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The general linear group over a finite field acting diagonally on projective triples has five orbits of a group action: all equal, three different positions for the repeated coordinate when exactly two are equal, and all distinct. The corresponding representative stabilizer subgroups are upper triangular, diagonal, and scalar matrices. Their orders are $p(p-1{)}^2$, $(p-1{)}^2$, and $p-1$, respectively; the orbit sizes are $p+1$, $p(p+1)$, and $p(p+1)(p-1)$. Transitivity follows from the action being sharply three-transitive on a projective line.

The general linear group over a finite field $G{L}_2({\mathbb{F}}_2)$ acts faithfully on the four vectors of ${\mathbb{F}}_2^2$, since a matrix fixing all vectors fixes the standard basis. The zero vector is fixed by every element; the induced action on the three nonzero vectors identifies this six-element group with $S_3$. Swapping the two coordinates fixes $(0,0)$ and $(1,1)$ and exchanges the other two vectors, so its four-point permutation is odd. This action therefore embeds the group into $S_4$ with nontrivial composite sign homomorphism.

A Hall subgroup for a set of primes $\pi$ is a subgroup $H\le G$ such that every prime divisor of $|H|$ belongs to $\pi$, while no prime divisor of $[G:H]$ belongs to $\pi$. Equivalently $|H|=|G{|}_{\pi }$, the product of the full prime powers in $|G|$ belonging to $\pi$.

The Hall conjugacy and embedding in finite soluble groups theorem says that, for every $\pi$, a finite soluble group has a Hall $\pi$-subgroup, any two such subgroups are conjugate, and every $\pi$-subgroup is contained in one. We prove all three assertions together by induction, including the needed complement argument rather than assuming it.

First establish coprime splitting over an elementary abelian normal subgroup. Suppose $V⊴E$ is an elementary Abelian $r$-group and $D=E/V$ has order $m$ prime to $r$. Write $V$ additively. Conjugation defines a well-defined action of $D$ on $V$ because $V$ is Abelian. Choose a section $s:D\to E$, with $s(1)=1$, and let its extension cocycle be defined by $s(x)s(y)=f(x,y)s(xy)$. Associativity givesSum over $z\in D$. With $S(x)={\sum }_{z}f(x,z)$ this becomesMultiplication by $m$ is invertible on $V$, so setting $b(x)=m^{-1}S(x)$ gives $f(x,y)=b(x)+xb(y)-b(xy)$. The new section $s^{\prime }(x)=(-b(x))s(x)$ has zero cocycle and is a homomorphism. Its image is a complement $K$ to $V$, proving $E=V⋊K$.

Every other complement has the form $\{h(x)s^{\prime }(x):x\in D\}$, with $h(xy)=h(x)+xh(y)$. Summing this identity over $y$ and putting $T={\sum }_{y}h(y)$ gives $mh(x)=T-xT$. Hence $h(x)=v-xv$, where $v=m^{-1}T$, and the other complement is $vK{v}^{-1}$. The same argument applies on any subgroup $D_0\le D$: a subgroup intersecting $V$ trivially is a graph of such a one-cocycle over its image $D_0$, so conjugation by an element of $V$ carries it into $K$. This proves the complement-conjugacy and embedding assertions needed below.

Now let $G\ne 1$ be finite soluble, and choose a minimal normal subgroup $V$. It is elementary Abelian of some prime characteristic $r$, by the minimal-normal argument in the series discussion. Induction gives Hall subgroups and the embedding and conjugacy assertions in $G/V$. Let $\overline{H}$ be a Hall $\pi$-subgroup there and $M$ its full preimage.

If $r\in \pi$, then $M$ itself is a Hall $\pi$-subgroup of $G$. Every Hall $\pi$-subgroup $H$ contains $V$, since $HV$ is a $\pi$-subgroup and $H$ already has the largest possible $\pi$-order. Thus $H$ is the full preimage of its image in $G/V$, and quotient conjugacy lifts to conjugacy in $G$. Given any $\pi$-subgroup $U$, induction puts its quotient image into some $\overline{H}$, whose preimage $M$ contains $U$.

If $r\notin \pi$, then $M/V=\overline{H}$ has order prime to $r$. The complement construction produces $K\le M$ of order $|\overline{H}|=|G{|}_{\pi }$, hence a Hall subgroup of $G$. Any Hall subgroup has trivial intersection with $V$ and a Hall image in $G/V$. Conjugating two Hall subgroups first aligns these images, after which both are complements to $V$ in the same $M$; the complement-conjugacy argument then conjugates them by an element of $V$. For an arbitrary $\pi$-subgroup $U$, quotient induction puts its image into some $\overline{H}$. Then $U\subseteq M$ and $U\cap V=1$, so the embedding argument conjugates $U$ into the chosen complement $K$. Equivalently $U$ lies in a conjugate of $K$. This completes the induction and all three parts of the Hall theorem for soluble groups.

For the specific general linear group over a finite field, counting ordered bases givesThere are seven lines and seven planes in ${\mathbb{F}}_2^3$. The group is transitive on each family, so line and plane stabilizers both have order $24$ and index seven. To show that these exhaust the index-seven subgroups, let $H$ have order $24$. It contains a Sylow $2$-subgroup of order eight, which can be conjugated toThe group orbits of $U$ on the seven nonzero vectors areof sizes $1,2,4$. An $H$-orbit is a union of these. The group $H$ cannot be transitive on seven points, since seven does not divide $24$. Therefore it has an orbit of size at most three. A size-one orbit fixes $e_1$. A size-two orbit must be $\{e_2,e_1+e_2\}$, whose vector sum $e_1$ is consequently fixed. A size-three orbit must be the union of the first two $U$-orbits, the nonzero part of the plane $⟨e_1,e_2⟩$, so that plane is invariant. In every case $H$ lies in a line or plane stabilizer, and equality follows from their equal orders.

The two stabilizer families are not conjugate. A line stabilizer fixes its nonzero vector and is transitive on the other six nonzero vectors: extend that fixed vector and either chosen other vector to a basis, and map one such basis to the other. Thus it has no invariant three-element subset of nonzero vectors and fixes no plane. In contrast a plane stabilizer fixes its defining plane. Conjugation sends line stabilizers to line stabilizers, so cannot send one to a plane stabilizer. We obtainThey are Hall subgroups for $\pi =\{2,3\}$ but are not conjugate. This explicitly shows why the solubility hypothesis of the Hall theorem for soluble groups matters.

Finally, an index-three subgroup would give a nontrivial transitive coset homomorphism $G{L}_3(2)\to S_3$. Its kernel is normal; the permitted simplicity of $G{L}_3(2)$ forces that kernel to be trivial, because the action is nontrivial. But an injective map from a group of order $168$ into one of order six is impossible. Hence $G{L}_3(2)$ has no subgroup of index three.

An element of the general linear group over a finite field is an invertible $2\times 2$ matrix. Its columns must form an ordered basis of ${\mathbb{F}}_p^2$. There are $p^2-1$ choices for the first column. Its linear span contains $p$ vectors, so there are $p^2-p$ choices for a second column outside that linear span. Thus the order of a general linear group over a finite field isThe determinant is a surjective group homomorphism $G{L}_2({\mathbb{F}}_p)\to {\mathbb{F}}_p^{\times }$: for each $a\ne 0$, the diagonal matrix $\mathrm{diag}(a,1)$ has determinant $a$. Its kernel of a group homomorphism is the special linear group over a finite field. The first isomorphism theorem gives $|G{L}_2|/|S{L}_2|=p-1$, hence

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

Use the standard implicit assumption that $p$ is prime, so ${\mathbb{F}}_p$ is a finite field. The matrices in question form $G{L}_2({\mathbb{F}}_p)$, the general linear group over a finite field. Products stay in the set because determinants multiply, the identity has determinant one, andhas entries in the finite field and nonzero determinant. Together with associativity this verifies the group axioms. The multiplication of matrices with column vectors is a group action because $Iv=v$ and $(gh)v=g(hv)$.

Let $C=⟨g⟩$ be the order-$p$ cyclic subgroup. By the orbit-stabilizer theorem, its group orbits on ${\mathbb{F}}_p^2$ have size either one or $p$. If $f$ is the number of fixed vectors, counting all $p^2$ vectors gives $f\equiv p^2\equiv 0(\mathrm{mod}p)$. Since the zero vector is fixed, $f\ge 1$ and hence $f\ge p$. There is a nonzero fixed vector $v$.

Extend $v$ to a basis $(v,w)$. In this basis $g$ has matrix $\left(\begin{array}{cc}1& b\\ 0& d\end{array}\right)$. From $g^p=I$ we get $d^p=1$; in the prime finite field, $d^p=d$, so $d=1$. Since $g$ has order $p>1$, it is not the identity, and $b\ne 0$. Replace $w$ by $b^{-1}w$. Then $g(b^{-1}w)=v+b^{-1}w$, and in the new basisThus all the order-p matrices in GL2 over the prime field form one conjugacy class. This proof also works at $p=2$. Primality is essential to the first assertion: for modulus four, $\mathrm{diag}(2,1)$ has nonzero determinant but no inverse, so the set defined using merely nonzero determinants would not be a group.

For subgroups $H,P\le G$, define $x\sim y$ when $y\in HxP$. The identity elements show reflexivity, inverses show symmetry, and multiplication shows transitivity. Thus the double cosets $HxP$ partition $G$.

Let $H$ act by left multiplication on the set of left cosets $G/P$. The orbit of $xP$ consists of the cosets contained in $HxP$, so its size is $|HxP|/|P|$. Its stabilizer isThe orbit-stabilizer theorem therefore gives

Suppose $P$ is a Sylow subgroup of $G$, with $|P|=p^a$, and write the largest power of $p$ dividing $|H|$ as $p^b$. If every $H\cap xP{x}^{-1}$ had order at most $p^{b-1}$, the displayed formula would make every double-coset size divisible by $p^{a+1}$. Their sum $|G|$ would then also be divisible by $p^{a+1}$, contradicting the choice of $P$. Hence some $H\cap xP{x}^{-1}$ has order $p^b$ and is a Sylow $p$-subgroup of $H$. This is the Sylow subgroup of a subgroup from double cosets argument.

To count the general linear group over a finite field $G{L}_n({\mathbb{F}}_p)$, choose its columns successively. There are $p^n-1$ choices for the first, $p^n-p$ for the second, and $p^n-p^k$ for column $k+1$. ConsequentlyNone of the factors $p^j-1$ is divisible by $p$, so the upper unitriangular group is a Sylow $p$-subgroup: it has one arbitrary field entry in each of the $n(n-1)/2$ positions above the diagonal and hence order $p^{n(n-1)/2}$. The permutation matrices form a subgroup isomorphic to the symmetric group $S_n$.

By Cayley theorem, every finite group $G$ embeds in $S_{|G|}$, and permutation matrices embed this symmetric group in $G{L}_{|G|}({\mathbb{F}}_p)$. The latter has the explicit Sylow $p$-subgroup just described, so the result proved in the first part, applied to the embedded copy of $G$, proves that every finite group has a Sylow $p$-subgroup.

The counting part of the Sylow theorems says that if $p^a$ is the largest power of $p$ dividing $|G|$, then the number $n_p$ of Sylow $p$-subgroups satisfies

Finally, let $|G|=pq$ with prime numbers $p>q$. The Sylow counts giveso $n_p=1$ and the Sylow $p$-subgroup is a normal subgroup. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal; their elements commute, so $G$ is their direct product and is abelian. In the nonabelian case one must therefore have $n_q=p$, whenceor equivalently $q\mid p-1$, as recorded by nonabelian group of order pq.