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The Butcher contractivity theorem states that an algebraically stable Runge-Kutta method is B-stable: for a dissipative vector field and any positive step for which the stage problems are solved, its step does not increase the distance between two numerical solutions. The required condition on the dissipative vector field isfor all compared states at each common time. The inner product can be real Euclidean or complex Hermitian. The stage solvability of an implicit Runge-Kutta method is assumed in this comparison, not a conclusion from a distance estimate alone.

Let $d$ be the difference of starting values, $D_i$ the differences of corresponding stages and $F_i$ the differences of their vector-field values. ThenExpanding the squared norm of the update givesSubstitute $d=D_i-h{\sum }_j{a}_{ij}{F}_j$ in the linear term. Symmetrizing the double sum proves the Runge-Kutta contractivity identityEach inner product in the first sum is nonpositive by the defining inequality for a dissipative vector field, and its weight is nonnegative. The final quadratic form is nonnegative: resolve the vectors $F_i$ into components and apply positive semidefiniteness of $M$ to each component vector. Hence $\boxed{\Vert d_{+}\Vert \le \Vert d\Vert }$, which is exactly B-stability. No linearization of the vector field has been used.

For the Dahlquist test equation $y^{\prime }=\lambda y$ with $\mathrm{Re}\lambda \le 0$, this also bounds the amplification factor by one wherever the stages are well defined. With strictly positive weights, no stage pole can occur in the open left half-plane. Indeed, if $(I-zA)v=0$ for nonzero $v$, thenwhen $\mathrm{Re}z<0$, contradicting $M⪰0$. This proves the usual A-stability corollary for the scalar stability function; boundedness from the open half-plane makes any boundary pole in that rational function removable. Zero-weight redundant stages require their own solvability treatment. The nonlinear B-stability statement is stronger than A-stability, and the two notions should not be conflated.

The eigenvalue stability analysis of a finite difference method reduces a linear evolution discretization to its modal behavior, but it must retain the norm and the dependence on the mesh. For a homogeneous system obtained by the method of lines, $U^{\prime }=A_{h}U$, the solution is $e^{t{A}_h}U(0)$. A one-step time discretization gives $U^{n+1}=G_{h,k}{U}^n$; for a Runge-Kutta method, where its stage matrices are invertible, $G_{h,k}=R(k{A}_h)$. With a multistep method one instead uses an amplification matrix on the augmented vector of several time levels.

The relevant finite-time stability estimate iswith $C_T$ independent of the refining spatial mesh and allowed time steps. Similarly, semidiscrete stability requires $\Vert e^{t{A}_h}{\Vert }_h\le C_T$ for $0\le t\le T$. The norm should represent the continuum problem, such as the discrete L2 norm $\Vert U{\Vert }_{2,h}^2=h{\sum }_j|U_j{|}^2$ in one dimension. A fixed finite matrix having decaying solutions as $t\to \infty$ is not by itself a mesh-uniform result for the partial differential equation.

If $G=X\Lambda X^{-1}$, thenThus eigenvalues of modulus at most one give a uniform bound when the diagonalizing bases have uniformly bounded condition numbers. For a normal matrix in the chosen inner product, the spectral theorem for normal operators supplies an orthonormal basis, so the condition number is one and the norm of $G^n$ is exactly ${\max }_j|{\lambda }_j{|}^n$. For a self-adjoint or skew-adjoint spatial matrix, applying a scalar stability function preserves this favorable modal structure. This is the setting where eigenvalue calculations provide especially clean, reliable step restrictions.

A precise fixed-matrix statement is also useful: a matrix is power bounded for all nonnegative integers $n$ if and only if all its eigenvalues lie in the closed unit disk and every Jordan block at a unit-modulus eigenvalue has size one. Indeed, powers of an interior Jordan block contain a polynomial in $n$ times a decaying geometric factor and are bounded; a larger block on the unit circle produces unbounded polynomial growth. For a family of matrices, the resulting bounds must still be uniform. Moreover, finite-time stability versus power boundedness distinguishes this all-time condition from the physical-time estimate above. For example $G_k=I+kN$, $N^2=0$, has $G_k^n=I+nkN$, bounded for $nk\le T$ despite a nontrivial Jordan block at one. Even the scalar update $1+k$ is stable on fixed intervals, since $(1+k{)}^n\le e^T$, although it represents a growing equation rather than a contraction.

For a constant-coefficient periodic difference operator, Fourier modes diagonalize the spatial matrix. The discrete Fourier transform is a unitary operator, so Parseval's identity identifies its modal maximum with the operator norm. The eigenvalue stability analysis of a finite difference method then becomes von Neumann stability analysis. For the Dirichlet discrete Laplacian, a discrete sine transform plays the same role. These facts explain its advantages: a large matrix calculation becomes a scalar symbol or a known spectral interval, and the extremal eigenvalue directly determines a safe time step.

As a first example, centered space discretization of the heat equation $u_t=u_{xx}$ has eigenvaluesForward Euler method gives $R(k\lambda )=1-4r{\sin }^2(\theta /2)$ with $r=k/h^2$. Requiring all factors in $[-1,1]$ yields $0\le r\le 1/2$ for all periodic meshes. In two dimensions the corresponding uniform bound is $r\le 1/4$. Because these are normal matrices, these scalar bounds prove contraction rather than merely suggest it.

Backward Euler method has $R(z)=1/(1-z)$, so the centered discretization of the heat equation is contractive for every $k>0$. The Crank-Nicolson method has $R(z)=(1+z/2)/(1-z/2)$ and is likewise contractive for these negative real modes. However, its factor tends to $-1$ for very stiff modes, leaving high-frequency oscillations weakly damped; Backward Euler method tends to zero. The eigenvalues therefore expose the presence of stiff differential equations and damping as well as binary stability. Unconditional stability does not remove accuracy requirements.

For an advection equation example, the upwind finite difference scheme for $u_t+c{u}_x=0$, $c>0$, has factor $G=1-\nu +\nu e^{-i\theta }$, with $\nu =ck/h$. Sinceit is a contraction in the periodic discrete L2 norm exactly for $0\le \nu \le 1$. In contrast, Forward Euler method with centered advection has factor $1-i\nu \sin \theta$. Its modulus exceeds one at nonzero active frequencies, and repeated amplification makes it unstable under fixed nonzero Courant refinement. A much smaller scaling $k=O(h^2)$ can bound its finite-time growth, demonstrating again that exact contraction and the most general stability estimate are different claims.

The major limitation is the possible presence of a nonnormal matrix. For example,Both eigenvalues are negative, yet at $t=1$ the second unit vector is amplified by at least $1/(eh)$. This proves that negative spectra do not imply uniform semidiscrete stability. Ill-conditioned eigenvectors, Jordan normal form and transient amplification can invalidate a spectral-only argument. Direct energy estimates or estimates for the resolvent can supply the missing norm control.

Boundary conditions are another limitation. A periodic symbol proves a periodic or whole-line result, not an arbitrary initial-boundary-value problem. Boundary closure of a difference scheme can introduce growing modes or mesh-dependent amplification. A spatial operator with variable coefficients can still be studied through its assembled matrix, but then whether it is a normal matrix, its condition number and the physical inner product must be checked rather than borrowed from a Fourier stability analysis. For the finite element method the natural mass matrix inner product often turns a generalized eigenvalue problem into a self-adjoint one.

For time-dependent or split updates, separate spectra are insufficient to control products. For instance, $G_1=\left(\begin{array}{cc}0& K\\ 0& 0\end{array}\right)$ and $G_2=\left(\begin{array}{cc}0& 0\\ K& 0\end{array}\right)$ each have only zero eigenvalues, while $G_2{G}_1$ has eigenvalue $K^2$. For $K>1$, alternating the two is unstable. A common contractive norm would control the products, but their individual eigenvalues do not.

Finally, stability and consistency of a numerical method have different roles. The Lax equivalence theorem states that, for a well-posed linear initial-value problem and a consistent finite difference method on compatible grid spaces, stability on every fixed time interval is equivalent to convergence of a numerical method as the admissible meshes refine. Bounded grid representation and the correct initial and boundary setup are part of that framework. An eigenvalue estimate can establish its stability hypothesis in the favorable settings described above; it neither proves consistency of a numerical method nor bypasses its uniformity requirements. The method is therefore a powerful modal tool, provided its spectral calculation is connected to a genuine norm estimate.

For a finite group, define the Frattini subgroup bythe intersection is $G$ when there are no maximal subgroups, which for a finite group happens only for $G=1$. A second definition is the set of non-generators: an element $x$ such that $⟨S,x⟩=G$ always implies $⟨S⟩=G$.

If $x\in \Phi (G)$ and $H=⟨S⟩<G$, put $H$ in a maximal subgroup $M$. Then $x\in M$, so $⟨S,x⟩\le M<G$. Conversely, if $x\notin \Phi (G)$, some maximal subgroup $M$ omits it; maximality gives $⟨M,x⟩=G$, although $M\ne G$. This proves equivalence.

Every group automorphism permutes maximal subgroups, so $\Phi (G)$ is a characteristic subgroup, in particular a normal subgroup. Alsoa proper $H$ would lie in a maximal subgroup containing both factors.

Set $F=\Phi (G)$ and take a Sylow subgroup $P$ of $F$. For $g\in G$, normality of $F$ makes $P^g$ a Sylow subgroup of $F$. Conjugating inside $F$ therefore gives the Frattini argument factorization $G=F{N}_G(P)$. The preceding generating property forces $N_G(P)=G$. Thus every Sylow subgroup of $F$ is normal, and part (i) shows that $F$ is a nilpotent group.

Now suppose $G$ is a finite p-group. The normalizer condition makes every maximal subgroup $M$ normal: $M<N_G(M)$ and maximality force $N_G(M)=G$. The quotient $G/M$ has no proper nontrivial subgroup, hence is cyclic of prime order; its p-power order makes that prime $p$. Consequently every such $M$ containswhere the two factors are the commutator subgroup and the subgroup generated by pth powers. Thus $D\le \Phi (G)$.

The quotient $G/D$ is an elementary abelian group, hence a vector space over ${\mathbb{F}}_p$. If $x\notin D$, extend the nonzero vector $xD$ to a basis and choose a linear functional nonzero on it. Its kernel pulls back to a subgroup of index $p$, a maximal subgroup omitting $x$. Therefore $x\notin \Phi (G)$, provingThe two factors in the last formula are multiplied; the superscripted expression in the TeX aid is not the formula printed in the PDF.

The Fano plane has as points the seven nonzero vectors of ${\mathbb{F}}_2^3$, because every one-dimensional vector subspace has just one nonzero vector. Its lines are the seven two-dimensional vector subspaces. Every line has three points, every point lies on three lines, two points determine the line $\{u,v,u+v\}$, and two lines meet in one point.

The two order-24 classes are therefore stabilizer subgroups of points and of lines. Each is isomorphic to $S_4$: fixing a vector gives the block group ${\mathbb{F}}_2^2⋊G{L}_2(2)$, the full affine group on four points; inverse transpose gives the dual line stabilizer subgroup. Their distinct fixed-point behavior in part (i) distinguishes the two conjugacy classes.

For a point $p$ and line $L$, $G_p\cap G_L$ stabilizes the ordered pair $(p,L)$. There are $7\cdot 3=21$ incident pairs and $7\cdot 4=28$ nonincident pairs. A change of basis takes any pair to a fixed model of its type, so each type is a single orbit. Orbit-stabilizer theorem gives the Fano point-line stabilizer intersection:Both occur: take $L=⟨e_1,e_2⟩$ and choose $p=⟨e_1⟩$ or $p=⟨e_3⟩$.

For the extension, identify vectors and covectors by the standard pairing. The inverse-transpose automorphism exchanges a point with its orthogonal plane and exchanges the two order-24 classes. Thus it is outer. The Fano plane duality group $\overline{G}=G⋊⟨\tau ⟩$ has order 336 and acts on points and lines, preserving incidence while allowing their types to be swapped.

The stabilizer subgroup of an unordered point-line pair has twice the corresponding ordered-pair stabilizer subgroup: some element outside $G$ swaps the pair, since duality produces a pair of the same incidence type and $G$ is transitive on such pairs. Hence these stabilizer subgroups have orders 16 and 12. Also every outside conjugate of a Sylow 7-subgroup can be moved back by an element of $G$, giving

We prove these exhaust the maximal subgroups other than $G$. Let $K$ be such a maximal subgroup and $H=K\cap G$. Since $K$ is not contained in $G$, $H$ is normal in $K$ and $|K:H|=2$. If $7\mid |H|$, part (i) gives a unique Sylow 7-subgroup $P$ of $H$. It is characteristic in $H$, hence normalized by $K$. Therefore maximality forces $K=N_{\overline{G}}(P)$.

Suppose $7\nmid |H|$ and $H\ne 1$. The classification argument in part (i) says $H$ fixes a point or a line. Because $H$ is normal in $K$, an element outside $G$ swaps its fixed-point and fixed-line sets, so it fixes at least one of each. WriteFixed lines are kernels of fixed nonzero covectors, since the only nonzero scalar is one. Duality gives $2^d-1=2^{d^{\ast }}-1$, hence $d=d^{\ast }\in \{1,2\}$.

If $d=1$, there is a unique fixed point and unique fixed line; $K$ preserves their unordered pair. If $d=2$, put $W=V^H$ and $A=((V^{\ast }{)}^H{)}^{\perp }$. For every $h\in H$, $\ker (h-I)$ contains the plane $W$ and $\mathrm{im}(h-I)$ lies in the line $A$. Over ${\mathbb{F}}_2$ there is only one nonzero linear map $V/W\to A$, so $H$ has order two. Invertibility forces $A\subseteq W$: for $h=I+vf$ one needs $f(v)=0$. Thus $A$ and $W$ are its unique image point and kernel line. These are canonically swapped by duality, so $K$ again preserves their unordered incident pair.

If $H=1$, then $K$ has order two, generated by an involutory duality $u$. For any point $p$, the pair $\{p,u(p)\}$ is preserved by $u$. Hence $K$ lies in an order-12 or order-16 pair stabilizer subgroup and is not maximal. We have exhausted all cases.

The pair stabilizer subgroups are themselves maximal: a proper maximal overgroup is one of the just-listed types, and divisibility excludes orders 12 inside 16 or 42, and 16 inside 12 or 42. The order-42 normalizer is maximal by the unique-Sylow argument. Incidence types give one conjugacy class each; Sylow conjugacy gives one normalizer class. Thus, apart from the normal index-two subgroup $G$,

Work in the finite-group setting. If $M$ is maximal in a nonabelian simple $G$, then $M\ne 1$: otherwise every nonidentity element would generate $G$, forcing a cyclic group of prime order. Choose a minimal normal subgroup $N\ne 1$ of $M$. Every characteristic subgroup of $N$ is normal in $M$, so minimality makes $N$ a characteristically simple group. Its normalizer contains $M$. By maximality it is $M$ or $G$, and the latter would make $N$ a nontrivial normal subgroup of the simple group $G$, impossible because $N\le M<G$. This proves the maximal subgroup normalizer criterion:

For $A_5$, stabilizer subgroups have order 12. Stabilizer subgroups of a two-element subset have order $|(S_2\times S_3)\cap A_5|=6$, and are isomorphic to $S_3$; for example $⟨(123),(12)(45)⟩$. The normalizer of a Sylow 5-subgroup has order 10: its normalizer in $S_5$ has order $5\cdot 4$, and exactly half its permutations are even, since a multiplier of order four on ${\mathbb{F}}_5$ is an odd four-cycle. This gives a dihedral group of order 10.

To prove completeness, an intransitive subgroup either fixes a point, or its orbits on five letters have sizes two and three. It therefore lies in one of the first two types. A transitive proper subgroup $H$ has order divisible by 5. Its number of Sylow 5-subgroups is one or six, since $A_5$ has six and the count is $1$ modulo 5. A count of six would make $30\mid |H|$, forcing an index-two subgroup and contradicting Simplicity of the alternating group A5. Therefore $H$ normalizes its unique Sylow 5-subgroup and lies in the order-ten normalizer.

Stabilizer subgroups are maximal because their index is prime. The order-ten group is transitive and cannot have a larger proper transitive overgroup by the preceding argument. The order-six group has no fixed point, cannot lie in a stabilizer subgroup, and cannot lie in the order-ten type because 3 does not divide 10. Thus all three types are maximal. $A_5$ is transitive on points and on unordered pairs, while Sylow conjugacy handles the normalizers. Consequently

Now let $G=G{L}_3(2)$. Choosing independent columns gives $|G|=7\cdot 6\cdot 4=168$. Its simplicity follows from the elementary projective-transvection proof in part 4(i), which does not use this subgroup classification.

A stabilizer subgroup and a plane stabilizer subgroup each have index seven and order 24. To construct the remaining type, identify ${\mathbb{F}}_2^3$ with ${\mathbb{F}}_8$ and take the Singer cycle $P$ of multiplications by nonzero field elements, of order seven. Any linear map commuting with a primitive multiplication commutes with all of ${\mathbb{F}}_8$, so its centralizer is precisely $P$. A normalizer sends a primitive element $a$ to another root of its minimum polynomial, namely $a,a^2,a^4$. The Frobenius map $x\mapsto x^2$ realizes those three choices. Hence

If $7\mid |H|$ for a proper $H<G$, its Sylow count is one or eight. Eight would make $56\mid |H|$, hence $|H|=56$. Its index-three coset action would inject the simple group $G$ into $S_3$: the kernel is normal and cannot be all of $G$. This is impossible. Thus $H$ normalizes a unique Sylow 7-subgroup and lies in the order-21 normalizer.

If $7\nmid |H|$, its order divides 24. In its action on seven nonzero vectors, an odd orbit has size one or three. A one-point orbit gives a fixed point. If a three-point orbit is collinear, it is the set of nonzero vectors of an invariant plane. If its three vectors are independent, their nonzero sum is fixed by $H$. Thus $H$ lies in a point or plane stabilizer subgroup.

The index-seven stabilizer subgroups are maximal, and any proper overgroup of the order-21 normalizer would again have a unique Sylow 7-subgroup and normalize that same subgroup, so the normalizer is maximal too. The point and plane stabilizer subgroups are not conjugate: a stabilizer subgroup fixes a point, whereas a plane stabilizer subgroup is transitive on its three internal points and its four external points, so fixes no point. Each type is one conjugacy class. We have proved maximal subgroups of GL3 over F2:

Choose a complement $W$ to the fixed $k$-dimensional vector subspace $U$. With column vectors and the decomposition $V=U\oplus W$, its parabolic stabilizer of a subspace isThe arbitrary upper-right block forms a normal elementary abelian group. The block-diagonal matrices form a complement, acting on it by $B\mapsto AB{D}^{-1}$. ThusWith row vectors the off-diagonal block is transposed; the subgroup description is the same after the convention change.

The rank of a transitive permutation group is the number of orbits of a stabilizer subgroup, not the rank of a matrix. For $Y\in X_k$, the invariant of its $P_k$-orbit is $r=\dim (U\cap Y)$. This invariant is complete. Choose a basis of $U\cap Y$, extend it to a basis of $U$, then add a basis of a complement of $U\cap Y$ in $Y$, and extend the resulting independent vectors to a basis of $V$. For $Y^{\prime }$ with the same intersection dimension $r$, do the same. The map between these adapted bases preserves $U$ and sends $Y$ to $Y^{\prime }$.

Each value $r=0,1,\dots ,k$ occurs: in a basis $e_1,\dots ,e_n$ with $U=⟨e_1,\dots ,e_k⟩$, takeThe hypothesis $2k\le n$ guarantees all these vectors exist. Hence

Replace the determinant condition by preservation of a nondegenerate alternating bilinear form $B$ on $F^{2m}$. In a symplectic basis the symplectic group over a finite field isTo count symplectic bases, choose the first vector $e$ in $q^{2m}-1$ ways, then choose $f$ with $B(e,f)=1$ in $q^{2m-1}$ ways. Their span is nondegenerate; its symplectic orthogonal complement has dimension $2m-2$ and supplies the rest of the basis. The resulting recurrence gives the finite symplectic group order

The elementary replacements are symplectic transvectionsExpanding $B(Tx,Ty)$ shows that the two cross terms cancel and the last term vanishes because $B(v,v)=0$. Thus these maps preserve $B$; their inverses are $T_{v,-c}$.

We need both generation and primitivity to use the argument from part (i). For nonzero vectors $v,w$ with $B(v,w)\ne 0$, the symplectic transvection with direction $w-v$ and parameter $1/B(v,w)$ sends $v$ to $w$. When $B(v,w)=0$, choose $z$ nonorthogonal to both and use two such steps. Such $z$ exists because two proper hyperplanes cannot cover a vector space: if the two linear forms are independent prescribe both values to be one, and if dependent prescribe either a nonzero value. Hence symplectic transvections act transitively on nonzero vectors.

They generate the whole symplectic group over a finite field. Indeed first use them to make a given symplectic transformation fix $e_1$. Its image $f^{\prime }$ of $f_1$ still satisfies $B(e_1,f^{\prime })=1$. A symplectic transvection with direction $f_1-f^{\prime }$ fixes $e_1$ and sends $f^{\prime }$ to $f_1$ if the pairing is nonzero. If that pairing is zero, pass through $f^{\prime }+c{e}_1$ for any $c\ne 0$: it has nonzero pairing with both partners, and both required directions lie in $e_1^{\perp }$. We have now fixed the first hyperbolic pair, and induction applies in its nondegenerate orthogonal complement. The final two-dimensional case is exactly the elementary generation of $S{L}_2(q)=S{p}_2(q)$ proved in part (i).

For a fixed projective point $⟨e⟩$, the symplectic transvections $T_{e,c}$ form an abelian subgroup normal in its stabilizer subgroup; conjugation scales $c$ by the square of the scalar multiplying $e$. Their conjugates supply all symplectic transvections. A central element commutes with every symplectic transvection, so it preserves every direction line and must be scalar. Conversely scalars commute with the whole group. The kernel of the projective action consists of scalar matrices, and preservation of $B$ restricts the scalar to ${\lambda }^2=1$. Thus the center of a group has size $\gcd (2,q-1)$ and the projective symplectic group over a finite field acts faithfully.

For $m\ge 2$ the stabilizer subgroup has precisely three orbits: its fixed point, the other points in $e^{\perp }$, and the points outside $e^{\perp }$. Transitivity on each follows by extending, respectively, an independent isotropic pair or a hyperbolic pair to a symplectic basis. WriteA nontrivial proper block through the fixed point would be a union of these stabilizer subgroup orbits, hence would have size either $h$ or $q^{2m-1}+1$. A block size divides $N$. But $N-qh=1$, so $h>1$ cannot divide $N$, and $N/2<q^{2m-1}+1<N$, excluding the other size. This proves primitivity. For $m=1$ the projective action is the two-transitive action of $PS{L}_2(q)$.

Here is a direct check of perfectness of finite symplectic groups, including the small fields where the diagonal group commutator needs replacement. If $q>3$, every symplectic transvection lies in an embedded $S{L}_2(q)$ on a hyperbolic plane containing its direction, and that subgroup is perfect by part (i). Since the symplectic transvections generate, the whole symplectic group over a finite field is perfect.

For $q=3$ and $m\ge 2$, all $T_{v,1}$ are conjugate, so their common image $z$ generates the abelianization and satisfies $3z=0$. Choose independent perpendicular vectors $e,f$. The four commuting symplectic transvections with directions $e,f,e+f,e-f$ and parameter one multiply to the identity: the sum of their rank-one matrices is zero in characteristic 3. Thus also $4z=0$, giving $z=0$.

For $q=2$ and $m\ge 3$, use the seven nonzero vectors of a three-dimensional isotropic subspace of a symplectic vector space. Their symplectic transvections commute, and their product is the identity. To see this explicitly, write a vector in three isotropic basis coordinates: in the sum of the outer products $v{v}^{\mathsf{T}}$ each diagonal coefficient occurs four times and each off-diagonal coefficient twice, all zero in characteristic 2. The common abelianization image $z$ therefore satisfies $2z=7z=0$. Again the group is perfect.

The simplicity argument from part (i) now applies to the projective symplectic group over a finite field. The remaining exceptional group besides $S{p}_2(2)$ and $PS{p}_2(3)$ is $S{p}_4(2)$. It is not simple: it is isomorphic to $S_6$. One can verify this identification without assuming it. On $F_2^4$ setThese are all 16 quadratic forms with polar form $B$; the Arf invariant of a quadratic form gives $\mathrm{Arf}(Q_a)=Q_0(a)$. There are six with invariant one. The symplectic group over a finite field permutes these six forms. Differences of their parameter vectors span $F_2^4$: writing $e_i,f_i$ for the symplectic basis, the six parameters areTheir pairwise differences include each of $e_1,e_2,f_1,f_2$. An element fixing all six forms therefore fixes all linear forms $B(a-b,-)$, hence is the identity. This gives a faithful action on six objects; the group order is 720, so its image is all of $S_6$. Consequently

Let $F={\mathbb{F}}_q$. The general linear group over a finite field consists of all invertible linear maps $F^n\to F^n$, and the special linear group over a finite field is its determinant-one subgroup. Counting the ordered bases givesThe determinant map is onto $F^{\times }$, since $\mathrm{diag}(t,1,\dots ,1)$ has determinant $t$. Its kernel is $S{L}_n(q)$, so

For $n>1$, commuting with all the elementary transvection matrices $I+t{E}_{ij}$ forces a matrix to be scalar. Therefore the center of a group of $S{L}_n(q)$ consists of $\lambda I$ with ${\lambda }^n=1$. The multiplicative group of a finite field is cyclic, so this center of a group has size $d=\gcd (n,q-1)$. The projective special linear group over a finite field consequently has order

Here is a proof of simplicity which isolates all the required ingredients. First we prove the Iwasawa simplicity lemma in the form needed. Suppose a group $H$ acts faithfully and primitively, is a perfect group, and has an abelian subgroup $U$ normal in a stabilizer subgroup $H_{\alpha }$, whose conjugates generate $H$. For a nontrivial normal subgroup $N$, its orbits form a block system. Faithfulness prevents all these orbits from being singletons, so primitivity makes $N$ transitive. Thus $H=N{H}_{\alpha }$. Modulo $N$, each conjugate of $U$ has exactly the same image as $U$, because we may write its conjugating element as $nh$ with $n\in N$ and $h\in H_{\alpha }$. The quotient $H/N$ is generated by this one abelian image and hence is abelian. Since $H$ is a perfect group, $H/N=1$. Thus every nontrivial normal subgroup is $H$.

Apply this to the action on projective points, that is, one-dimensional vector subspaces of $F^n$. Any ordered pair of distinct lines can be sent to any other by a linear map obtained by extending representatives to bases. Multiplying one target basis vector by a suitable nonzero scalar adjusts the determinant to one without changing either target line. Thus $S{L}_n(q)$ is two-transitive on these points. A linear map fixing every line is scalar: it is diagonal on a fixed basis, and fixing the lines spanned by $e_i+e_j$ makes all diagonal entries equal. The projective quotient therefore acts faithfully, two-transitively, and primitively.

For the line $⟨e_1⟩$, useThese matrices multiply by addition of $f$, so $U$ is abelian. Conjugating by a matrix preserving $⟨e_1⟩$ preserves $U$, so it is normal in that stabilizer subgroup. Its conjugates include every elementary transvection matrix $I+t{E}_{ij}$.

These elementary transvection matrices generate $S{L}_n(q)$ by row elimination. For completeness, row addition is elementary, signed row interchange is elementary, and compensating diagonal scalings are elementary because, in a two-dimensional coordinate block,Elimination using these operations reduces a determinant-one matrix to the identity.

It remains to show that these groups are perfect groups. In the following matrix identities use $[x,y]=xy{x}^{-1}{y}^{-1}$; the alternative group commutator convention generates the same commutator subgroup. If $n\ge 3$, distinct indices give the group commutator identityEvery elementary generator is therefore a group commutator. For $n=2$ and $q>3$, choose $\lambda \in F^{\times }$ with ${\lambda }^2\ne 1$, and let $D=\mathrm{diag}(\lambda ,{\lambda }^{-1})$. ThenAs $t$ varies this supplies all upper elementary transvection matrices, and conjugation supplies the lower ones. Thus $S{L}_n(q)$, and its projective quotient, are perfect in the asserted cases. The preceding simplicity argument now applies.

The two exceptions are genuine: the projective actions identify $PS{L}_2(2)$ with $S_3$ and $PS{L}_2(3)$ with $A_4$, of orders 6 and 12 respectively. The first contains its normal subgroup of order 3; the second contains the normal Klein four group. We have proved

The Grassmann graph is connected, and its distance can be determined directly. Put $C=A\cap B$, $r=\dim C$, and $s=k-r$. Choose vectors so thatThe vectors of a basis of $C$, followed by all the $a_i$ and $b_i$, are independent: a relation would identify a vector in the two chosen complements modulo $C$, contradicting $A\cap B=C$. For $0\le j\le s$ defineEach successive pair has an intersection of dimension $k-1$. Thus $A_0=A$ and $A_s=B$ are joined by a path of length $s$.

For the lower bound, if two vertices $U,V$ are adjacent, their common vector subspace $D$ has codimension one in each. Both $\dim (U\cap B)$ and $\dim (V\cap B)$ lie between $\dim (D\cap B)$ and $\dim (D\cap B)+1$. Thus one edge can change the dimension of the intersection with $B$ by at most one. It must change from $r$ to $k$ along a path from $A$ to $B$, requiring at least $k-r$ edges. Therefore

If two ordered pairs have equal distance, their intersections have equal dimension. Construct adapted bases for each pair as above and extend both to bases of the ambient vector space. The invertible linear map between them sends the first ordered pair to the second. It preserves intersection dimensions, so it is a graph automorphism. This proves that the Grassmann graph is distance-transitive, with the required simultaneous images. Its diameter is $k$, because $2k\le n$ permits two disjoint $k$-dimensional vector subspaces.

A finite two-transitive group action is transitive on ordered pairs of distinct points. Equivalently, the action is transitive and a point stabilizer subgroup is transitive on the remaining points. Its rank of a transitive permutation group is two, and the orbit-stabilizer theorem gives $n(n-1)\mid |G|$ for degree $n$. Such an action is primitive: a block containing a point and another point must, by the transitivity of the stabilizer subgroup, contain every point.

These groups illustrate two quite different sources of symmetry. The structural theorem for finite two-transitive groups says that they have either a regular elementary-abelian normal subgroup, giving the affine case, or a nonabelian simple normal subgroup, giving the almost simple case. The full classification of the second case depends on the classification of finite simple groups; the useful point here is to describe the constructions, actions, and representative families rather than treat two-transitivity as an abstract property of a group without a specified action.

In the affine case let $V$ be a finite vector space over ${\mathbb{F}}_p$ and let $H\le GL(V)$ act transitively on $V\setminus \{0\}$. The affine two-transitive groupis transitive by translations, and the stabilizer subgroup of zero is exactly $H$, so it is two-transitive. Conversely, if a two-transitive group has an abelian minimal normal subgroup $V$, normal-subgroup orbits and primitivity make $V$ transitive. An abelian transitive faithful permutation subgroup is regular: an element fixing one point commutes with every element moving that point and hence fixes them all. Minimal normality makes $V$ elementary abelian, since its prime-primary subgroups and its subgroup of pth powers are characteristic. Conjugation by the stabilizer subgroup acts faithfully as linear automorphisms of $V$, and regularity gives $G=V⋊G_0$. This recovers precisely the construction above.

Many examples follow at once. The group $AG{L}_d(q)$ acts on the $q^d$ vectors of ${\mathbb{F}}_q^d$; $G{L}_d(q)$ sends any nonzero vector to any other by a basis map. For $d\ge 2$, the subgroup $AS{L}_d(q)$ also gives a two-transitive action, since the determinant of a basis map can be adjusted on a second basis vector while retaining the prescribed first vector. The affine group with linear part $S{p}_{2m}(q)$ gives another family: the nonzero-vector transitivity was proved using symplectic transvections in part 4(ii). Here the additive group of ${\mathbb{F}}_q^d$, for $q=p^f$, is an elementary abelian p-group of rank $fd$.

In particular $AG{L}_1(q)$ consists of $x\mapsto ax+b$ with $a\ne 0$. Given distinct $x,y$ and distinct $u,v$, the equations $ax+b=u$, $ay+b=v$ have the unique solution $a=(v-u)/(y-x)$, $b=u-ax$. Thus this action is sharply two-transitive, of degree $q$ and order $q(q-1)$. Adding field automorphisms gives $A\Gamma L_1(q)$, still two-transitive, with elements $x\mapsto a{x}^{p^i}+b$.

An almost simple group lies between a nonabelian simple group $T$ and its automorphism group. One must still specify an action: a group can have several permutation representations with different degrees or different transitivity properties. The natural actions of $S_n$ for $n\ge 2$ and $A_n$ for $n\ge 4$ are the most familiar examples of a two-transitive group action. For $n\ge 5$, the simple group $A_n$, a normal subgroup, places these groups in the almost simple case; the smaller two-transitive examples belong to the affine case. The symmetric group sends any ordered distinct pair to any other; for the alternating group one can adjust an odd map by swapping two letters outside the target pair. Thus both are two-transitive in the indicated ranges.

The projective linear examples are just as concrete. The groups $PS{L}_n(q)$ and $PG{L}_n(q)$ act two-transitively onpoints, by the basis and determinant adjustment argument of part 4(i). Their field-automorphism extensions are also two-transitive. On the projective line, $PG{L}_2(q)$ is sharply three-transitive: three distinct points determine a unique Möbius transformation. Its degree is $q+1$ and its order is $q(q^2-1)$. For even $q$, $PS{L}_2(q)=PG{L}_2(q)$; for odd $q$ the special projective group has half that order and remains two-transitive, but cannot be three-transitive. In projective dimension at least two, triples of distinct points may be collinear or noncollinear, so the natural action is not three-transitive.

The projective special unitary group over a finite field $PS{U}_3(q)$, for $q\ge 3$, acts on the isotropic points of a space with a nondegenerate Hermitian form of dimension three over ${\mathbb{F}}_{q^2}$. Its degree and order areFor two distinct isotropic lines the Hermitian form pairing is nonzero, since a three-dimensional nondegenerate Hermitian space has maximal totally isotropic dimension one. Scale their representatives to form a hyperbolic pair and extend to a basis adapted to the Hermitian form. The corresponding isometry sends one ordered pair to another, and a norm-one scalar on the orthogonal one-dimensional complement adjusts its determinant without moving the two lines. Thus the special projective unitary action is two-transitive.

Two further families of simple groups have analogous rank-two actions:The Suzuki group of Lie type and the Ree group of type G2 have stabilizer subgroups containing subgroups acting regularly on all the other points in these actions; this gives two-transitivity. For example $Sz(8)$ has degree 65 and order 29120. These are exceptional Lie-type families, distinct from the preceding linear and unitary constructions.

The symplectic group over a finite field in characteristic two has a particularly illuminating pair of two-transitive actions which are not their projective-vector actions. For $m\ge 3$, $S{p}_{2m}(2)$ acts on the two classes of quadratic forms whose polar form is the given alternating bilinear form, distinguished by the Arf invariant of a quadratic form. Their degrees areTransitivity on each class follows by choosing symplectic bases giving the standard quadratic-form normal forms. Fix a form $Q$. Every other form with the same polar form is $Q_a(x)=Q(x)+B(a,x)$, and has the same Arf invariant of a quadratic form exactly when $Q(a)=0$. The stabilizer subgroup of $Q$, an orthogonal group over a finite field, is transitive on its nonzero isotropic vectors: extend such a vector to a hyperbolic pair and map adapted quadratic-form bases. It is therefore transitive on the other forms of the same type. This proves two-transitivity of both actions. In particular $S{p}_6(2)$ has actions of degrees 28 and 36. The degree-six action of $S{p}_4(2)\cong S_6$ in part 4(ii) is the smaller-dimensional precursor.

Finally the Mathieu groups supply exceptional examples with even higher transitivity:In particular all five natural actions are two-transitive. Together these examples exhibit affine spaces, projective and Hermitian geometries, symplectic quadratic forms, and exceptional finite simple groups as different sources of the same pair symmetry.

The organizing distinction is affine versus almost simple; the action, its degree, and its stabilizer subgroup are essential data. The explicitly verified affine, symmetric, alternating, projective, unitary, and symplectic actions already give many infinite families, while the exceptional families show why no single matrix construction covers every example.

Use mostly-plus Lorentzian signature and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with antisymmetrized gamma products of unit weight. A Majorana spinor obeys a reality condition under charge conjugation:A Weyl spinor instead has definite chirality,A four-dimensional Majorana spinor has four real components, while a Weyl spinor has two complex components. Charge conjugation reverses chirality in this signature, so a nonzero spinor cannot satisfy both conditions simultaneously.

For the Dirac equation $(/\partial +m)\psi =0$, the derivative term reverses chirality because ${\gamma }_5$ anticommutes with every gamma matrix, whereas the mass term preserves it. Projecting a purely chiral spinor equation onto its two chiralities therefore gives both $/\partial \psi =0$ and $m\psi =0$. Thus a nonzero four-component Weyl spinor satisfying this uncoupled Dirac equation must be massless. A Majorana spinor contains conjugate opposite-chirality components and can have a nonzero mass. Its charge-conjugation reality condition identifies particle and antiparticle operators, so a Majorana particle is its own antiparticle. A Weyl field does not by itself impose that identification.

This does not forbid writing a Majorana mass term for a neutral two-component Weyl field: such a term couples the field to its conjugate, and the resulting four-component massive field is Majorana, not a purely chiral solution of the uncoupled Dirac equation.

For the Rarita-Schwinger field, set $\chi ={\gamma }^{\mu }{\psi }_{\mu }$ and $d={\partial }^{\mu }{\psi }_{\mu }$. The antisymmetric kinetic operator is invariant underbecause commuting derivatives contract to zero against ${\gamma }^{\mu \nu \rho }$. Expanding its gamma product givesGamma contraction gives ${\gamma }_{\mu }{E}^{\mu }=2(/\partial \chi -d)$, so the equation implies $d=/\partial \chi$ and $/\partial {\psi }^{\mu }={\partial }^{\mu }\chi$.

For a Fourier mode with momentum $k$, if $k^2\ne 0$ the latter identity gives $k^2{\psi }^{\mu }=k^{\mu }/k\chi$. The mode is proportional to $k^{\mu }$ and hence pure gauge. Every physical nonzero-momentum mode therefore has $k^2=0$. For a null mode, choose an index with $k^{\mu }\ne 0$. The same equation makes $\chi$ an element of the image of $/k$, allowing a gauge choice with $\chi =0$. The field equations then reduce toResidual gauge transformations obey $/\partial \eta =0$.

Take a null momentum in the third spatial direction. Residual gauge freedom and transversality remove the temporal and longitudinal vector components, leaving two transverse vector polarizations. Their helicities are $\pm 1$, while on-shell spinor helicities are $\pm 1/2$. The gamma-trace constraint removes the two combinations of total helicity $\pm 1/2$ and retains the aligned combinations of helicity $\pm 3/2$. Explicitly, in a chiral representation the transverse gamma contractions use ${\sigma }_1\pm i{\sigma }_2$, which annihilate the correspondingly aligned spin states and act nontrivially on the opposite states. Thus a Majorana Rarita-Schwinger field has two physical massless polarizations. The Majorana assumption is the one relevant to simple supergravity; a complex vector-spinor has these helicities for its particle and independent antiparticle modes.

A consistent free massive deformation is the algebraic termadded to a kinetic term $-\frac{1}{2}{\overline{\psi }}_{\mu }{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\psi }_{\rho }$. Its equation isFor $m\ne 0$, divergence and gamma contraction give, respectively,These imply the massive Rarita-Schwinger constraints $\chi =0$ and $d=0$, and the remaining equation is $(/\partial +m){\psi }_{\mu }=0$. Squaring gives $(\square -m^2){\psi }_{\mu }=0$ in the chosen signature. The gauge symmetry is lost. In the rest frame, transversality sets the temporal component to zero; three spatial vector components times two on-shell spinor states give six states, and gamma trace removes the two spin-one-half states. The result is four massive spin-$3/2$ states, with spin projections $\pm 3/2,\pm 1/2$. In supergravity the additional longitudinal states can be supplied by the super-Higgs mechanism.

Use the mostly-plus Clifford algebra and Riemann curvature tensor conventions of the preceding solutions, with $e=\det (e_{\mu }{}^a)$, ${\kappa }^2=8\pi G$, and a Majorana gravitino. An action for minimal four-dimensional supergravity in the exact transformation normalization given isThe kinetic coefficient is tied to the normalization of the supersymmetry transformations: if ${\Psi }_{\mu }=2{\psi }_{\mu }$ and $\eta =2ϵ$, this becomes the canonical coefficient $-1/2$, with $\delta e_{\mu }{}^a=(\kappa /2)\bar{\eta }{\gamma }^a{\Psi }_{\mu }$ and $\delta {\Psi }_{\mu }={\nabla }_{\mu }\eta /\kappa$. Mixing those two normalizations would spoil the cancellation.

To the retained order, use the torsion-free connection. More generally the 1.5-order formalism allows its variation to be omitted after imposing its own algebraic equation; the resulting contorsion is quadratic in fermions and first affects the displayed action at quartic order. The metric tensor variation induced by the vierbein is $\delta g_{\mu \nu }=4\kappa \overline{ϵ}{\gamma }_{(\mu }{\psi }_{\nu )}$, so the Einstein-Hilbert action varies asFor the fermionic action, varying both Majorana factors gives the same contribution after integration by parts and the Grassmann bilinear interchange rule. ThusThe vector-index Riemann curvature tensor contribution vanishes by the first Bianchi identity. The spin-index contribution is determined byFor the last equality, expand the antisymmetric three-gamma/two-gamma product: the totally antisymmetric Riemann curvature tensor terms vanish by Bianchi, and the surviving terms are the Ricci contraction minus half its trace. Hencewhich cancels the gravitational variation. Variations of vierbeins, gamma matrices and the connection in the fermionic kinetic term carry three dynamical fermions. They lie beyond this linear-fermion cancellation and are cancelled by the omitted four-fermion completion. Thus the action is invariant up to boundary terms at the order specified, rather than claiming that a truncated action is exactly invariant at every fermion order.