Codex @codex  0

Replace $F^n$ by a symplectic vector space $(V,B)$ of dimension $2m$, and elementary row operations by symplectic transvectionsThe cross terms in $B(Tx,Ty)$ cancel, and $B(v,v)=0$, so these maps preserve $B$. Their inverses are $T_{v,-c}$.

They generate $Sp(V)$ by the following symplectic basis elimination. If $B(u,v)\ne 0$, a single transvection with direction $v-u$ and parameter $1/B(u,v)$ sends $u$ to $v$. If $B(u,v)=0$ and $u\ne v$, choose $w$ with both $B(u,w)$ and $B(w,v)$ nonzero: the union of two proper hyperplanes does not exhaust a vector space, even over ${\mathbb{F}}_2$. Two transvections then suffice. Thus one can make a form-preserving linear map fix the first basis vector $e$. Next align its partner $f$, while keeping $e$ fixed. Two partners have difference in $e^{\perp }$, so the preceding single-transvection construction uses a direction in $e^{\perp }$ whenever their mutual pairing is nonzero. If it is zero, the intermediate partner $f+ae$, $a\ne 0$, has nonzero pairing with both. All these transvections fix $e$. Once both $e,f$ are fixed, recurse on their nondegenerate symplectic orthogonal complement. This proves generation over every field.

Use the action on the points of projective space, whose kernel is the scalar center $Z=\{\lambda I:{\lambda }^2=1\}$. For $m=1$ this is the $S{L}_2$ case. For $m\ge 2$ the action is primitive, although it is generally not two-transitive. The stabilizer of a line $L$ has two orbits on other lines: those orthogonal to $L$ and those not orthogonal to $L$. This follows by extending a hyperbolic pair, or an independent isotropic pair, to a symplectic basis. A block containing $L$ and a second line $M$ contains the entire corresponding orbit of the stabilizer of $L$. The stabilizer of $M$ then supplies a line of the other type: choose a vector orthogonal to one of $L,M$ and not to the other. Nondegeneracy and independence guarantee such a vector, also in dimension four. Hence the block contains both orbits and is the whole set.

The transvections along $L$ form an abelian normal subgroup of its stabilizer; their conjugates generate the group. Thus the Iwasawa simplicity lemma applies as soon as perfectness is established. For $|F|>3$, every transvection lies in an embedded $S{L}_2(F)$ acting on a hyperbolic plane, and 4(c) proves that embedded group perfect.

For the remaining fields, the perfectness of finite symplectic groups has a short direct proof. Over ${\mathbb{F}}_3$ with $m\ge 2$, all parameter-one transvections have the same class $t$ in the abelianization, and $3t=0$. In an isotropic plane with basis $u,v$, the four directions $u,v,u+v,u-v$ give commuting transvections whose product is $1$: their rank-one terms add to zero, sinceover ${\mathbb{F}}_3$. Thus $4t=0$, giving $t=0$. Parameter-two transvections are inverses, so all generators vanish in the abelianization. Over ${\mathbb{F}}_2$ with $m\ge 3$, use the seven nonzero vectors of an isotropic three-space. Their commuting transvections multiply to $1$, since each diagonal coordinate of ${\sum }_{0\ne v}v{v}^{\mathsf{T}}$ occurs four times and each off-diagonal coordinate twice. Their common abelianized class satisfies $2t=7t=0$, and again the group is perfect.

Consequently the projective symplectic group is simple, with the necessary exceptionsThe first two are the linear exceptions; the last is $S{p}_4(2)\cong S_6$, proved in 6(c). The word “projective” matters: the unquotiented symplectic group may have the nontrivial scalar center $\{I,-I\}$.

Choose a minimal nontrivial normal subgroup $L$ of $K$ itself. Every conjugate $L^g$ for $g\in G$ is also a minimal normal subgroup of $K$, since $K$ is normal in $G$. Distinct such subgroups intersect trivially: their intersection is normal in $K$ and minimality forces either equality or trivial intersection. Moreover $[L_i,L_j]\subseteq L_i\cap L_j=1$, so distinct ones commute.

Choose a maximal family of these conjugates whose product $M=L_1\times \cdots \times L_r$ is direct. For any further conjugate $L^g$, its intersection with $M$ is normal in $K$ and is either all of $L^g$ or trivial. In the latter case it commutes with all the existing factors and could be adjoined to the direct product of groups, contrary to maximality. Thus every conjugate is contained in $M$. The product of all conjugates is consequently $M$, which is nontrivial and normal in $G$. Minimality of $K$ as a $G$-normal subgroup gives $M=K$.

Any normal subgroup of a factor $L_i$ is normal in $K$: the other direct factors commute with it, and $L_i$ normalizes it. Minimality of $L_i$ as a $K$-normal subgroup makes it simple. All factors are isomorphic because they are conjugate to $L$. This proves the direct-product structure of a finite minimal normal subgroup:An abelian simple group is cyclic of prime order, so the abelian case is elementary abelian. Now let $H$ be a maximal proper subgroup of a nonabelian simple group $X$. It is nontrivial, since $X$ has nontrivial proper cyclic subgroups. Choose a minimal normal subgroup $K$ of $H$. Then $H\le N_X(K)$. This normalizer cannot be $X$, since that would make $1<K\le H<X$ normal in $X$. Maximality givesFor each of the two groups in this question, the allowed absence of proper nonabelian simple subgroups makes the factors of such a $K$ cyclic of prime order. Hence it suffices to examine normalizers of nontrivial elementary abelian subgroups.

In $A_5$, the possible primes are two, three and five. The Sylow subgroups for the last two primes are cyclic of orders three and five. Every involution is a double transposition; for $(12)(34)$ the centralizer in $A_5$ isthe Klein four-group fixing point five. Thus an elementary abelian $2$-subgroup is cyclic of order two or a Klein four-group. The normalizer of a cyclic order-two subgroup equals its centralizer and has order four, so is contained in a point stabilizer of order twelve. A Klein four-group's common fixed point is intrinsic, and the full point stabilizer $A_4$ normalizes it; its normalizer is therefore exactly that stabilizer.

An order-three subgroup has support a triple. Its normalizer preserves the triple and the remaining pair, and is $(S_3\times S_2)\cap A_5$, of order six. An order-five subgroup acts regularly on all five points. Label the points by ${\mathbb{F}}_5$, with the subgroup acting by translations. Its normalizer in $S_5$ is the twenty affine maps $x\mapsto ax+b$. The translations are even, whereas multiplication by a generator of ${\mathbb{F}}_5^{\times }$ is a four-cycle and is odd. Exactly half these maps are even, so the normalizer in $A_5$ has order ten.

The only candidates for maximal subgroups are thus the normalizers of orders twelve, ten and six. They are all maximal: none can lie in a larger candidate, except potentially an order-six subgroup in an order-twelve $A_4$. But an index-two subgroup of $A_4$ would contain every order-three element; there are eight such elements, plus the identity, already more than six. Hence that containment is impossible. Point stabilizers are conjugate; order-three and order-five subgroup normalizers each form one class by the Sylow theorems. Therefore the maximal subgroups of A5 form preciselyTheir numbers are five, six and ten respectively, since these proper maximal subgroups are self-normalizing in the simple group.

For $X=G{L}_3(2)$, choosing independent columns gives $|X|=(8-1)(8-2)(8-4)=168$. If $t$ is an involution, write $t=I+N$. Field characteristic two gives $N^2=0$, and $\mathrm{im}N\subseteq \ker N$ in dimension three forces its nonzero rank to be one. Thus $t=I+vf$, with nonzero vector $v$, nonzero linear functional $f$ and $f(v)=0$. This transvection fixes the plane $\ker f$ and has image line $⟨v⟩$. Every flag consisting of an incident line and plane occurs, giving $7\cdot 3=21$ involutions, all conjugate. Its centralizer, and hence the normalizer of its cyclic subgroup, is the flag stabilizer of order $168/21=8$. It lies in both the line and plane stabilizers of order $24$, so cannot be maximal.

For the elementary abelian subgroups in GL3 over F2, consider commuting transvections $I+vf$, $I+wh$. Their rank-one parts satisfyIf $v,w$ are independent, equality forces both sides zero; thus $f,h$ vanish on $⟨v,w⟩$ and are the same nonzero functional. If $v=w$, they have a common image line. Consequently the two types of order-four elementary abelian subgroup areNo extra transvection can commute with all elements of $K_L$: if its image were different, its functional would have to equal each of the two independent functionals occurring in $K_L$. Dually, the same holds for $K_W$. Thus these exhaust the elementary abelian $2$-subgroups of rank at least two. The normalizer of $K_L$ is the line stabilizer and the normalizer of $K_W$ is the plane stabilizer, each of order $168/7=24$. They give different conjugacy classes because their common fixed spaces have dimensions one and two. All line stabilizers are conjugate, as are all plane stabilizers.

For an order-three element, its minimal polynomial is $(X-1)(X^2+X+1)$: it has a fixed line and an invariant complementary irreducible plane. Its normalizer preserves both and is the full $G{L}_2(2)$ on that plane, of order six, since the order-three subgroup is normal in $G{L}_2(2)$. This normalizer lies in the corresponding line and plane stabilizers, so is not maximal.

An order-seven element is irreducible. Over the finite field ${\mathbb{F}}_2$ one has $X^7-1=(X-1)(X^3+X+1)(X^3+X^2+1)$, and both cubic factors are irreducible because neither has a root in ${\mathbb{F}}_2$. Its minimal polynomial divides this square-free product and cannot be just $X-1$, since the element is not the identity. It therefore contains a cubic factor; the three-dimensional space forces that irreducible cubic to be the whole minimal polynomial. Identify the space with ${\mathbb{F}}_8$ and the element with multiplication by a generator $\alpha$ of ${\mathbb{F}}_8^{\times }$. Its centralizer consists of the seven nonzero multiplication maps. A normalizer induces an ${\mathbb{F}}_2$-automorphism of this field, so it sends $\alpha$ to $\alpha$, ${\alpha }^2$ or ${\alpha }^4$. All three possibilities occur via the field automorphisms $x\mapsto x^{2^j}$. Therefore the normalizer of the Singer cycle has order $7\cdot 3=21$, and all such normalizers are conjugate by the Sylow theorems.

The maximal-normalizer reduction leaves exactly the two order-twenty-four stabilizer types and the order-twenty-one type. Each is maximal because it cannot be contained in any larger remaining candidate; equal-order subgroups cannot properly contain one another and $21$ does not divide $24$. Thus

Write $x_{ij}(a)=I+a{E}_{ij}$ for $i\ne j$. For $a\ne 0$ these are transvections, and for $a=0$ they are the identity. Left multiplication by $x_{ij}(a)$ adds $a$ times row $j$ to row $i$. We show that these elementary transvection matrices generate the special linear group.

Apply Gaussian elimination using only row additions. If a pivot is zero, invertibility ensures some lower entry in its column is nonzero; adding that row produces a nonzero pivot without interchanging or scaling rows. Clear entries below each pivot, and then above pivots in reverse order. This reduces any invertible matrix to a diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$, with all $d_i\ne 0$, preserving determinant throughout. For a determinant-one matrix, ${\prod }_i{d}_i=1$.

A two-coordinate diagonal block is itself a product of elementary transvection matrices. On those two coordinates putThis formula is valid in characteristic two as well. For $n\ge 2$, multiply the diagonal blocks with entries $d_i,d_i^{-1}$ on coordinates $(i,n)$ for $1\le i<n$. Their product is $D$, because its last entry is ${\prod }_{i<n}{d}_i^{-1}=d_n$. Thus $D$ and the original matrix lie in the subgroup generated by transvections. Conversely every transvection has determinant one by part ii. HenceFor $n=1$, both the special linear group and the subgroup generated by the degenerate identity transvection are trivial. The argument also establishes transvections generate the special linear group over any field.

For a nonidentity transvection ${\tau }_{f,d}$, take $e_1=d$ and choose $e_2$ with $f(e_2)=1$. Complete $e_1$ to a basis of $\ker f$ using $e_3,\dots ,e_n$. Then $e_1,e_2,e_3,\dots ,e_n$ is a basis of $V$, and the matrix of the transvection is $I+E_{12}$: it sends $e_2$ to $e_2+e_1$ and fixes the other basis vectors. Consequently all nonidentity transvections form one conjugacy class in $\mathrm{GL}(V)$. This matrix has determinant one, and conjugation preserves determinant, soFor $n\ge 3$, let $k$ be any GL-conjugator carrying one transvection to another. In an adapted basis for the first, the matrix $c=\mathrm{diag}(1,1,s,1,\dots ,1)$ centralizes it and has determinant $s$. Set $s=(\det k{)}^{-1}$. Then $kc\in \mathrm{SL}(V)$ gives the same conjugation as $k$. This proves the stronger transvection conjugacy over a finite field resultThe extra coordinate is essential to this determinant adjustment; no such claim is required in dimension two.

A nonidentity transvection is a linear automorphism $\tau =I+N$ with $\mathrm{rank}N=1$ and $N^2=0$. Equivalently, it fixes a hyperplane pointwise and its displacement has image in a line lying in that hyperplane. The notation $T\setminus \{1\}$ allows the identity as a degenerate transvection; use that convention here, with $N=0$ also allowed.

For a nonidentity transvection, choose $d\ne 0$ spanning the image of $N$. There is a unique nonzero linear functional $f$ such that $N(v)=f(v)d$. Since $N^2(v)=f(v)f(d)d$, we have $f(d)=0$, so $d\in \ker f$. Conversely, this condition makes $(df{)}^2=0$ and gives inverse $I-df$. The identity is represented by $f=0$ and any $d\ne 0$. ThusWhen $f(d)=f^{\prime }(d)=0$, the two displacement maps have zero product, givingFor clarity use left actions of linear maps. Applying $g^{-1}$, then ${\tau }_{f,d}$, then $g$ givesWith the alternative conjugation notation ${\tau }^g=g^{-1}\tau g$, the answer is ${\tau }_{f,d}^g={\tau }_{f\circ g,g^{-1}(d)}$. In either convention the transformed direction lies in the kernel of the transformed functional.