Omega| = 60/10 = 6$ (orbit stabilizer) so $G_0 = \langle a, b \rangle = \{1,a,b,ab,ba,aba=bab\}$ and the $G_0$ orbits are:
so take $\infty \not\in \Omega$ and set $\Omega^+ = \Omega \cup \{\infty\}$ then choose $$x = (\infty\;0)(2\;3)(4\;6)(8\;9)$$ then
Theorem $L$ is simple.
$1$ and $x$ are $(G,G)$-double coset reps in $L$. Take $\omega \not \in \Omega^+$ and form $\Omega^\star = \Omega^+ \cup \{\omega\}$. $$y=(\omega\;\infty)(1\;4)(2\;5)(3\;6)$$ then
Theorem $M_{11}$ is simple.
From Jordan's theorem we see that we cannot perform another one point extension.
- 0: $\{\delta_1,\delta_2\}$, 1: $\{\delta_1,\delta_3\}$, 2: $\{\delta_1,\delta_4\}$
- 3: $\{\delta_1,\delta_5\}$, 4: $\{\delta_2,\delta_3\}$, 5: $\{\delta_2,\delta_4\}$
- 6: $\{\delta_2,\delta_5\}$, 7: $\{\delta_3,\delta_4\}$, 8: $\{\delta_3,\delta_5\}$
- 9: $\{\delta_4,\delta_5\}$
- $a = (\delta_1\;\delta_2)(\delta_3\;\delta_5) = (1\;6)(2\;5)(3\;4)(7\;9)$
- $b = (\delta_1\;\delta_2)(\delta_4\;\delta_5) = (1\;4)(2\;6)(3\;5)(7\;8)$
- $g_2 = (\delta_2\;\delta_3)(\delta_4\;\delta_5) = (0\;1)(2\;3)(5\;8)(6\;7)$
- $g_3 = (\delta_1\;\delta_4)(\delta_2\;\delta_3) = (0\;7)(1\;5)(3\;9)(6\;8)$
The stabilizer of 0 has size $|G_0| = |G|/|\Omega| = 60/10 = 6$ (orbit stabilizer) so $G_0 = \langle a, b \rangle = \{1,a,b,ab,ba,aba=bab\}$ and the $G_0$ orbits are:
- $\{0\}$
- $\{1,2,3,4,5,6\}$
- $\{7,8,9\}$
gap> a:=(1,2)(3,5);;b:=(1,2)(4,5);;
gap> DoubleCosets(Group((1,2,3,4,5),(1,2,3)),Group(a,b),Group(a,b));
[ DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(),Group(
[ (1,2)(3,5), (1,2)(4,5) ] )),
DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(2,3)(4,5),Group(
[ (1,2)(3,5), (1,2)(4,5) ] )),
DoubleCoset(Group( [ (1,2)(3,5), (1,2)(4,5) ] ),(1,3)(2,4),Group(
[ (1,2)(3,5), (1,2)(4,5) ] )) ]
so take $\infty \not\in \Omega$ and set $\Omega^+ = \Omega \cup \{\infty\}$ then choose $$x = (\infty\;0)(2\;3)(4\;6)(8\;9)$$ then
- $x^2 = 1 \in G_0$
- $a^x = b$ and (since $x$ has order 2) $b^x = a$ so $G_0^x = G_0$.
- $g_2^x = (\infty\;1)(2\;3)(5\;9)(4\;7)=x^{g_2} \in G x G$.
- $g_3^x = (\infty\;7)(1\;5)(2\;8)(4\;9)=x^{g_3} a^b \in G x G$.
Theorem $L$ is simple.
$1$ and $x$ are $(G,G)$-double coset reps in $L$. Take $\omega \not \in \Omega^+$ and form $\Omega^\star = \Omega^+ \cup \{\omega\}$. $$y=(\omega\;\infty)(1\;4)(2\;5)(3\;6)$$ then
- $y^2 = 1 \in G = L_{\infty}$
- $G^y = G$ since $G = \langle a,b,g_2,g_3 \rangle$ and $a^y = a$, $b^y = b$, $g_2^y = g_2^b$, $g_3^y = g_3 a^b$.
- $x^y = y^x a^b \in L y L$
Theorem $M_{11}$ is simple.
From Jordan's theorem we see that we cannot perform another one point extension.
A_n is simple!
Lemma $A_5$ is perfect.
proof: $A_5$ is generated by $(1\;2\;3)$ and $(1\;2\;3\;4\;5)$ both are commutators:
Theorem $A_5$ is simple.
proof: The most basic proof using cycles directly can be found in Goodman.
proof: The conjugacy classes have sizes: 1, 15, 20, 12, 12. No sum of these that includes 1 is a divisor of 60 so there are no normal subgroups (which would necessarily be a union of conjugacy classes).
proof: A perfect group is not solvable, and every smaller group whose order divides $|A_5| = 60$ is solvable so $A_5$ has no normal subgroups (else it would be solvable too!)
Theorem $A_n$ is simple.
proof: Induction on $n$ with base case $5$. $A_n$ is $n-2$ transitive in the natural action (by the multiple-transitivity section) so for $n > 5$ this action is (at least) 2-transitive so primitive (by primitivity section) which by the powerful corollary about transitivity with regular normal subgroups tells us that a regular normal subgroup would have to be $C_2^2$ in the case of $A_6$ and there isn't one otherwise but $C_2^2$ doesn't have enough elements to be transitive on 6 points so it can't be regular - so $A_n$ has no regular normal subgroups: therefore by the proposition in that section it's simple.
proof: $A_5$ is generated by $(1\;2\;3)$ and $(1\;2\;3\;4\;5)$ both are commutators:
gap> a := (1,5,2);; b := (4,2,3);; a^(-1)*b^(-1)*a*b; (1,2,3) gap> a := (1,2,3)*(3,4,5);; b := (1,4,2)*(3,5,2);; a^(-1)*b^(-1)*a*b; (1,2,3,4,5)
Theorem $A_5$ is simple.
proof: The most basic proof using cycles directly can be found in Goodman.
proof: The conjugacy classes have sizes: 1, 15, 20, 12, 12. No sum of these that includes 1 is a divisor of 60 so there are no normal subgroups (which would necessarily be a union of conjugacy classes).
proof: A perfect group is not solvable, and every smaller group whose order divides $|A_5| = 60$ is solvable so $A_5$ has no normal subgroups (else it would be solvable too!)
gap> List(AllSmallGroups(2), StructureDescription); [ "C2" ] gap> List(AllSmallGroups(2^2), StructureDescription); [ "C4", "C2 x C2" ] gap> List(AllSmallGroups(2*3), StructureDescription); [ "S3", "C6" ] gap> List(AllSmallGroups(2*5), StructureDescription); [ "D10", "C10" ] gap> List(AllSmallGroups(2*3*5), StructureDescription); [ "C5 x S3", "C3 x D10", "D30", "C30" ] gap> List(AllSmallGroups(2^2*3), StructureDescription); [ "C3 : C4", "C12", "A4", "D12", "C6 x C2" ] gap> List(AllSmallGroups(2^2*5), StructureDescription); [ "C5 : C4", "C20", "C5 : C4", "D20", "C10 x C2" ]
Theorem $A_n$ is simple.
proof: Induction on $n$ with base case $5$. $A_n$ is $n-2$ transitive in the natural action (by the multiple-transitivity section) so for $n > 5$ this action is (at least) 2-transitive so primitive (by primitivity section) which by the powerful corollary about transitivity with regular normal subgroups tells us that a regular normal subgroup would have to be $C_2^2$ in the case of $A_6$ and there isn't one otherwise but $C_2^2$ doesn't have enough elements to be transitive on 6 points so it can't be regular - so $A_n$ has no regular normal subgroups: therefore by the proposition in that section it's simple.
Tuesday, 26 February 2013
Iwasawa's lemma
Lemma (Iwasawa) If $(G,\Omega)$ is a primitive permutation group with $G$ perfect and for some $\alpha \in \Omega$, $G_\alpha$ has a normal abelian subgroup $A$ whose conjugates generate $G$, then $G$ is simple.
proof: Suppose $1 \not = N \unlhd G$, we will gradually show that $N$ must be the whole group. By primitivity $N$ is transitive on $\Omega$ and $G_\alpha$ is a maximal subgroup, so $N \not \le G_\alpha$ and $N G_\alpha = G$. Any $g$ may be written $n x$ for some $n \in N$, $x \in G_\alpha$ so $A^g = A^{nx} = A^x \le AN$ and these conjugates cover the whole group so $AN = G$. Now $G/N \simeq A/(A \cap N)$ is abelian, but $G$ is perfect so $N = G$.
proof: Suppose $1 \not = N \unlhd G$, we will gradually show that $N$ must be the whole group. By primitivity $N$ is transitive on $\Omega$ and $G_\alpha$ is a maximal subgroup, so $N \not \le G_\alpha$ and $N G_\alpha = G$. Any $g$ may be written $n x$ for some $n \in N$, $x \in G_\alpha$ so $A^g = A^{nx} = A^x \le AN$ and these conjugates cover the whole group so $AN = G$. Now $G/N \simeq A/(A \cap N)$ is abelian, but $G$ is perfect so $N = G$.
Projective Spaces and Groups
We aim to "fix" the following two issues: GL and SL are not 2-transitive because they can't linearly dependent vectors to linearly independent ones. SL is not simple (even though it's perfect) because it has a center. Let $V=V_n(q)$ and $n \ge 2$ throughout.
We define an equivalence relation $R$ on $V^\#$ by $vRw$ iff $v = \lambda w$ for some nonzero $\lambda \in \mathbb F_q$.
Definition We then have the projective space $\mathbb P(V)$ of projective vectors. Write $\mathbb P^{n-1}(q)$.
For a subspace $U \subseteq V$ the set of equivalence classes (projective vectors) $[U]$ (the image of $U^\#$) is a subspace of $\mathbb P(V)$ so it inherits geometric structure. The dimension of $[U]$ is the dimension of $U$ minus 1. A point is a class $[v]$ for some vector $v$, and a line is the projective class of a 2D subspace.
Given $g \in GL(V)$, $v \in V^\#$ and $\lambda \in \mathbb F_q^\#$ we have $(\lambda v)g = \lambda (vg) \in [vg]$ so we can (well) define an action by $[v]g = [vg]$. In this way $GL(V)$ and $SL(V)$ act on $\mathbb P(V)$, but not faithfully.
Lemma Let $G$ be $GL(V)$ or $SL(V)$, the kernel of the action of $G$ on $\mathbb P(V)$ is $Z(G)$.
proof: From the previous post we have seen what $Z(G)$ is: scalar multiples of the identity. If $gs = 1$ then clearly $g$ acts trivially on $\mathbb P(V)$. Conversely if $g \in GL(V)$ is in the kernel of the action then $[vg]=[v]$ for all $[v] \in \mathbb P(V)$, then for every vector $V$ we have $v g = \lambda_v v$ for some $\lambda_v \in \mathbb F^\#$ and the proof concludes in the same way as before.
Definition The projective general linear group and projective special linear group are defined by $PGL(V) = GL(V)/Z(GL(V))$ and $PSL(V) = SL(V)/Z(SL(V))$. They have faithful action on $\mathbb P(V)$. (Note: PSL might not be a subgroup of PGL anymore).
We call $[g] \in PGL$ or $PSL$ the image (???) if $g \in GL$ or $SL$ if $[v][g] = [vg]$ for $v\in \mathbb P(V)$. $PGL_n(q)$ and $PSL_n(q)$ act on $\mathbb P^{n-1}(q)$ by the order calculations in the previous post we find $|PGL_n(q)| = q^{n(n-1)/2}\prod_{i=2}^n(q^i-1)$ and $|PSL_n(q)| = \frac{q^{n(n-1)/2}}{(n,q-1)}\prod_{i=2}^n(q^i-1)$. Thus $|PGL_2(q)| = (q+1)q(q-1)$.
Proposition The permutation group $PGL_2(q)$ is sharply 3-transitive on $\mathbb P^1(q)$.
proof: For $g = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in GL_2(q)$ and $v \in V_2(q)^\#$, $vg = (a \lambda_1 + c \lambda_2, b \lambda_1 + d \lambda_2)$. These work out exactly as the mobius transformations when regarding $\mathbb P^1(q)$ as $\mathbb F_q \cup \{\infty\}$. We have already shown this one point extension is genrated by the mobius transforms!
Proposition Both $PGL(V)$ and $PSL(V)$ act 2-transitively on $\mathbb P(V)$.
proof: Given $([e_1],[e_2])$, $([e'_1],[e'_2])$ distinct then extend to a basis and find a map between them (why doesn't this work for all n-transitivity?)
We define an equivalence relation $R$ on $V^\#$ by $vRw$ iff $v = \lambda w$ for some nonzero $\lambda \in \mathbb F_q$.
Definition We then have the projective space $\mathbb P(V)$ of projective vectors. Write $\mathbb P^{n-1}(q)$.
For a subspace $U \subseteq V$ the set of equivalence classes (projective vectors) $[U]$ (the image of $U^\#$) is a subspace of $\mathbb P(V)$ so it inherits geometric structure. The dimension of $[U]$ is the dimension of $U$ minus 1. A point is a class $[v]$ for some vector $v$, and a line is the projective class of a 2D subspace.
Given $g \in GL(V)$, $v \in V^\#$ and $\lambda \in \mathbb F_q^\#$ we have $(\lambda v)g = \lambda (vg) \in [vg]$ so we can (well) define an action by $[v]g = [vg]$. In this way $GL(V)$ and $SL(V)$ act on $\mathbb P(V)$, but not faithfully.
Lemma Let $G$ be $GL(V)$ or $SL(V)$, the kernel of the action of $G$ on $\mathbb P(V)$ is $Z(G)$.
proof: From the previous post we have seen what $Z(G)$ is: scalar multiples of the identity. If $gs = 1$ then clearly $g$ acts trivially on $\mathbb P(V)$. Conversely if $g \in GL(V)$ is in the kernel of the action then $[vg]=[v]$ for all $[v] \in \mathbb P(V)$, then for every vector $V$ we have $v g = \lambda_v v$ for some $\lambda_v \in \mathbb F^\#$ and the proof concludes in the same way as before.
Definition The projective general linear group and projective special linear group are defined by $PGL(V) = GL(V)/Z(GL(V))$ and $PSL(V) = SL(V)/Z(SL(V))$. They have faithful action on $\mathbb P(V)$. (Note: PSL might not be a subgroup of PGL anymore).
We call $[g] \in PGL$ or $PSL$ the image (???) if $g \in GL$ or $SL$ if $[v][g] = [vg]$ for $v\in \mathbb P(V)$. $PGL_n(q)$ and $PSL_n(q)$ act on $\mathbb P^{n-1}(q)$ by the order calculations in the previous post we find $|PGL_n(q)| = q^{n(n-1)/2}\prod_{i=2}^n(q^i-1)$ and $|PSL_n(q)| = \frac{q^{n(n-1)/2}}{(n,q-1)}\prod_{i=2}^n(q^i-1)$. Thus $|PGL_2(q)| = (q+1)q(q-1)$.
Proposition The permutation group $PGL_2(q)$ is sharply 3-transitive on $\mathbb P^1(q)$.
proof: For $g = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \in GL_2(q)$ and $v \in V_2(q)^\#$, $vg = (a \lambda_1 + c \lambda_2, b \lambda_1 + d \lambda_2)$. These work out exactly as the mobius transformations when regarding $\mathbb P^1(q)$ as $\mathbb F_q \cup \{\infty\}$. We have already shown this one point extension is genrated by the mobius transforms!
Proposition Both $PGL(V)$ and $PSL(V)$ act 2-transitively on $\mathbb P(V)$.
proof: Given $([e_1],[e_2])$, $([e'_1],[e'_2])$ distinct then extend to a basis and find a map between them (why doesn't this work for all n-transitivity?)
Saturday, 23 February 2013
Transvections
Assume $n \ge 2$ throughout,
Definition A linear functional on $V$ is a linear map from $V$ to $\mathbb F_q$. The set of these is the dual space $V^*$. Given $f \in V^*$ write $V_f$ for the kernel of $f$, if $f \not = 0$ then $V_f$ is a subspace of dimension $n-1$ (any such space of codimension 1 we call a hyperplane).
Lemma If you have $f,f' \in V$ with the same hyperplane then there exists $\lambda$ such that $f' = \lambda f$.
proof: The result is clear if $V_f=V$ assume not so $f,f'\not = 0$. Take $v \in V \setminus V_f$ (i.e. not in the common hyperplane) so $vf, vf' \not = 0$ and put $\lambda = (vf')(vf)^{-1}$ then $f' - \lambda f \in V^*$ and its kernel is contained in $\langle V_f, v \rangle = V$ so it equals zero.
Definition A linear automorphism $\tau \in GL(V)$ is called a transvection with direction $d \in V^\#$ if $\tau$ fixes $d$ and $$v \tau - v$$ is a scalar multiple of $v$ for all $v$.
Clearly $1$ is a transvection of any direction.
Lemma If $\tau$ is a transvection with direction $d$ then the vectorspace $\operatorname{fix}(\tau)$ is a hyperplane containing $d$.
proof: Define $f : V \to \mathbb F_q$ by $(vf)d = v \tau - v$ i.e. $f$ is that scalar multiple which a transvection defines. Since $\tau$ is linear $f \in V^*$. The result is clear for $\tau = 1$ and if not $f \not = 0$ so $d \in \operatorname{fix}(\tau) = V_f$.
Corollary Any transvection can be written as $\tau_{f,d}$ mapping $v$ to $v + (vf)d$ for some $f \in V^*$, $d \in (V_f)^\#$.
Lemma For $f,f' \in V^*$, $d \in V^\#$, $g \in GL(V)$ we have
Definition For a direction $d \in V^\#$ set $\mathscr T(d) = \{\tau_{f,d}|f \in V^*, d \in V_f\}$ and $\mathscr T$ the union over all directions (all transvections).
Proposition $\mathscr T^\#$ is a single conjugacy class in $GL(V)$ and lies in $SL(V)$. If $n\ge 3$ then $\mathscr T^\#$ is a single conjugacy class in $SL(V)$.
proof: By the calculation lemma previous, we know that $\mathscr T^\#$ is closed under conjugation. Let $\tau_{f,d},\tau_{f',d'} \in \mathscr T^\#$ and write $e_1=d,e'_1=d'$ then take bases $e_1,\ldots,e_{n-1}$ and $e'_1,\ldots,e'_{n-1}$ of the hyperplanes $V_f,V_{f'}$, choose $e_n,e'_n$ such that $e_n f = e_n' f' = 1$. Now we have bases for $V$ so there is a GL map $g$ from one to the other.
For $i < n$ we have $e'_i(g^{-1} \circ f) = e_i f = 0$ so $V_{g^{-1} \circ f}$ is the space spanned by $\langle e'_1,\ldots,e'_{n-1} \rangle = V_{f'}$ so they are scalar multiples of each other, let $\lambda$ be such that $f' = \lambda (g^{-1} \circ f)$ and $$1 = e'_n f' = \lambda e'_n (g^{-1} \circ f) = \lambda e_n f = \lambda$$ so since $dg = d'$ it follows that $\tau_{f,d}^g = \tau_{f',d}$!
If they are all conjugate they all have the same determinant $\delta$, now $\det(\tau_{f,d}\tau_{f',d}) = \det(\tau_{f+f',d})$ so $\delta^2 = \delta$ proves they lie in $SL$.
For $n \ge 3$ we can use the $\mu$ trick as before to get them in $SL$.
Proposition If $d \in V^\#$ then $\mathscr T(d)$ is an abelian normal subgroup of the stabilizer $SL(V)_d$; $\mathscr T(d)$ are all conjugates in $SL(V)$.
proof: Certainly elements of $\mathscr T(d)$ stabilize $d$, by the computational lemma before we see that it is an abelian group (commutative and closed, therefore has identity and inverses). If $g \in SL(V)$ with $dg=d$ then $\mathscr T(d)^g = \mathscr T(d)$ (by the previous) so $\mathscr T(d) \unlhd SL(V)_d$. Given $d,d' \in V^\#$ by the transitivity lemma (that requires dimension > 1) exists $g \in SL(V)$ which takes $d$ to $d'$ then $\mathscr T(d)^g = \mathscr T(d')$.
Proposition The set $\mathscr T$ generates $SL(V)$.
proof: Elementary matrices.
Definition A group if perfect if $G' = G$. This is equivalent to there being no nontrivial abelian quotients: Clearly if $G/[G,G] \not = 1$ then $G\not = [G,G]$. Conversely if $G/N \simeq A = \{Ng\}$ then we always have $Ngg' = Ng'g$ i.e. there is some $n$ such that $gg' = ng'g$. So every commutator $[g,g']$ is an element of $N$.
Proposition If $n \ge 2$ the group $SL_n(q)$ is perfect provided $(n,q)$ isn't $(2,2)$ or $(2,3)$.
proof: We just need to show that each $\tau \in \mathscr T^\#$ is a commutator since that set generates our group. If $\tau$ has direction $d$ take some $\sigma \in \mathscr T(d)^\#$ not equal to $\tau^{-1}$, then $\sigma \tau \in \mathscr T(d)^\#$ so there is a $g \in SL_n(q)$ that conjugates $\sigma \tau = \sigma^g$, whence $\tau = \sigma^{-1} g^{-1} \sigma g= [\sigma,g]$.
For $n=2$ we will use $2 \times 2$ matrices directly, in some basis $\tau = \begin{pmatrix} 1 & \gamma \\ 0 & 1 \end{pmatrix}$ (nonzero $\gamma$), now for any nonzero $\lambda$ and $\mu \in \mathbb F_q$ we have $$\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}\begin{pmatrix} 1 & \mu \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \lambda^{-1} & 0 \\ 0 & \lambda \end{pmatrix}\begin{pmatrix} 1 & -\mu \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & \mu(\lambda^2-1) \\ 0 & 1 \end{pmatrix}$$ so for $q>3$ take $\lambda \not = 0,1,-1$ then $\lambda^2-1\not = 0$ so let $\mu = \gamma(\lambda^2-1)^{-1}$.
Proposition
Definition A linear functional on $V$ is a linear map from $V$ to $\mathbb F_q$. The set of these is the dual space $V^*$. Given $f \in V^*$ write $V_f$ for the kernel of $f$, if $f \not = 0$ then $V_f$ is a subspace of dimension $n-1$ (any such space of codimension 1 we call a hyperplane).
Lemma If you have $f,f' \in V$ with the same hyperplane then there exists $\lambda$ such that $f' = \lambda f$.
proof: The result is clear if $V_f=V$ assume not so $f,f'\not = 0$. Take $v \in V \setminus V_f$ (i.e. not in the common hyperplane) so $vf, vf' \not = 0$ and put $\lambda = (vf')(vf)^{-1}$ then $f' - \lambda f \in V^*$ and its kernel is contained in $\langle V_f, v \rangle = V$ so it equals zero.
Definition A linear automorphism $\tau \in GL(V)$ is called a transvection with direction $d \in V^\#$ if $\tau$ fixes $d$ and $$v \tau - v$$ is a scalar multiple of $v$ for all $v$.
Clearly $1$ is a transvection of any direction.
Lemma If $\tau$ is a transvection with direction $d$ then the vectorspace $\operatorname{fix}(\tau)$ is a hyperplane containing $d$.
proof: Define $f : V \to \mathbb F_q$ by $(vf)d = v \tau - v$ i.e. $f$ is that scalar multiple which a transvection defines. Since $\tau$ is linear $f \in V^*$. The result is clear for $\tau = 1$ and if not $f \not = 0$ so $d \in \operatorname{fix}(\tau) = V_f$.
Corollary Any transvection can be written as $\tau_{f,d}$ mapping $v$ to $v + (vf)d$ for some $f \in V^*$, $d \in (V_f)^\#$.
Lemma For $f,f' \in V^*$, $d \in V^\#$, $g \in GL(V)$ we have
- $\tau_{f,g}^g = \tau_{g^{-1}\circ f,dg}$
- $\tau_{f,d} \tau_{f',d} = \tau_{f+f',d}$
Definition For a direction $d \in V^\#$ set $\mathscr T(d) = \{\tau_{f,d}|f \in V^*, d \in V_f\}$ and $\mathscr T$ the union over all directions (all transvections).
Proposition $\mathscr T^\#$ is a single conjugacy class in $GL(V)$ and lies in $SL(V)$. If $n\ge 3$ then $\mathscr T^\#$ is a single conjugacy class in $SL(V)$.
proof: By the calculation lemma previous, we know that $\mathscr T^\#$ is closed under conjugation. Let $\tau_{f,d},\tau_{f',d'} \in \mathscr T^\#$ and write $e_1=d,e'_1=d'$ then take bases $e_1,\ldots,e_{n-1}$ and $e'_1,\ldots,e'_{n-1}$ of the hyperplanes $V_f,V_{f'}$, choose $e_n,e'_n$ such that $e_n f = e_n' f' = 1$. Now we have bases for $V$ so there is a GL map $g$ from one to the other.
For $i < n$ we have $e'_i(g^{-1} \circ f) = e_i f = 0$ so $V_{g^{-1} \circ f}$ is the space spanned by $\langle e'_1,\ldots,e'_{n-1} \rangle = V_{f'}$ so they are scalar multiples of each other, let $\lambda$ be such that $f' = \lambda (g^{-1} \circ f)$ and $$1 = e'_n f' = \lambda e'_n (g^{-1} \circ f) = \lambda e_n f = \lambda$$ so since $dg = d'$ it follows that $\tau_{f,d}^g = \tau_{f',d}$!
If they are all conjugate they all have the same determinant $\delta$, now $\det(\tau_{f,d}\tau_{f',d}) = \det(\tau_{f+f',d})$ so $\delta^2 = \delta$ proves they lie in $SL$.
For $n \ge 3$ we can use the $\mu$ trick as before to get them in $SL$.
Proposition If $d \in V^\#$ then $\mathscr T(d)$ is an abelian normal subgroup of the stabilizer $SL(V)_d$; $\mathscr T(d)$ are all conjugates in $SL(V)$.
proof: Certainly elements of $\mathscr T(d)$ stabilize $d$, by the computational lemma before we see that it is an abelian group (commutative and closed, therefore has identity and inverses). If $g \in SL(V)$ with $dg=d$ then $\mathscr T(d)^g = \mathscr T(d)$ (by the previous) so $\mathscr T(d) \unlhd SL(V)_d$. Given $d,d' \in V^\#$ by the transitivity lemma (that requires dimension > 1) exists $g \in SL(V)$ which takes $d$ to $d'$ then $\mathscr T(d)^g = \mathscr T(d')$.
Proposition The set $\mathscr T$ generates $SL(V)$.
proof: Elementary matrices.
Definition A group if perfect if $G' = G$. This is equivalent to there being no nontrivial abelian quotients: Clearly if $G/[G,G] \not = 1$ then $G\not = [G,G]$. Conversely if $G/N \simeq A = \{Ng\}$ then we always have $Ngg' = Ng'g$ i.e. there is some $n$ such that $gg' = ng'g$. So every commutator $[g,g']$ is an element of $N$.
Proposition If $n \ge 2$ the group $SL_n(q)$ is perfect provided $(n,q)$ isn't $(2,2)$ or $(2,3)$.
proof: We just need to show that each $\tau \in \mathscr T^\#$ is a commutator since that set generates our group. If $\tau$ has direction $d$ take some $\sigma \in \mathscr T(d)^\#$ not equal to $\tau^{-1}$, then $\sigma \tau \in \mathscr T(d)^\#$ so there is a $g \in SL_n(q)$ that conjugates $\sigma \tau = \sigma^g$, whence $\tau = \sigma^{-1} g^{-1} \sigma g= [\sigma,g]$.
For $n=2$ we will use $2 \times 2$ matrices directly, in some basis $\tau = \begin{pmatrix} 1 & \gamma \\ 0 & 1 \end{pmatrix}$ (nonzero $\gamma$), now for any nonzero $\lambda$ and $\mu \in \mathbb F_q$ we have $$\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}\begin{pmatrix} 1 & \mu \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \lambda^{-1} & 0 \\ 0 & \lambda \end{pmatrix}\begin{pmatrix} 1 & -\mu \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & \mu(\lambda^2-1) \\ 0 & 1 \end{pmatrix}$$ so for $q>3$ take $\lambda \not = 0,1,-1$ then $\lambda^2-1\not = 0$ so let $\mu = \gamma(\lambda^2-1)^{-1}$.
Proposition
- $Z(GL(V)) = \{\lambda 1 | \lambda \in \mathbb F_q^\# \}$
- $Z(SL(V)) = \{\lambda 1 | \lambda \in \mathbb F_q^\#, \lambda^n = 1\}$
Finite Fields and Finite Vector Spaces
Definition An Affine transformation of $\mathbb F_q$ is a map $f_{a,b}$ taking $\lambda$ to $a \lambda + b$ where $a \in \mathbb F_q^\#$ is nonzero. The group of such maps is called $A(\mathbb F_q)$.
Proposition $A(\mathbb F_q)$ is sharply 2-transitive of order $q(q-1)$.
proof: Let $\alpha,\beta$ distinct, the system of equations $\alpha = a 0 + b$, $\beta = a 1 + b$ has a unique solution.
Corollary By a general lemma about sharply 2-transitive groups this group must have a regular characteristic subgroup, this group is $\{f_1,b|b\in\mathbb F_q\}$.
Proposition $A(\mathbb F_q)$ has a one point extension which is sharply $3$-transitive of degree $q+1$.
proof: This is a straightforward application of the one point extension theorem, adjoin $\infty$ to $\mathbb F_q$ and define $x$ on $\mathbb F_q \cup \{\infty\}$ to swap $0$ and $\infty$ and invert all other elements $\lambda x = \lambda^{-1}$. Clearly $x^2=1$. Let $G_0 = \{f_{a,0}\mid a \in \mathbb F_q^\# \}$ and note $f_{a,0}^x$ fixes $0$ and $\infty$ while for $\lambda \in \mathbb F_q^\#$ we $f_{a,0}^x = f_{a^{-1},0}$ so $G_0^x = G_0$. Finally a system for the double cosets is given by just $1$ and any other element e.g. $f = f_{-1,1}$ will do (so $\lambda f = 1 - \lambda$ and $f^2=1$). We see that $xf$ acts on the $\infty, 1, 0$ by cycling them and for the remaining elements $\lambda (xf)^3 = 1 - \frac{1}{1-\frac{1}{\lambda}}=\lambda$ and (apparently...) $x^f = f^x \in GxG$ so we have a one point extension.
Definition $V_n(q)$ is the $n$-dimensional vector space over $\mathbb F_q$, clearly $|V_n(q)|=q^n$.
Definition If $V$ is a vector space then a linear automorphism of $V$ is a bijective linear map $V \to V$. The group of these is called the general linear group $GL(V)$ or $GL_n(q)$ when $V=V_n(q)$.
Definition The special linear group $SL(V)$ of linear automorphisms of determinant 1.
Lemma If $V$ is a vector space over $\mathbb F_q$ then $SL(V)$ is a normal subgroup of $GL(V)$ and the index is $q-1$: $|GL(V):SL(V)|=q-1$.
proof: SL is just the kernel of the surjective determinant map from GL to $\mathbb F_q^\times$. As a consequence $GL/SL \simeq \mathbb F_q^\#$ so $|GL:SL| = q-1$.
Lemma The group $GL(V)$ acts transitively on $V^\#$ and if the dimension of $V$ is $> 1$ the same is true of $SL(V)$.
proof: Take two nonzero vectors $e_1,f_1$ then to get a map between them choose bases $e_1,\ldots,e_n$ and $f_1,\ldots,f_n$ this gives $g \in GL(V)$ mapping between them. If $n > 1$, since we don't necessarily have $\det(g)=1$ let $\det(g)=\mu$ and replace $e_n$ by $\mu^{-1} e_n$. Now $g'$ mapping between these bases has determinant $1$. (did I get this right?)
Proposition $$|GL_n(q)| = q^{n(n-1)/2} \prod_{i=1}^n (q^i-1)$$ and $$|SL_n(q)| = q^{n(n-1)/2} \prod_{i=2}^n (q^i-1).$$
proof: $GL_n(q)$ acts regularly on ordered bases of $V_n(q)$, so the size of $GL_n(q)$ is equal to the number of ordered bases: $q^n-1$ choices for the first element, and having chosen $e_1,\ldots,e_i$ (which spans a $q^i$ sized space) already there are $q^n-q^i$ choices for the next. The size of $SL$ comes from the lemma before the previous.
Proposition $A(\mathbb F_q)$ is sharply 2-transitive of order $q(q-1)$.
proof: Let $\alpha,\beta$ distinct, the system of equations $\alpha = a 0 + b$, $\beta = a 1 + b$ has a unique solution.
Corollary By a general lemma about sharply 2-transitive groups this group must have a regular characteristic subgroup, this group is $\{f_1,b|b\in\mathbb F_q\}$.
Proposition $A(\mathbb F_q)$ has a one point extension which is sharply $3$-transitive of degree $q+1$.
proof: This is a straightforward application of the one point extension theorem, adjoin $\infty$ to $\mathbb F_q$ and define $x$ on $\mathbb F_q \cup \{\infty\}$ to swap $0$ and $\infty$ and invert all other elements $\lambda x = \lambda^{-1}$. Clearly $x^2=1$. Let $G_0 = \{f_{a,0}\mid a \in \mathbb F_q^\# \}$ and note $f_{a,0}^x$ fixes $0$ and $\infty$ while for $\lambda \in \mathbb F_q^\#$ we $f_{a,0}^x = f_{a^{-1},0}$ so $G_0^x = G_0$. Finally a system for the double cosets is given by just $1$ and any other element e.g. $f = f_{-1,1}$ will do (so $\lambda f = 1 - \lambda$ and $f^2=1$). We see that $xf$ acts on the $\infty, 1, 0$ by cycling them and for the remaining elements $\lambda (xf)^3 = 1 - \frac{1}{1-\frac{1}{\lambda}}=\lambda$ and (apparently...) $x^f = f^x \in GxG$ so we have a one point extension.
Definition $V_n(q)$ is the $n$-dimensional vector space over $\mathbb F_q$, clearly $|V_n(q)|=q^n$.
Definition If $V$ is a vector space then a linear automorphism of $V$ is a bijective linear map $V \to V$. The group of these is called the general linear group $GL(V)$ or $GL_n(q)$ when $V=V_n(q)$.
Definition The special linear group $SL(V)$ of linear automorphisms of determinant 1.
Lemma If $V$ is a vector space over $\mathbb F_q$ then $SL(V)$ is a normal subgroup of $GL(V)$ and the index is $q-1$: $|GL(V):SL(V)|=q-1$.
proof: SL is just the kernel of the surjective determinant map from GL to $\mathbb F_q^\times$. As a consequence $GL/SL \simeq \mathbb F_q^\#$ so $|GL:SL| = q-1$.
Lemma The group $GL(V)$ acts transitively on $V^\#$ and if the dimension of $V$ is $> 1$ the same is true of $SL(V)$.
proof: Take two nonzero vectors $e_1,f_1$ then to get a map between them choose bases $e_1,\ldots,e_n$ and $f_1,\ldots,f_n$ this gives $g \in GL(V)$ mapping between them. If $n > 1$, since we don't necessarily have $\det(g)=1$ let $\det(g)=\mu$ and replace $e_n$ by $\mu^{-1} e_n$. Now $g'$ mapping between these bases has determinant $1$. (did I get this right?)
Proposition $$|GL_n(q)| = q^{n(n-1)/2} \prod_{i=1}^n (q^i-1)$$ and $$|SL_n(q)| = q^{n(n-1)/2} \prod_{i=2}^n (q^i-1).$$
proof: $GL_n(q)$ acts regularly on ordered bases of $V_n(q)$, so the size of $GL_n(q)$ is equal to the number of ordered bases: $q^n-1$ choices for the first element, and having chosen $e_1,\ldots,e_i$ (which spans a $q^i$ sized space) already there are $q^n-q^i$ choices for the next. The size of $SL$ comes from the lemma before the previous.
Subscribe to:
Posts (Atom)