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A subset $U\subseteq {\mathbb{R}}^n$ is an open set when every $x\in U$ has some radius $r>0$ for which $B(x,r)\subseteq U$. If a continuous function takes only the values zero and one, continuity at $x$ with tolerance $1/2$ gives a neighborhood on which $|f(y)-f(x)|<1/2$. The two possible values differ by one, so $f(y)=f(x)$ throughout that neighborhood. Thus $f$ is locally constant, and the zero linear map is its derivative, since the difference $f(x+h)-f(x)$ is exactly zero for all sufficiently small $h$. Hence $D{f}_x=0$ everywhere.

For the bound on $g$, parametrize the segment by $\gamma (t)=b+t(a-b)$, $0\le t\le 1$. The chain rule givesby the definition of operator norm. The mean value theorem applied to the real function $g\circ \gamma$ provesIf the endpoints agree the estimate is immediate. No continuity of the derivative is required.

For the finite-line complement, each removed line ${\ell }_i$ is closed, so $V$ is open. Because $a,b\notin {\ell }_i$, there is a unique affine plane $P_i$ containing $a$ and ${\ell }_i$, and a unique affine plane $Q_i$ containing $b$ and ${\ell }_i$. Choose $c$ outside the union of these finitely many planes, using the permitted fact about finite plane covers. In particular $c\in V$. If $[a,c]$ met ${\ell }_i$ at a point $d$, then $d\ne a$, and the line through $a,d$ lies in $P_i$. Since $c$ is on that same line, it would belong to $P_i$, a contradiction. The same argument using $Q_i$ applies to $[c,b]$. Thus $a$ and $b$ can be joined in $V$ by these two segments, proving polygonal connectivity of a finite-line complement.

Finally a continuous $f:V\to \{0,1\}$ has zero derivative by the first argument. Apply the proved segment estimate with $M=0$ separately on $[a,c]$ and $[c,b]$. It gives $f(a)=f(c)=f(b)$. Since $a,b$ were arbitrary, $f$ is constant on $V$.

A function is Fréchet differentiable at $x$ when there is a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThis linear map, which is unique, is the derivative $Df(x)$.

For the chain rule, if $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$, then $g\circ f$ is differentiable at $x$ and $D(g\circ f)(x)=Dg(f(x))Df(x)$. Indeed write $f(x+h)-f(x)=Ah+r(h)=v(h)$ and $g(f(x)+v)=g(f(x))+Bv+s(v)$, where $r(h)=o(|h|)$, $s(v)=o(|v|)$. Since $v(h)=O(|h|)$, substitution giveswhich proves the assertion, including paths on which $v(h)=0$.

For $g_1$, write $q=x^2-y^2$. Away from the two diagonal lines, $q\ne 0$ and the formula is a composition of smooth functions, hence differentiable. At the origin, $|g_1(x,y)|\le |x^2-y^2|\le x^2+y^2$, so its ratio to the displacement length tends to zero. Its derivative there is therefore zero: this is quadratic damping restores differentiability at the origin.

At a nonzero point $(a,b)$ with $a=\pm b$, necessarily $a\ne 0$. Along the $x$ direction,The denominator inside the sine is locally an invertible function of $t$, with nonzero derivative $2a$. Choose sequences tending to zero for which its reciprocal equals $2\pi n+\pi /2$ and $2\pi n+3\pi /2$. The quotients tend to $2a$ and $-2a$. Thus even this directional derivative fails to exist, and the function is not differentiable there. ConsequentlyFor $g_2$, the formula is smooth away from zero. At zero, $|g_2(x,y)|\le x^2+y^2$, so the same remainder estimate gives zero derivative. HenceNo limit of the sine factor itself is needed for either derivative at the origin.

Two real symmetric bilinear forms are congruent if an invertible linear map takes one to the other; in matrices, $B=S^{\mathsf{T}}AS$ for some invertible $S$. The spectral theorem for real symmetric matrices gives an orthogonal eigenbasis. Rescaling each basis vector with nonzero eigenvalue then reduces any such form toSylvester's law of inertia asserts that $(p,q,r)$ is independent of the congruence reduction. To prove it, $p$ is the greatest dimension of a subspace on which the form is positive definite: the positive coordinate subspace attains it, whereas any subspace of dimension greater than $p$ meets the nonpositive coordinate subspace of dimension $n-p$ and therefore cannot be positive definite. Applying the same argument to the negative of the form determines $q$. Then $r=n-p-q$ is determined too.

The rank of a quadratic form is $p+q$, and its signature of a quadratic form is $p-q$ under the signed-difference convention. They determine $p,q,r$ and hence the congruence class. Counting all pairs $p,q\ge 0$ with $p+q\le n$ givesFor $n\ge 1$, exactly two classes have every level set bounded: the positive-definite and negative-definite classes. Their nonempty nonzero levels are ellipsoids, and their zero level is the origin. A degenerate form has an entire null line in its zero level. An indefinite nondegenerate form also has a nonzero isotropic vector, for example the sum of one positive and one negative normalized basis vector, whose scalar multiples make the zero level unbounded.

For the displayed three-dimensional form,Its leading principal minors are $3,14,32$, all positive. Sylvester's criterion makes it positive definite, so the surface is a bounded ellipsoid. Directly, its quadratic expression iswhich also proves positivity and boundedness of the unit level.

For finite-dimensional Complex projective space ${\mathbb{CP}}^r$, use the filtration by its linear projective subspaces. The difference ${\mathbb{CP}}^j\setminus {\mathbb{CP}}^{j-1}$ is ${\mathbb{C}}^j$, so there is one topological cell in every even dimension $0,2,\dots ,2r$, and none in odd dimensions. Every cellular boundary vanishes for degree reasons. The integral cohomology is therefore $\mathbb{Z}$ in those even degrees and zero elsewhere.

For the cohomology ring, let $x\in H^2({\mathbb{CP}}^r;\mathbb{Z})$ be the Poincare dual of a projective hyperplane, using the complex orientation. Generic hyperplanes meet transversely, and a collection of $j$ of them intersects in ${\mathbb{CP}}^{r-j}$. The geometric interpretation of the cup product identifies $x^j$ with the dual of that intersection. Restricting to a linear ${\mathbb{CP}}^j$, the $j$ hyperplanes meet in one point, with positive sign because the transverse normal directions are complex lines. HenceThus $x^j$ is a primitive generator in degree $2j$, not merely a nonzero multiple. There are no classes beyond degree $2r$, soThe point case $r=0$ has ring $\mathbb{Z}$. Taking the union of the finite projective spaces gives $H^{\ast }({\mathbb{CP}}^{\infty };\mathbb{Z})=\mathbb{Z}[x]$, since each fixed cohomological degree stabilizes. This proves the cohomology ring of complex projective space computation.

For the rank bound for a nonsingular complex bilinear map, write $a={\dim }_{\mathbb{C}}U$, $b={\dim }_{\mathbb{C}}V$, $R=\mathrm{im}\varphi$, and $s={\dim }_{\mathbb{C}}R$. Assume $a,b\ge 1$. Injectivity with either nonzero factor fixed implies $\varphi (u\otimes v)\ne 0$ whenever $u,v\ne 0$. Thereforeis well defined and continuous. Scaling either vector only scales the output, and in local projective charts the formula is given by bilinear coordinates with no simultaneous zero. Let $h$ be the target's degree-two generator and $x,y$ the pullbacks of the two factor generators. The Künneth theorem givesFixing $[v]$ or $[u]$ makes $F$ a projective linear embedding induced by an injective linear map. Such an embedding pulls back the hyperplane class to the factor's hyperplane class. Since the two restrictions determine $H^2$ of the product, $F^{\ast }h=x+y$, with the generator of a point factor interpreted as zero.

Put $d=a+b-2$. In the source cohomology ring, every term in $(x+y{)}^d$ vanishes except the one with exponents $a-1,b-1$, andThe binomial coefficient is a positive integer and the final monomial generates the top integral cohomology. If $s\le d$, the class $h^d$ in ${\mathbb{CP}}^{s-1}$ would be zero. This contradiction provesWhen $a=b=1$, the nonzero pure-tensor image already gives $s\ge 1$. A positivity convention is genuinely necessary: if $U=0$, $V={\mathbb{C}}^2$, and $W=0$, all restrictions to the zero tensor subspaces are injective, but the literal bound would read $0\ge 1$. The displayed theorem is therefore the nonzero-factor version. If a nonzero factor is infinite-dimensional, either factor restriction already bounds the image dimension below by that factor dimension; the sum of two positive cardinal dimensions, with at least one infinite, is their maximum, giving the corresponding cardinal bound.

A smooth differential form of degree $p$ is a smooth section of ${\Lambda }^p{T}^{\ast }M$, soIt is equivalently a smooth alternating covariant $p$-tensor field. In a coordinate chart $x^1,\dots ,x^n$ it has a unique expressionwith smooth coefficients. At $p=0$ these are the smooth functions, and for $p>n$ the space is zero.

Define the exterior derivative in a chart byTo check that it is well-defined, let $y^1,\dots ,y^n$ be other smooth coordinates. The expression $d{y}^a={\sum }_i({\partial }_i{y}^a)d{x}^i$ satisfiesbecause the coefficients are symmetric and the wedge products antisymmetric. The coordinate formula also satisfies the graded Leibniz rule, and for a function $b$ the chain rule gives $d_{x}b={\sum }_a(\partial b/\partial y^a)d{y}^a=d_{y}b$. Applying these two facts to $\omega ={\sum }_I{b}_{I}d{y}^I$ yields $d_x\omega ={\sum }_I{d}_y{b}_I\wedge d{y}^I=d_y\omega$. Thus the local operators agree on overlaps and define a global linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$.

A second application in any chart givesagain by symmetry of mixed derivatives and antisymmetry of the wedge product of differential forms. Consequently every exact differential form is closed. Put ${\Omega }^{-1}(M)=0$ and define the de Rham cohomology byThe identity $d^2=0$ is exactly what puts the denominator inside the numerator, making this quotient meaningful. Since ${\Omega }^p(M)=0$ for $p>\dim M$, $H_{\mathrm{dR}}^p(M)=0$ in those degrees.

A left-invariant vector field $X$ on a Lie group satisfiesor equivalently $(L_g{)}_{\ast }X=X$ for every $g$. For $\xi \in T_{e}G$, defineThis is smooth because multiplication is smooth and its differential depends smoothly on the base points. The identity $L_g\circ L_h=L_{gh}$ gives $(d{L}_g{)}_h{X}_h^{\xi }=X_{gh}^{\xi }$, so $X^{\xi }$ is a left-invariant vector field. Conversely, left invariance forces any such field to have this form with $\xi =X_e$. Evaluation at $e$ and the construction above are inverse linear maps. HenceThe Lie bracket of vector fields is their commutator as derivations on smooth functions: $[X,Y]f=X(Yf)-Y(Xf)$. It is another vector field because the cross terms cancel in the Leibniz rule for its action on a product. The bracket is preserved by a diffeomorphism: conjugating the derivation operators by pullback of a smooth function preserves their commutator. Thereforefor left-invariant $X,Y$. These fields are consequently closed under the bracket.

Transport the bracket to the tangent space byBilinearity and antisymmetry come from the commutator. The Jacobi identity follows by expanding the three nested commutators of derivation operators: their six terms cancel in pairs. Thus $T_{e}G$ is a Lie algebra, and the displayed correspondence is a Lie algebra isomorphism onto the algebra of left-invariant vector fields.

Let $D=\dim {\mathcal{H}}_A$ and introduce an isomorphic reference system $R$. Choose a basis $\{|j⟩{\}}_{j=1}^D$ and the maximally entangled stateComplete positivity makesa positive density operator. This is the normalized Choi matrix, with the reference placed first. Its spectral decomposition can be writtenThe number $r$ is at most $D^2$.

Here an index state is a chosen basis vector $|j{⟩}_R$ in the reference. The corresponding relative state of $|v_{\ell }⟩$ is the generally unnormalized vectorThus $|v_{\ell }⟩={\sum }_j|j{⟩}_R|{\eta }_{\ell j}{⟩}_A$. Define a linear map $A_{\ell }$ by its columns:Then $(I_R\otimes A_{\ell })|\Omega ⟩=\sqrt{{\lambda }_{\ell }}|v_{\ell }⟩$, soExpand both sides in reference matrix units. The block with reference indices $i,j$ on the left is $D^{-1}\Phi (|i⟩⟨j|)$; on the right it is $D^{-1}{\sum }_{\ell }{A}_{\ell }|i⟩⟨j|A_{\ell }^{†}$. Since matrix units span all operators and $\Phi$ is linear,This constructs the Kraus representation explicitly from the relative states, rather than assuming it.

Finally, trace preservation implies for every density operator $\rho$ thatTaking every rank-one orthogonal projection as $\rho$ forces the Hermitian operator in brackets to have the same quadratic form as the identity. ThereforeThe argument is finite-dimensional, as appropriate to the quantum-information setting; a channel with a different output dimension has the same construction with rectangular Kraus operators.

The printed statement needs $A\ne \varnothing$: for the empty set every linear subspace has cardinality at least one, so the requested bound is impossible. We prove the intended nonempty version of the Freiman-Ruzsa theorem over a finite field. Necessarily $C\ge 1$, because any one translate of $A$ inside $A+A$ has size $|A|$.

We first prove all sumset estimates needed below. For finite nonempty $X$ and finite $U,V$ in an abelian group, choose one representation $z=u_z-v_z$ for each $z\in U-V$. The mapis injective: subtracting the outputs recovers $z$, hence its chosen representation and then $x$. This proves the Ruzsa triangle inequality

Choose nonempty $X\subseteq B$ minimizing $c^{\prime }=|X+B|/|X|\le C$, where $B$ is any set with $|B+B|\le C|B|$. We prove the Petridis minimal-growth lemma $|X+B+D|\le c^{\prime }|X+D|$ for every finite $D$. List $D=\{d_1,\dots ,d_s\}$, and let $X_i$ consist of those $x\in X$ for which $x+d_i$ was not already in $X+\{d_1,\dots ,d_{i-1}\}$. The disjoint pieces $X_i+d_i$ partition $X+D$.

Every $(X\setminus X_i)+B+d_i$ was already covered in $X+B+\{d_1,\dots ,d_{i-1}\}$. The new part at step $i$ therefore has size at mostHere minimality of $X$ applies to every nonempty subset $X\setminus X_i$, and the empty case has size zero. Summing proves the lemma. Iteration gives $|X+kB|\le C^k|X|$. Apply the Ruzsa triangle inequality with $U=kB$, $V=\ell B$ to obtainThis derives the required Plünnecke-Ruzsa inequality. In characteristic two, subtraction is addition, so $|rB|\le C^r|B|$ for every positive integer $r$.

Fix $a\in A$ and translate to $B=A+a$, which contains zero, has size $|A|$, and has the same doubling constant. Choose $m=\lceil {\log }_2(2|12B|)\rceil$. A uniformly random linear map $L:{\mathbb{F}}_2^N\to G^{\prime }={\mathbb{F}}_2^m$ kills each specified nonzero vector with probability $2^{-m}$. The union bound shows that some $L$ satisfiesBecause $0\in B$, all lower iterated sumsets lie in $12B$. In particular $L$ is injective on $B$, since the difference of two elements belongs to $2B$. For $B^{\prime }=L(B)$, the density $\alpha =|B^{\prime }|/|G^{\prime }|$ is at least $1/(4C^{12})$. This proves injective linear modelling of a small sumset.

We next prove the Finite-field Bogolyubov lemma in this model. Write $f=1_{B^{\prime }}$ and use normalized real Fourier coefficients on a finite abelian group $\hat{f}(\xi )={\mathbb{E}}_{x}f(x)(-1{)}^{\xi \cdot x}$. LetBy Parseval identity, $|S|\le 2{\alpha }^{-2}$. Thus $W$ is a vector subspace of codimension at most $D=\lceil 32C^{24}\rceil$.

For $w\in W$, Fourier inversion of the fourfold normalized convolution givesAll terms with $\xi \in S$ are nonnegative, and the zero-frequency term is ${\alpha }^4$. The absolute contribution from the remaining frequencies is at mostHence the convolution is positive and $W\subseteq 4B^{\prime }$. Also $|W|\ge 2^{-D}|G^{\prime }|\ge 2^{-D}|B|$.

For each $w\in W$, choose its lift $\sigma (w)\in 4B$. It is unique because two lifts differ by an element of $8B\cap \ker L$, which is zero. For $w_1,w_2\in W$, the discrepancybelongs to $12B\cap \ker L$, so it vanishes. Also $\sigma (0)=0$. Therefore $\sigma$ is a linear map and its image $\tilde{W}\subseteq 4B$ is a vector subspace with $|\tilde{W}|=|W|\ge 2^{-D}|B|$. This establishes lifting a subspace through a Freiman model without assuming that arbitrary choices of preimages preserve addition.

Finally, $B+\tilde{W}\subseteq 5B$ is a union of $\tilde{W}$-cosets. The number of these cosets is at mostChoose representatives $b_1,\dots ,b_t$ from $B$. Then $B\subseteq \tilde{W}+\{b_1,\dots ,b_t\}$, so adjoining these representatives and $a$ to $\tilde{W}$ gives a vector subspace $V$ containing $A$. Over ${\mathbb{F}}_2$, adjoining each vector multiplies the size by at most two. Since $|\tilde{W}|\le |4B|\le C^4|B|$, the Freiman-Ruzsa bound from Bogolyubov and linear modelling yieldsThis bound depends only on $C$, as required, and every sumset estimate used in its derivation has been proved.

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:

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.

True. In the norm topology on the unit ball, the set $\{x\in B:\Vert x\Vert <1/2\}$ is a neighbourhood of zero. If the relative weak topology agrees with it, there are finitely many continuous linear functionals $f_1,\dots ,f_m$ and positive ${ϵ}_j$ such thatIf ${\bigcap }_j\ker f_j$ contained a nonzero vector, rescaling it to have norm $3/4$ would put it in the left-hand set but not the right-hand set. Therefore ${\bigcap }_j\ker f_j=\{0\}$. The linear mapis injective into a finite-dimensional scalar space, soThis proof works for either the open or closed unit-ball convention and does not require completeness.

Let $v=T(0)$ and $B(x)=T(x)-v$. The Euclidean isometry $B$ fixes the origin and preserves norms. The polarization identity therefore gives $⟨Bx,By⟩=⟨x,y⟩$. In particular, $B{e}_1,B{e}_2$ are an orthonormal basis. The coefficients of $Bx$ in this basis are $⟨Bx,B{e}_j⟩=x_j$, so $Bx=x_{1}B{e}_1+x_{2}B{e}_2$. Thus $B$ is a linear map represented by an orthogonal matrix, and $T(x)=Bx+v$. Conversely, every such map preserves distances because $\Vert B(x-y)\Vert =\Vert x-y\Vert$.

When $\det B=-1$, the real eigenvalues of the two-dimensional orthogonal matrix are $1,-1$. Choose corresponding unit eigenvectors $e,n$, and write $v=ae+bn$. In these coordinates $T(s,t)=(s+a,-t+b)$. This is reflection in the line $t=b/2$, followed by translation $ae$ along that line. It is a Euclidean reflection if $a=0$, and a glide reflection if $a\ne 0$.

The vector triple product identity gives $n\times (x\times n)=x-(n\cdot x)n$. Decompose $x=x_{\perp }+x_{\parallel }$, where $x_{\parallel }=(n\cdot x)n$ and $x_{\perp }\cdot n=0$. ThenThe linear map acts as the identity on $n^{\perp }$ and as multiplication by $\alpha$ along $n$. Consequently it is invertible exactly when $\alpha \ne 0$, withApplying $\Phi$ to this expression recovers the perpendicular and parallel components of $y$ separately.

For a binary linear code $C\le {\mathbb{F}}_2^n$, its rank is its dimension of a vector space $r$. Use the weight enumerator conventionwhere $w(c)$ is the Hamming weight. This convention is essential here: $W_C(1,0)$ counts words of weight $n$, rather than zero words. There are $2^r$ codewords, so $\boxed{W_C(1,1)=2^r}$.