The symplectic group consists of the invertible linear maps preserving the alternating bilinear form:We use row-vector action in this question, which is the convention compatible with its printed upper-triangular flag stabilizer subgroup. Thus matrices preserve a form matrix $J$ by $gJ{g}^T=J$.

Count ordered symplectic bases. There are $q^{2m}-1$ choices for the first nonzero vector $e_1$. Nondegeneracy makes the equation $(e_1,f_1)=1$ a nonzero linear-functional equation, with $q^{2m-1}$ solutions. Their span is a nondegenerate plane; its orthogonal complement is symplectic of dimension $2m-2$. Repeating there givesEach symplectic basis is the image of a fixed one under exactly one form-preserving map, justifying the count as a group order. If $q=p^a$, every factor $q^{2i}-1$ is prime to $p$, so the exact $p$-part is $q^{m^2}$.

Iwasawa's simplicity lemma states the following. Suppose $G$ acts faithfully and primitively on a set, and a stabilizer subgroup $H$ has an abelian normal subgroup $A$ whose $G$-conjugates generate $G$. Then every nontrivial normal subgroup $N$ contains $G^{\prime }$. In particular, if $G$ is nontrivial and perfect, then $G$ is simple.

Indeed a nontrivial normal subgroup in a faithful primitive action is transitive, so $G=NH$. Since $H$ normalizes $A$, all conjugates of $A$ have the same image in $G/N$. Those images generate the quotient, which is therefore abelian. This gives $G^{\prime }\le N$ and proves the stated conclusion.

Put $n=|\Omega |\ge 2$, and let $H=G_{\alpha }$. Sharp two-transitivity gives $|G|=n(n-1)$ and $|H|=n-1$. A nonidentity element fixes at most one point. Counting the nonidentity elements in the $n$ stabilizer subgroups shows that there are $n-1$ fixed-point-free elements. Let $N$ be this set together with the identity. We first prove it is a normal subgroup, rather than presuming that fixed-point-free elements are closed under multiplication.

For $n=2$, $G=C_2$ and $N=G$ directly. Otherwise use complex characters of a finite group. Let $\pi$ be the permutation character and $\psi =\pi -1$, the character of the permutation representation with its constant line removed. For every nontrivial irreducible character $\theta$ of $H$, form the virtual characterIts values are $\theta (1)$ at the identity and at every fixed-point-free element. At an element with one fixed point, conjugate it to $h\in H\setminus \{1\}$; induction gives ${\phi }_{\theta }(g)=\theta (h)$, because there is exactly one fixed coset.

The identity and the $n-1$ fixed-point-free elements together contribute $n\theta (1)\overline{\eta (1)}$ to the inner product. The remaining elements are partitioned into the nonidentity parts of the $n$ stabilizer subgroups. Hence character orthogonality givesA virtual character of norm one is plus or minus an irreducible character: its coefficients in the irreducible-character basis are integers whose squares sum to one. Its positive degree $\theta (1)$ selects the plus sign. Thus each ${\phi }_{\theta }$ is an actual irreducible character.

For a group representation, $\chi (g)=\chi (1)$ holds exactly on its kernel: make the representation unitary and compare the sum of its unit-modulus eigenvalues with its dimension. All of $N$ therefore lies in the intersection of the kernels of the ${\phi }_{\theta }$. Conversely, a nonidentity element fixing a point gives $h\ne 1$ in $H$. Some nontrivial irreducible character of $H$ has $\theta (h)\ne \theta (1)$, since otherwise the regular representation of $H$ would not vanish at $h$. ConsequentlyThis proves normality and subgroup closure. It has order $n$ and no nonidentity element fixing a point, so it is a regular permutation subgroup.

Now $H$ acts transitively by conjugation on $N\setminus \{1\}$: identify an element of $N$ with its image of $\alpha$ and use transitivity of $H$ on the remaining points. Thus all nonidentity elements of $N$ have the same order. Taking a suitable power of one element shows this common order is a prime $p$. By Cauchy's theorem no other prime divides $|N|$, so $N$ is a $p$-group. Its nontrivial center is $H$-invariant, so transitivity forces the center to be all of $N$. Therefore $N$ is elementary abelian of order $p^d=n$.

Since $|H|=p^d-1$ is prime to $p$, $N$ is the unique Sylow $p$-subgroup of $G$. Uniqueness makes it characteristic under every group automorphism. We have provedThe character argument supplies the regular kernel of a finite sharply two-transitive group; the final Sylow argument establishes the stronger characteristic assertion.

For $1\le t\le |\Omega |$, a group is sharply t-transitive when any two ordered $t$-tuples of distinct points are related by exactly one group element. Equivalently its action on the set of such tuples is regular. In the finite caseand the stabilizer subgroup of an ordered $t$-tuple is trivial. The condition includes both existence and uniqueness, not just transitivity.

Here is the double-coset criterion for a one-point extension. Let $H=G_{\alpha }$, and suppose $x$ swaps $\omega$ and $\alpha$. Then $⟨G,x⟩$ is a one-point extension if and only ifFor necessity, in an extension $H$ fixes both $\omega$ and $\alpha$, so $x$ normalizes it and $x^2$ lies in it. Since $G$ is transitive on $\Omega$, the extension has exactly two double cosets relative to $G$: $G$ and $GxG$. For $g\notin H$, $xgx$ moves $\omega$ into $\Omega$ and is in the latter double coset.

For sufficiency, the displayed conditions make $G\cup GxG$ closed under multiplication. Products with middle element in $H$ reduce using $xhx=(xh{x}^{-1})x^2\in G$; those with middle element outside $H$ remain in $GxG$. A finite nonempty multiplication-closed set of permutations containing the identity is a group. It contains $G$ and $x$, hence equals $⟨G,x⟩$. Every element in $GxG$ moves $\omega$, while $G$ fixes it, giving the required stabilizer subgroup. The group is transitive because $G$ is transitive on $\Omega$ and $x$ moves the additional point.

Extend each element of $G$ to fix $\omega$. A one-point extension of a permutation group is a transitive permutation group $G^{+}\le S_{{\Omega }^{+}}$ such thatwith the induced action on $\Omega$ equal to the prescribed action. The stabilizer subgroup equality is the substantive condition: merely adjoining an element that moves $\omega$ does not suffice. If $|\Omega |=n$, the orbit-stabilizer theorem gives $|G^{+}|=(n+1)|G|$.

Use the Sylow theorems. The number $n_p$ divides $qr$ and is congruent to one modulo $p$. Since $q,r<p$, it is either one or $qr$. Suppose $n_p=qr$. Different subgroups of prime order intersect trivially, so their nonidentity elements occupy $qr(p-1)$ places, leaving only $qr-1$ nonidentity elements of other prime orders.

If neither the Sylow $q$-subgroup nor the Sylow $r$-subgroup is normal, then $n_q\ge p$ and $n_r\ge q$. Indeed $n_q$ divides $pr$, and its possible divisor $r<q$ cannot satisfy the Sylow congruence; the smallest remaining nontrivial possibility is at least $p$. Likewise any nontrivial divisor of $pq$ is at least $q$. Their elements would require at leastplaces, since the excess is $(p-1)(q-1)>0$. Hence some Sylow subgroup of order $s\in \{q,r\}$ is normal.

In the quotient by this normal subgroup, the largest prime $p$ has a normal Sylow subgroup: for a group of order $pt$ with $p>t$ its Sylow count divides $t<p$ and so equals one. Pulling back gives a normal subgroup of order $ps$. Inside it, the subgroup of order $p$ is again the unique Sylow $p$-subgroup. It is characteristic in that normal subgroup and therefore normal in $G$, contradicting $n_p=qr$. Thus $\boxed{n_p=1}$.

Let $P$ be this normal Sylow subgroup. In $G/P$, of order $qr$, its subgroup of order $q$ is normal by the same argument. Its preimage $K$ is a normal Hall $\{p,q\}$-subgroup. The serieshas factors of orders $p,q,r$, hence cyclic and abelian. Therefore $\boxed{G\text{ is soluble}.}$ This establishes solubility before using any Hall-existence conclusion that itself assumes solubility.

We prove Hall subgroup existence in soluble groups by induction on $|G|$. The trivial group is immediate. Choose a nontrivial minimal normal subgroup $N$, elementary abelian of order $p^a$ by part (c). Induction gives a Hall $\pi$-subgroup of $G/N$; let $K$ be its full preimage.

If $p\in \pi$, then $K$ itself is the required subgroup. If $p\notin \pi$ and $K<G$, apply induction inside the soluble subgroup $K$ to obtain a Hall $\pi$-subgroup $H$ of $K$. Since $[G:K]$ and $[K:H]$ are both ${\pi }^{\prime }$-numbers, $H$ is Hall in $G$ too.

It remains to treat $p\notin \pi$ and $K=G$. Then $G/N$ is a $\pi$-group. If $G=N$, the subgroup one works. Otherwise choose a minimal normal subgroup $M/N$ of $G/N$, an elementary abelian $q$-group with $q\in \pi$. Let $Q$ be a Sylow $q$-subgroup of $M$. As $|M|=|N|q^b$ and $p\ne q$, we have $M=NQ$. The permitted Frattini argument givesThe last equality uses $Q\le N_G(Q)$ and the normality of $N$.

If $T=N_G(Q)<G$, then $[G:T]=[N:N\cap T]$ is a power of $p$, hence a ${\pi }^{\prime }$-number. Apply induction to $T$ and multiply indices as before. If $T=G$, then $Q$ is a nontrivial normal $\pi$-subgroup. Induction in $G/Q$ gives a Hall $\pi$-subgroup whose full preimage in $G$ is Hall, since its additional factor $|Q|$ is a $\pi$-number. These cases exhaust the possibilities, proving existence for every prime set. No unproved complement theorem or conjugacy theorem for Hall subgroups was inserted into the proof.

A Hall subgroup for a prime set $\pi$ is a subgroup $H\le G$ whose order has only prime divisors in $\pi$ and whose index has no prime divisor in $\pi$:Here one is allowed in either class, and ${\pi }^{\prime }$ is the complementary set of primes. Equivalently $|H|$ contains the complete prime-power contribution to $|G|$ for each prime in $\pi$.

Let $N\ne 1$ be a minimal normal subgroup. Its commutator subgroup $N^{\prime }$ is characteristic in $N$, hence normal in $G$. Minimality makes $N^{\prime }=1$ or $N^{\prime }=N$. The latter would prevent the soluble group $N$ from having a terminating derived series, so $N^{\prime }=1$ and $N$ is abelian.

Choose a prime $p$ dividing $|N|$. In a finite abelian group its Sylow $p$-subgroup is characteristic, so minimal normality makes this subgroup all of $N$. The subgroup ${\Omega }_1(N)=\{n\in N:n^p=1\}$ is nontrivial, characteristic and hence normal in $G$. Minimality again makes it all of $N$. Thusan elementary abelian p-group. Both abelianness and minimal normality are essential to the two characteristic subgroup arguments.

If $H\le G$, induction gives $H^{(i)}\le G^{(i)}$, because commutators of elements of a subgroup are also commutators in the larger group. Thus termination of the derived series of $G$ forces termination for $H$.

For a normal subgroup $N$, the quotient map sends $[x,y]$ to the commutator of their images. Surjectivity then givesConsequently subgroups and quotient groups of a soluble group are soluble. The assertion about a quotient uses a normal subgroup; it is not a quotient by an arbitrary subgroup.

The derived series is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$, where the bracket denotes the commutator subgroup. The group is soluble ifEquivalently it has a finite series with abelian factors. The trivial group is included.

An invertible time-series representation recovers the driving white noise from current and past observations. In the inverse series the support condition is thereforeAgain the series must converge. Stable invertibility uses ${\sum }_{r\ge 0}|b_r|<\infty$, which ensures mean-square convergence when $X_t$ has finite variance. Merely writing a bilateral inverse is not invertibility in this one-sided sense: it may require future observations. For a general correlated input $X$, square summability of $b_r$ alone is not the same sufficient condition as it is for a white noise input.

A causal time-series representation uses only the present and past driving white noise. Thus the coefficient condition isThe series must have its stated convergence meaning. For centered white noise of positive finite variance, ${\sum }_{r\ge 0}|a_r{|}^2<\infty$ is sufficient and necessary for mean-square convergence. In the usual stable-filter convention one imposes the stronger ${\sum }_{r\ge 0}|a_r|<\infty$. A bilateral stationary linear process need not be causal: terms with $r<0$ involve future driving values.