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The special linear group over a finite field $S{L}_n(q)$ consists of the determinant-one invertible matrices over ${\mathbb{F}}_q$. Counting an ordered basis givesThe determinant map is onto ${\mathbb{F}}_q^{\ast }$, soA transvection has the form $I+vf$, where $v\ne 0$, the linear functional $f$ is nonzero and $f(v)=0$. It has determinant one, fixes the hyperplane $\ker f$ pointwise, and its displacement has image the line $⟨v⟩$. In particular $E_{ij}(t)=I+t{e}_{ij}$ for $i\ne j$ is an elementary transvection when $t\ne 0$.

Row addition is multiplication by such a matrix. Gaussian elimination reduces a determinant-one matrix to diagonal form using row additions and determinant-one pivot exchanges. The latter, and all remaining diagonal factors, are themselves products of transvections: within two coordinates,A diagonal matrix with product of entries one is a product of these two-coordinate diagonal matrices. Thus transvections generate $S{L}_n(q)$. This generation statement actually holds for every $n\ge 2$ and every finite field; the exclusions in the question are needed for the subsequent simplicity assertion.

Commuting with all elementary matrices forces the group center to consist of scalar matrices $\lambda I$ with ${\lambda }^n=1$, so its order is $d=\gcd (n,q-1)$. The projective quotient acts faithfully on the one-dimensional subspaces. Its action is two-transitive: send an ordered pair of independent representative vectors to any other pair and adjust the determinant by rescaling a representative without changing its line. Therefore it is primitive.

The Iwasawa simplicity lemma says that in a faithful primitive action, if a point stabilizer has an abelian normal subgroup whose conjugates generate the group, every nontrivial normal subgroup contains the derived subgroup. In particular a nontrivial perfect group satisfying these conditions is simple. No proof of the lemma is required here.

For the point $\ell =⟨v⟩$, the subgroupis abelian, is normal in the line stabilizer, and its conjugates contain all transvections. Its image in the projective quotient has the same properties. Perfectness follows explicitly from elementary commutators. If $n\ge 3$, choose a third index $r$ to obtainIf $n=2,q>3$, choose $t$ with $t^2\ne 1$ and set $h(t)=\mathrm{diag}(t,t^{-1})$. ThenEvery upper transvection is a commutator, and conjugation gives the lower ones. The group and its projective quotient are perfect. Iwasawa's simplicity lemma now proves

The order of $PS{L}_2(4)$ is $4(4^2-1)=60$. Its faithful projective-line action has degree five, so its image is an index-two subgroup of $S_5$, necessarily $A_5$. Thus $PS{L}_2(4)\cong A_5$.

For $PS{L}_2(5)$ the order is $5(5^2-1)/2=60$. We construct a faithful degree-five action. In $S{L}_2(5)$ takeThey satisfy $i^2=j^2=-I$ and $ij=-ji$. Their quaternion subgroup projects to a Sylow 2-subgroup $V\cong C_2^2$. Its projective centralizer is $V$: a lift of a centralizing element conjugates each of $i,j$ to itself or its negative. In the matrix basis $I,i,j,ij$, those four sign choices each leave a one-dimensional space, and determinant one leaves just its two quaternion representatives. The matrixhas order three and cyclically permutes the three nonidentity elements of $V$. The normalizer quotient embeds in $S_3$ and has order dividing both $6$ and $60/4=15$, so it has order exactly three. Hence $|N(V)|=12$ and there are five Sylow 2-subgroups. Their conjugation action is nontrivial and transitive, and simplicity makes it faithful. Its image again has order $60$ in $S_5$, proving

Finally,In characteristic two an involutory matrix has form $I+N$ with $N^2=0$. Its Jordan blocks have size at most two. A projective involution in $PS{L}_3(4)$ has a unique involutory lift to $S{L}_3(4)$: its square is central, and squaring is an automorphism of the scalar center of order three, allowing a unique scalar adjustment. The nonidentity Jordan form in dimension three is only $(2,1)$. This GL-class remains a single SL-class, since the centralizer contains $\mathrm{diag}(I_2,c)$ of every determinant $c\in {\mathbb{F}}_4^{\ast }$. Thus $PS{L}_3(4)$ has one involution class.

In $PS{L}_4(2)=S{L}_4(2)=G{L}_4(2)$ there are two forms, $(2,1,1)$ and $(2,2)$, so there are two involution classes. More concretely, the first group's single class has $21\cdot 5\cdot 3=315$ elements, counting image line, kernel plane through it and nonzero map between the quotient and image. The rank-one class in dimension four over ${\mathbb{F}}_2$ has $15\cdot 7=105$ elements; the rank-two class has $35\cdot |G{L}_2(2)|=210$. Isomorphisms preserve conjugacy classes, so

A transitive permutation group is primitive when it preserves no nontrivial partition into blocks. A point stabilizer is maximal precisely when the action is primitive. The orbits of any normal subgroup form a block system, so every nontrivial normal subgroup of a faithful primitive group is transitive. This turns the classification of primitive groups into a study of their minimal normal subgroups and their group socle.

The socle of a finite group is generated by its minimal nontrivial normal subgroups. A minimal normal subgroup is characteristically simple, hence a direct power of a simple group. Distinct minimal normal subgroups commute. If a transitive normal subgroup is abelian it is regular: its point stabilizer, being the same at all points, lies in the group homomorphism kernel. In that case it is elementary abelian, and the ambient group is affine. If two minimal normal subgroups occur, each centralizes the other; the centralizer of a transitive permutation group is semiregular, so both are regular and the two regular groups are mutual left/right partners. These observations are central to the O'Nan–Scott theorem.

The theorem organizes finite primitive groups into the following socle/action types. In the affine type HA, the group socle is $C_p^d$, degree $p^d$, and $G=C_p^d⋊H$ with $H\le G{L}_d(p)$ irreducible. In almost-simple type AS the group socle is one nonabelian simple group $T$ and $T\le G\le \mathrm{Aut}(T)$; a maximal subgroup not containing $T$ specifies the primitive coset action.

In simple-diagonal type SD the group socle is $T^r$, $r\ge 2$, and its intersection with a point stabilizer is a full diagonal copy of $T$. The degree is $|T{|}^{r-1}$, and the action includes the appropriate permutation of the simple factors. Compound-diagonal type CD replaces this one diagonal by a product of diagonals in disjoint collections of factors; it is a product version of diagonal type. In product-action type PA one embeds in a wreath product with a primitive nonregular almost-simple base, acting on a Cartesian power ${\Delta }^r$, so the degree is $|\Delta {|}^r$ and the group socle is a corresponding power of $T$.

Types HS and HC have two regular minimal normal subgroups, each isomorphic respectively to $T$ or to $T^r$ with $r>1$. Their socle consists of the left and right regular groups; the degrees are $|T|$ and $|T{|}^r$. Finally TW, twisted-wreath type, has a unique regular nonabelian minimal normal subgroup $T^r$ and a point stabilizer acting through a twisted permutation of its factors. This has degree $|T{|}^r$; primitivity imposes restrictive conditions, in particular $r\ge 6$. These eight modern labels refine the older affine/almost-simple/diagonal/product description. The theorem reduces the problem to simple groups, their maximal subgroups, irreducible linear actions and the specified product constructions; it does not make every primitive group maximal in a symmetric group.

For maximal subgroups of $S_n$ there are first two elementary families. An intransitive maximal subgroup is the stabilizer $S_a\times S_{n-a}$ of an unordered distinguished subset, with $1\le a<n/2$; replacing the subset by its complement gives the same class. When $a=n/2$, this subgroup sits properly inside the stabilizer of the two equal blocks. A transitive imprimitive maximal subgroup is a full block-system stabilizer $S_a\wr S_b$, with $ab=n$, $a,b>1$, in its imprimitive action.

The alternating group $A_n$ is itself maximal. A remaining primitive maximum must contain odd permutations. The primitive part of O'Nan–Scott places it among full affine, diagonal, product-action, or almost-simple overgroups. Full affine groups have degree $p^d$; full diagonal groups have degree $|T{|}^{r-1}$ and socle $T^r$; full product-action wreath groups have degree $d^r$. Almost-simple cases require the actual simple-group permutation degree and its full permutation normalizer. Regular nonabelian types are contained in the corresponding holomorph/diagonal or product overgroups, so do not supply extra maximal families. Parity and exceptional containments must be checked before calling any of these candidates maximal.

In degree sixteen the complete list, up to $S_{16}$-conjugacy, isThe three wreath products here act on blocks, not Cartesian coordinates. The other proper primitive groups of this degree are affine and lie in $AG{L}_4(2)$. A nonzero translation of its sixteen points is eight transpositions, and a generating linear transvection is four transpositions, so the full affine group lies in $A_{16}$. Even the tempting product-action $S_4\wr S_2$ on $4^2$ points is contained in this affine group: identify each four-point coordinate with ${\mathbb{F}}_2^2$. Alternatively, each base transposition repeats four times and coordinate exchange makes six transpositions, both even. Thus none is an additional maximum of $S_{16}$. There are eleven conjugacy classes in the displayed list.

In degree sixty the complete list isThe notation $D(A_5,2)$ denotes the full diagonal normalizer, not just $A_5\times A_5$. It is generated on the set $A_5$ by $x\mapsto ax{b}^{-1}$, an outer automorphism induced by odd conjugation in $S_5$, and $x\mapsto x^{-1}$. Its order isIts two alternating factors form the group socle. The point stabilizer intersects them in the diagonal $A_5$, of index sixty. Odd conjugation by a transposition of $S_5$ fixes its six commuting elements in $A_5$, hence makes $(60-6)/2=27$ transpositions and is odd. Inversion fixes the identity and the fifteen involutions, so makes twenty-two transpositions and is even. This verifies that the full diagonal group is not lost inside $A_{60}$.

The almost-simple maximum $PG{L}_2(59)$ has order $59(59^2-1)=205320$ and its usual two-transitive projective-line action. Multiplication by a primitive element of ${\mathbb{F}}_{59}^{\ast }$ makes one 58-cycle and fixes zero and infinity, so it too contains odd permutations. The simple subgroup $PS{L}_2(59)$ alone is not maximal. Degree sixty is neither a prime power nor a proper power, excluding affine and product/compound types; its diagonal possibility is $T=A_5,r=2$. The almost-simple primitive possibilities, besides the natural alternating/symmetric groups, have socle $PS{L}_2(59)$. Their full normalizer is the displayed $PG{L}_2(59)$. These degree-specific classification facts supply the exhaustive primitive part of the list. Altogether there are forty-two conjugacy classes of maximal subgroups of $S_{60}$: twenty-nine intransitive, ten imprimitive, the alternating group, and two further primitive classes.

Suppose $x$ is a quadratic nonresidue. On the nonzero residue classes modulo $p$, define the involutionIt is an involution because $T(T(a))=a$. A fixed point would satisfy $a^2=x$, which is excluded by the quadratic nonresidue hypothesis. Thus it partitions the $p-1$ classes into $(p-1)/2$ disjoint pairs $\{a,x{a}^{-1}\}$, each with product $x$. ConsequentlyNow evaluate the same product by pairing every class with its own modular inverse. The only classes equal to their modular inverses solve $(a-1)(a+1)=0$ and hence are $1$ and $-1$. All other pairs have product one. The complete product is therefore $-1$, soTogether with the preceding part, this is Euler's criterion; the pairing argument establishes the converse without assuming it.

Let $N$ be a normal subgroup and let $\pi :G\to G/N$ be the quotient group homomorphism. By the first part, $G=⟨a,b⟩$ for involutions $a,b$, so the quotient group is generated by $\pi (a),\pi (b)$, whose squares are the identity. If both images are nonidentity, they are involutions and the second part applies. If only one image is nonidentity, it generates $C_2$; choose it twice. Both images cannot be trivial because the quotient group is nontrivial. Hence every nontrivial quotient is dihedral in the stated extended convention. This is the nontrivial-quotient closure of two-involution groups.

Let $a,b$ be the generating involutions. If $a\ne b$, set $x=ab$ and $y=b$. Then $x\ne e$, so its order of a group element is at least two or infinite, andTherefore $x,y$ generate the group and satisfy the required dihedral inversion relation. If $a=b$, the group is $C_2$ and $x=y=a$ gives its permitted degenerate description. This proves the two-involution characterization of a dihedral group without excluding either $C_2$ or the infinite dihedral group.

The two-involution characterization of a dihedral group uses the extended convention that includes $C_2$ and the infinite dihedral group; the two generators need not be distinct. Put $a=xy$ and $b=y$. The inversion relation givesIf $a\ne e$, both $a,b$ are involutions and generate the original group. If $a=e$, then $x=y$ and the group is $C_2$; choose the same nonidentity element twice. Thus two elements of order two generate the group, including the degenerate case allowed here. The distinction between square equal to the identity and order of a group element exactly two is essential.

Lagrange's theorem for finite groups states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$. For $x\in G$, two powers among $1,x,\dots ,x^{|G|}$ coincide, giving $x^m=1$ for some positive $m$. The least such exponent $d$ is the order of a group element, and the cyclic subgroup generated by $x$ consists of the $d$ distinct powers $1,x,\dots ,x^{d-1}$. Apply Lagrange's theorem for finite groups to obtain $d\mid |G|$.

Cauchy's theorem for finite groups states that if a prime $p$ divides the order of a finite group, the group contains an element of order $p$.

The classification of a group of order eight isHere $D_8$ denotes the dihedral group of order eight, not a group of order sixteen. To prove exhaustiveness, every element order is $1,2,4$ or $8$. An element of order eight generates the whole group, giving $C_8$.

If no element has order four or eight, every nonidentity element is an involution. Since $(ab{)}^{-1}=ab$ while $a^{-1}=a$ and $b^{-1}=b$, we have $ab=ba$. Choose an involution $a$, another $b$ outside $⟨a⟩$, and then $c$ outside the four-element subgroup $⟨a,b⟩$. The factors commute and each added factor has trivial intersection with the preceding subgroup. The internal direct product theorem gives $G\cong C_2^3$, an elementary abelian group.

Otherwise choose $a$ of order four and set $A=⟨a⟩$. It has index two, so is normal. Choose $b\notin A$. The two cosets show that $a,b$ generate $G$ and that $b^2\in A$. There is no element of order eight, so $b^2$ is either $1$ or $a^2$.

If $G$ is abelian and $b^2=1$, the internal direct product theorem yields $G=A\times ⟨b⟩\cong C_4\times C_2$. If instead $b^2=a^2$, let $c=a^{-1}b$; commutativity gives $c^2=1$ and $c\notin A$, producing the same direct product.

If $G$ is nonabelian, conjugation by $b$ is a nonidentity group automorphism of $A$: otherwise $a,b$ would commute and hence so would all of $G$. Thus $ba{b}^{-1}=a^{-1}$. With $b^2=1$ these are the relations of $D_8$; with $b^2=a^2$ they are the relations of the quaternion group $Q_8$. In either presentation every word reduces to $a^i{b}^j$, $0\le i<4$, $0\le j<2$. The standard eight-element model maps onto $G$, and equal orders make that group homomorphism an isomorphism.

Finally group isomorphisms preserve element orders. The following numbers of elements distinguish all five possibilities:

Thus the list is both exhaustive and pairwise non-isomorphic.

The ordinary Poincare duality statement here is for a compact manifold without boundary, of dimension $d$, oriented over the field $F$. Its fundamental class $[M{]}_F$ induces isomorphismsfor every $q$. For a manifold with boundary the appropriate statement is Poincare-Lefschetz duality with relative groups; the ordinary pairing need not be nondegenerate. Over a field, the universal coefficient theorem for cohomology identifies $H^{d-q}(M;F)$ with the full dual of $H_{d-q}(M;F)$. The cap-cup evaluation identity and duality therefore makea perfect pairing. Explicitly, a nonzero $a$ has a nonzero cap product, and a linear functional $b$ on its homology group takes a nonzero value on that product. The same argument in the other variable proves nonsingularity. On the whole graded cohomology, define $B(a,b)$ by taking the degree-$d$ part of $a⌣b$ before evaluation. For a nonzero component $a_q$ choose $b$ homogeneous of degree $d-q$ to pair nontrivially with it. All other components contribute zero in degree $d$. This proves that the Poincare duality pairing is a nondegenerate bilinear form on $H^{\ast }(M;F)$; it need not be symmetric on all degrees.

Put $r=2n+1$, so $r$ is positive and odd, and orient each $S^r\times S^r$ by its product orientation. The Künneth theorem gives integral cohomology $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^2$ in degree $r$, and zero elsewhere. Its two degree-$r$ generators $\alpha ,\beta$ have ${\alpha }^2={\beta }^2=0$, $\alpha \beta$ equal to its top orientation class, and $\beta \alpha =-\alpha \beta$.

For a connected sum of oriented manifolds, excision and the long exact sequence for deleting a ball show that deleting a ball removes the top homology class and leaves all lower positive homology groups unchanged. The boundary sphere represents zero in the punctured manifold: it is the boundary of its relative fundamental chain. In the Mayer–Vietoris sequence for the two punctured pieces joined along the separating sphere, the orientation class of the connected sum maps onto that sphere's class. The intermediate positive groups are consequently the direct sums of the groups of the two original manifolds. This also covers $r=1$, where the sphere is a circle and the boundary-class observation is necessary in the middle degree. Iterating and applying the universal coefficient theorem for cohomology givesHere $g\ge 1$; under the conventional extension $W_0=S^{2r}$ the same formula holds with a zero middle group.

Let $\omega$ be the top cohomological orientation class. The connected-sum pinch map to the wedge of the $g$ sphere products gives degree-one projections to each summand. Pull back the two factor classes from summand $i$ to obtain ${\alpha }_i,{\beta }_i$. Their product is $\omega$, since the projection has degree one. Classes from different wedge summands have zero positive-degree cup products, and graded commutativity of the cup product supplies the reversed sign. Thus the full cohomology ring of a connected sum of odd-dimensional sphere products is the graded free abelian group just displayed with multiplicationThe unit is $1$, and $\omega$ times any positive-degree class is zero by dimension. This describes all products, including ${\omega }^2=0$.

The smooth involution is a diffeomorphism, since it is its own inverse. Its fixed set is closed; discreteness and compactness therefore make it finite. The supplied positivity of $\det (I-D{f}_x)$ makes every fixed point nondegenerate with local index $+1$. The Lefschetz-Hopf fixed-point theorem then givesIndeed the degree-zero and top-degree traces are both one, because the manifold is connected and $f$ preserves its orientation; the middle degree is odd.

On $V=H^r(W_g;\mathbb{R})$, the form $\Omega (a,b)=⟨a⌣b,[W_g]⟩$ is a nondegenerate alternating bilinear form by Poincare duality and the oddness of $r$. Thus it is a symplectic vector space of dimension $2g$. Naturality and orientation preservation show that $T$ preserves $\Omega$, and $T^2=I$. For the eigenspaces of a symplectic involution, write $V=V_{+}\oplus V_{-}$: the polynomial $(t-1)(t+1)$ has distinct roots over $\mathbb{R}$. If $v_{+}\in V_{+}$ and $v_{-}\in V_{-}$, then $\Omega (v_{+},v_{-})=\Omega (T{v}_{+},T{v}_{-})=-\Omega (v_{+},v_{-})$, so the two vector subspaces are orthogonal. Each restricted form is nondegenerate, since a vector annihilating its own vector subspace also annihilates the other and hence all of $V$. Their dimensions are therefore even, say $\dim V_{+}=2a$, $\dim V_{-}=2b$, with $a+b=g$. ConsequentlyThis proves the fixed-point congruence for an involution on an odd-sphere connected sum:

Use the double covering map $\pi :S^n\to {\mathbb{RP}}^n$ and its deck involution, the antipodal map $a(x)=-x$. The given sphere cohomology vanishes outside degrees $0,n$, so only those degrees can survive as invariant classes. The degree-zero action is the identity because the sphere is connected for $n\ge 1$.

For the top degree, orient $S^n$ as the boundary of the unit ball. A positive tangent basis $(v_1,\dots ,v_n)$ at $x$ means that $(x,v_1,\dots ,v_n)$ has positive determinant in ${\mathbb{R}}^{n+1}$. Under the antipodal map every one of these $n+1$ vectors changes sign, so the orientation sign is $(-1{)}^{n+1}$. Equivalently, the sphere volume form isIts cohomology class is nonzero by the integral argument and therefore generates the one-dimensional top de Rham cohomology. Thus the top class is invariant precisely when $n$ is an odd integer. Applying the quotient result yieldsIn particular ${\mathbb{RP}}^1$ has one copy of $\mathbb{R}$ in each of degrees zero and one. These are real cohomology groups; integral torsion is not detected by de Rham cohomology.

Let $R$ interchange the two complete graphs, sending $v_i$ to $v_i^{\prime }$ and fixing $w$. This involution preserves the transition matrix of the simple random walk. A mirror coupling under an involution starts $X$ at $v_i$, keeps $Y_t=R(X_t)$ until the hitting time $\tau$ of $w$, and then makes the two walks move identically. Each coordinate has the correct transition probabilities, and they coalesce precisely at $\tau$. The coupling inequality for total variation gives

There is a genuine error in the printed expected hitting time bound, as well as an unspecified initial law. For $n\ge 2$, write $a={\mathbb{E}}_{v_1}\tau$ and $b={\mathbb{E}}_{v_i}\tau$ for $i\ne 1$. From every clique vertex there is a positive chance of reaching $w$ within two steps, uniformly over the finite state space for each fixed $n$, so $\tau$ has finite expected value. First-step analysis, using the degree of a vertex $n$ at $v_1$ and $n-1$ at the other clique vertices, givesSolving these linear equations yieldsIn particular even the attachment vertex exceeds the printed bound by one. The valid uniform replacement is ${\max }_x{\mathbb{E}}_x\tau \le n^2$, the hitting time of the bridge vertex between two cliques formula.

We must also handle arbitrary starting states, including $w$, before applying the mirror argument. Identify related vertices and consider the lumped Markov chain $Z$ on $\{1,\dots ,n,w\}$. Its transition matrix $Q$ has $Q(1,j)=1/n$ for $j\ne 1$, $Q(1,w)=1/n$, $Q(i,j)=1/(n-1)$ for $i\ne 1$ and $j\ne i$, and $Q(w,1)=1$, with all other entries zero. Here the row for $i\ne 1$ only ranges over $j\in \{1,\dots ,n\}$.

For $n\ge 3$ and $j\in \{2,\dots ,n\}$, direct two-step calculation givesAll these entries are at least $1/(2n)$. Thus $Q^2(z,\cdot )\ge \alpha \nu (\cdot )$ for every $z$, where $\nu$ is the uniform distribution on a finite set $\{2,\dots ,n\}$ and $\alpha =(n-1)/(2n)\ge 1/3$. This is a Doeblin condition. At each two-step block couple the two quotient endpoints to the same sample from $\nu$ with probability $\alpha$, using the residual probability distributions otherwise. The first successful block has expected value at most $1/\alpha$, so the alignment time $A$ has $\mathbb{E}A\le 6$. Lift each endpoint coupling of probability distributions to the original walks by their conditional two-step path laws; this preserves every marginal path law. The success test uses fresh block randomness and so permits restarting the walks with their original transition matrix at the aligned endpoints.

At alignment the original walks are equal or related. Use identical transitions in the first case and the mirror coupling under an involution in the second. The resulting coalescing coupling has coalescence time $T$ satisfying $\mathbb{E}T\le 6+n^2$ for every pair of initial states. Therefore Markov inequality and the pairwise mixing diameter implyFor the hint's particular starting states, the quotient one-step laws from two different clique indices have a common component of mass at least $1-2/n$: for two ordinary indices it is $(n-2)/(n-1)$; for one attachment index it is $(n-2)/n$. A maximal coupling therefore aligns those indices with probability $1-O(1/n)$, as suggested. The two-step argument above additionally covers $w$.

The nonlazy walk is an aperiodic Markov chain for $n\ge 3$, because the connected graph contains a triangle. For $n=1,2$ it is a periodic Markov chain on a bipartite graph; each bipartition class has stationary mass $1/2$, and the walk stays on a single class at each time. Hence $d(t)\ge 1/2$ and the $1/4$ mixing time is infinite. The asymptotic conclusion is valid for $n\ge 3$.

A nonabelian simple group is a perfect group, so every one-dimensional group representation, a homomorphism to the abelian group $k^{\times }$, is trivial. Thus a nontrivial irreducible Brauer character cannot have degree $1$.

Suppose instead that its representation has dimension $2$, working over a splitting extension if necessary. Its determinant is a one-dimensional representation and is therefore trivial. Its kernel is a normal subgroup, and the representation is nontrivial, so simplicity makes it faithful. Consequently $G$ embeds in ${\mathrm{SL}}_2(k)$.

We use the Feit–Thompson theorem: every finite group of odd order is solvable. Therefore this nonabelian simple group has even order, and Cauchy's theorem for finite groups supplies an involution $t$. In odd characteristic, its representing matrix $A$ satisfies $A^2=I$ and is diagonalizable with eigenvalues in $\{1,-1\}$. Since $\det A=1$, it is $I$ or $-I$. Faithfulness excludes $I$, so $t$ acts as the scalar matrix $-I$. It commutes with the whole image; faithfulness then makes $t$ central in $G$, contradicting nonabelian simplicity. Hence $\boxed{\chi (1)>2}$.

The odd order theorem is the deep standard group-theoretic input in this proof; its use is explicit rather than hidden in an unsupported assertion that $G$ has an involution.

Let $\dim V=2n$ and let $V_{+}=\ker (S-I)$ and $V_{-}=\ker (S+I)$ be the two eigenspaces. The involution identity gives a direct sum, sinceFor $u,v$ in either one of these eigenspaces, the anti-symplectic involution identity impliesso both are isotropic subspaces of a symplectic vector space. Such a subspace has dimension at most $n$: $W\subset W^{\Omega }$ and the nondegenerate bilinear form $\Omega$ gives $\dim W^{\Omega }=2n-\dim W$, where $W^{\Omega }$ is the symplectic orthogonal complement. As their dimensions add to $2n$, both have dimension $n$. Therefore

The map $x\mapsto -x(\mathrm{mod}K)$ is a bijection and an involution, so its permutation matrix satisfiesIn particular $S^{†}S=I$, which proves that $S$ is a unitary operator. Equivalently, it permutes the computational basis and therefore preserves every inner product.

The map $x\mapsto -x$ is an involution, soIf $\lambda$ is an eigenvalue of ${\mathrm{QFT}}_N$, then ${\lambda }^4=1$. Hence every eigenvalue belongs toBecause $X^4-1$ has distinct roots, the transform is diagonalizable; which fourth roots occur and their multiplicities depend on $N$.

Assume for contradiction that the involution $\varphi$ is fixed-point-free. The finite group action is then a covering space action, so the quotient mapis a double covering and $Y$ is an $n$-manifold.

Apply the long exact sequence in homology to the transfer chain map of a double covering. Since ${\mathbb{R}}^n$ is contractible, its positive-dimensional mod-two homology vanishes. The degree-zero portion, together with the fact that $q_{\ast }:H_0({\mathbb{R}}^n;{\mathbb{F}}_2)\to H_0(Y;{\mathbb{F}}_2)$ is an isomorphism, givesIn every higher degree the same exact sequence givesThus $H_i(Y;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ for every $i\ge 0$. This contradicts homology above the dimension of a manifold, which gives $H_i(Y;{\mathbb{F}}_2)=0$ for $i>n$. Therefore $\varphi$ has a fixed point.