Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 1 (Topics in Group Theory)

First prove the primitive three-cycle criterion directly. The useful generation fact is that adding a three-cycle whose support meets a set $S$, with $|S|\ge 3$, to the alternating group $A_S$ generates the alternating group on the enlarged set. If the support is $\{a,b,x\}$ with $a,b\in S$ and $x\notin S$, conjugation by $A_S$ and taking inverses produces every three-cycle with $x$ and two points of $S$. For $|S|\ge 4$ this uses two-transitivity of $A_S$; for $|S|=3$, its cyclic action on pairs together with inversion gives both pair orientations. These cycles, together with the cycles supported inside $S$, are all the three-cycles on $S\cup \{x\}$ and generate $A_{S\cup \{x\}}$. Every even permutation is a product of three-cycles: pair its transpositions, and use $(ab)(bc)=(abc)$ and $(ab)(cd)=(acb)(acd)$, with rightmost factors acting first.

If instead the new support is $\{a,x,y\}$ with just $a\in S$, choose $b\in S\setminus \{a\}$ and conjugate $(axy)$ by an element of $A_S$ carrying $a$ to $b$. This gives $(bxy)$, andThe two-point-overlap case first adjoins $x$; the original cycle then adjoins $y$. This proves connected triple supports generate an alternating group by successively adjoining intersecting supports.

Now take the supports of every conjugate in $G$ of a given three-cycle, and make these triples the edges of a hypergraph on the permutation domain. Its connected components are permuted by $G$, so they form a block system. Since at least one component contains three points, primitivity forces the single component to be the entire domain. Start with one triple and adjoin intersecting triples until all points have been reached. The generation argument shows that the subgroup generated by the conjugate cycles is $A_n$. Hence

To classify the proper maximal subgroups containing a three-cycle, distinguish the three possible types of action. A primitive maximal subgroup contains $A_n$, and thus equals $A_n$, since $[S_n:A_n]=2$. An intransitive maximal subgroup lies in the full stabilizer $S_k\times S_{n-k}$ of one of its orbits, and maximality makes it that full stabilizer. When $k=n-k$, this stabilizer lies properly inside the subgroup that can interchange the two parts, so it is not maximal. Take the smaller part to obtain $1\le k<n/2$. Such an unequal-part stabilizer is maximal: its transpositions connect all points within each part, and any element outside it carries at least one within-part transposition to a cross-part transposition. The resulting connected graph of transpositions generates $S_n$. To see that last assertion, transpositions along a path generate the transposition of its endpoints by successive conjugations. The stabilizer contains a three-cycle exactly when $n-k\ge 3$.

A transitive imprimitive maximal subgroup preserves a nontrivial partition into $b$ blocks of size $a$, with $n=ab$. It equals the full permutation wreath product $S_a\wr S_b$. This full block stabilizer is maximal: any permutation not preserving the partition sends some same-block pair to a different-block pair, giving a cross-block transposition by conjugation. The block stabilizer is transitive on cross-block pairs, so all cross-block transpositions follow; together with its within-block transpositions they generate $S_n$. A three-cycle in this group must act trivially on the blocks, since any nontrivial block permutation moves at least two entire blocks, and a block three-cycle moves $3a$ points. Thus a single three-cycle exists exactly when $a\ge 3$. The full classification, up to conjugation, is thereforeThis is the maximal symmetric-group subgroups containing a three-cycle classification. For $n<3$ there are no such subgroups.

For the first family of counterexamples to extending the criterion to longer cycles, let $G=PS{L}_2(p)$ act on the projective line ${\mathbb{F}}_p\cup \{\infty \}$. The determinant-one matricesact respectively as $x\mapsto x+1$ and $x\mapsto -1/x$. The first fixes infinity and is one $p$-cycle on the finite points. Its translations make the infinity stabilizer transitive on the other points; the second moves infinity, so the full action is two-transitive, hence primitive. The action is faithful after quotienting by scalar matrices. Counting a nonzero first column and the $p$ second columns with determinant one gives $|S{L}_2(p)|=p(p^2-1)$, and the scalar group center has order two. Thussince the ratio on the right to the left is $(p-2)!>1$. Consequently $PS{L}_2(p)$ is the required example for every prime $p>3$.

For the final example take $p=59$, and let the direct product of groups $A_5\times A_5$ act on the sixty elements of $A_5$ byThe action is transitive and the stabilizer of the identity is the diagonal subgroup $\Delta =\{(z,z):z\in A_5\}$. To prove that the diagonal is a maximal subgroup, no classification theorem is needed. If $\Delta <K\le A_5\times A_5$, choose $(x,y)\in K\setminus \Delta$. Multiplying by $(y^{-1},y^{-1})$ gives $(x{y}^{-1},1)\in K$ with $x{y}^{-1}\ne 1$. The subgroup $L=\{z:(z,1)\in K\}$ is normal in $A_5$, by diagonal conjugation, and is nontrivial. Simplicity of $A_5$ gives $L=A_5$, and then $K=A_5\times A_5$. Thus the coset action is primitive. It is also faithful: if $xt{y}^{-1}=t$ for every $t$, then $x=y$ by taking $t=1$, and $x$ is central in the centerless simple group $A_5$, so $x=y=1$. Finally,The prime $59$ is odd, and its primality follows by testing the primes $2,3,5,7$, the only ones below its square root. This is a faithful diagonal primitive permutation group whose order is not divisible by the degree minus one.

A subnormal series is a finite chain $1=G_r⊴G_{r-1}⊴\cdots ⊴G_0=G$, in which each term is normal in the preceding term. Its factors are the quotient groups $G_i/G_{i+1}$. A normal series of a group imposes the stronger condition that every term be normal in $G$. These conditions should not be conflated: inserting a subgroup normal in an intermediate term need not preserve normality in the whole group. A refinement inserts additional terms, and two series are equivalent when their nontrivial factors can be paired by group isomorphisms.

The basic refinement mechanism is the Zassenhaus butterfly lemma. For $A^{\prime }⊴A$ and $B^{\prime }⊴B$, let $C=A\cap B$ and $D=(A^{\prime }\cap B)(A\cap B^{\prime })$. Both factors of $D$ are normal in $C$, so $D⊴C$. The map from $C$ onto $A^{\prime }C/A^{\prime }(A\cap B^{\prime })$ has kernel $D$: if an element of $C$ equals $a^{\prime }{b}^{\prime }$ with $a^{\prime }\in A^{\prime }$ and $b^{\prime }\in A\cap B^{\prime }$, then $a^{\prime }=c{b}^{\prime -1}\in A^{\prime }\cap B$. Conversely each product in $D$ is in the kernel. The first isomorphism theorem therefore givesThe denominators are normal in their numerators: conjugation by $C$ preserves them, while conjugation by $A^{\prime }$ or $B^{\prime }$ changes their elements only by a factor already in that normal subgroup.

Given two subnormal series $(G_i)$ and $(H_j)$, refine the first by $G_{i+1}(G_i\cap H_j)$, and the second by $H_{j+1}(H_j\cap G_i)$. The endpoints are $G_i,G_{i+1}$ and $H_j,H_{j+1}$ respectively, and the Zassenhaus butterfly lemma pairs the successive factors indexed by $(i,j)$ and $(j,i)$. This proves the Schreier refinement theorem.

A composition series is a strict subnormal series all of whose factors are simple groups. Every finite group has one: choose a maximal proper normal subgroup, repeat inside it, and use decreasing orders to terminate. A simple factor cannot admit a nontrivial refinement. Applying the preceding refinement theorem to two composition series proves the Jordan–Hölder theorem: their lengths and multisets of simple factors agree, up to group isomorphism. The factors are the Jordan–Hölder factors, and their number is the composition length. If $N⊴G$, lifting a composition series of $G/N$ and adjoining one for $N$ gives

These factors do not determine the whole group, because they do not determine how its group extensions are assembled. For example, $C_4$ and the Klein four-group both have two factors $C_2$, but one contains an element of order four and the other does not. The symmetric group $S_3$ has a composition series $1<C_3<S_3$, with factors $C_3,C_2$, whereas the simple group $A_5$ has the single factor $A_5$.

A chief series is a maximal strict normal series of a group; each chief factor is a minimal nontrivial normal subgroup in the appropriate quotient of $G$. Finite chief factors are characteristically simple. To justify their structure, let $N$ be any nontrivial finite characteristically simple group and choose a minimal nontrivial normal subgroup $M$ of $N$. Its distinct images under automorphisms of $N$ are minimal normal subgroups, so distinct ones intersect trivially and commute. Choose a maximal family giving an internal direct product of groups. If any other image were not contained in that product, its intersection with the product would be trivial by minimality, and it could be added, a contradiction. Hence this product is characteristic and equals $N$. Each factor $M$ is simple: a proper nontrivial normal subgroup of $M$ would be normal in the whole direct product $N$, contrary to the choice of $M$. Thus $N$ is a direct product of groups of isomorphic finite simple groups. This applies to chief factors by conjugation from $G$.

Chief and composition factors need not coincide. In $A_4$, the chain $1<V_4<A_4$ is a chief series, since conjugation cycles the three order-two subgroups of $V_4$. A composition series instead inserts any one of those subgroups, giving factors $C_2,C_2,C_3$. Its inserted order-two subgroup is normal in $V_4$ but not in $A_4$. Chief series retain the ambient group's conjugation information that composition series discard.

The derived series gives a canonical criterion for a group to be soluble: put $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. Its terms are characteristic, and each factor is Abelian. A finite group is soluble exactly when this series reaches the identity, equivalently when all its composition factors are cyclic of prime order. For one direction, the derived series supplies Abelian factors, whose simple refinements are cyclic of prime order. Conversely an extension of two soluble groups is soluble: if $(G/N{)}^{(r)}=1$, then $G^{(r)}\subseteq N$; if $N^{(s)}=1$, then $G^{(r+s)}=1$. Induction up a composition series with cyclic prime factors proves the other direction. In a finite soluble group, each minimal normal subgroup is elementary Abelian: its characteristic commutator subgroup must be trivial, its primary components force a single prime, and its characteristic subgroup of $p$th powers must be trivial. Thus every chief factor is a vector space over some ${\mathbb{F}}_p$. In a general finite group the Jordan–Hölder factors can instead be nonabelian simple groups.

The central series demand more than Abelian factors: each factor must be central in the corresponding quotient of the whole group. Define the upper central series by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, and the lower central series by ${\gamma }_1=G$ and ${\gamma }_{i+1}=[{\gamma }_i,G]$. A group is nilpotent exactly when the upper series reaches $G$, equivalently when the lower series reaches the identity. Indeed, for any ascending central series $1=C_0\le \cdots \le C_c=G$, induction gives $C_i\subseteq Z_i$ and ${\gamma }_{c+1-i}\subseteq C_i$; reversing the lower series gives an ascending central series when it terminates. The least such $c$ is the nilpotency class. A central series has Abelian factors, so every nilpotent group is soluble; the converse fails, as $S_3$ is soluble but has trivial group center.

For a finite p-group, the class equation shows that the group center is nontrivial, since every noncentral conjugacy class has size divisible by $p$. Applying the same argument to successive central quotients makes the upper central series reach the group. More generally a finite group is nilpotent exactly when it is the direct product of groups of its Sylow subgroups. Here is the nontrivial direction. A central series gives the normalizer condition for nilpotent groups: for a proper subgroup $H$, choose the first $Z_i$ not contained in $H$. Then $[Z_i,H]\subseteq Z_{i-1}\subseteq H$, so $Z_i\subseteq N_G(H)$ and $N_G(H)>H$. If a Sylow subgroup $P$ had proper normalizer $N$, then $N_G(N)>N$; but $P$ is the unique Sylow subgroup of $N$, hence characteristic there, and $N_G(N)\subseteq N_G(P)=N$, a contradiction. Thus all Sylow subgroups are normal, different ones commute because their intersection is trivial, and they form an internal direct product. Conversely each finite Sylow subgroup is nilpotent, and products of their central series give a central series for their direct product.

Composition series record simple building blocks, chief series record ambient normal structure, derived series detect solubility, and central series detect nilpotence. Their associated factors and extension data explain why the same group can have very different useful series.

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.

For the transfer homomorphism, choose representatives $T$ of the right cosets $H\backslash G$. For $t\in T$ and $g\in G$, write $tg=h(t,g)t_g$, where $h(t,g)\in H$ and $t_g\in T$. The permutation $t\mapsto t_g$ represents right multiplication on the cosets. DefineThe product has no order ambiguity in the abelianization $H/H^{\prime }$. Replace each representative $t$ by $a_{t}t$, with $a_t\in H$. The new factor is $a_{t}h(t,g)a_{t_g}^{-1}$. In the Abelian quotient, the extra products cancel because $t\mapsto t_g$ permutes $T$. Thus the transfer is independent of the transversal. Moreover,Multiplying and using the same permutation of representatives gives $V(g_1{g}_2)=V(g_1)V(g_2)$. This proves that group transfer is a well-defined group homomorphism.

For the Burnside transfer theorem, let $P\subseteq Z(N_G(P))$ be a Sylow $p$-subgroup. In particular $P$ is Abelian. We first prove the fusion fact that two elements of $P$ which are conjugate in $G$ must be equal. Suppose $u\in P$ and $v=g^{-1}ug\in P$. Both $P$ and $g^{-1}Pg$ centralize $v$, and they are Sylow subgroups of $C_G(v)$: they have the maximal $p$-order possible even in $G$. By Sylow theorems inside this centralizer, choose $h\in C_G(v)$ with $h^{-1}{g}^{-1}Pgh=P$. Thus $gh\in N_G(P)$ andBut $gh$ centralizes $P$ by hypothesis, so $v=u$.

Evaluate transfer to $P$ on $u\in P$. Partition the right cosets of $P$ into cycles under multiplication by $u$. For a cycle of length $r$ beginning at $Pt$, the product of its transfer factors telescopes to $t{u}^r{t}^{-1}\in P$. It is conjugate to $u^r\in P$, so the fusion fact makes it equal to $u^r$. The cycle lengths sum to $m=[G:P]$. Since $P$ is Abelian, all factors commute andThe integer $m$ is prime to $p$, and this power map is an automorphism of the finite Abelian $p$-group $P$: choose an inverse exponent modulo the exponent of $P$. Therefore $V_{G,P}:G\to P$ is surjective. Its kernel $K$ is normal and the first isomorphism theorem givesThus $K$ is the required normal p-complement; it also has trivial intersection with $P$.

Now suppose $P$ is cyclic of order $p^a$, where $p$ is the smallest prime divisor of $|G|$. Conjugation embeds $N_G(P)/C_G(P)$ into $\mathrm{Aut}(P)$, whose order is $p^{a-1}(p-1)$. Since $P\subseteq C_G(P)$ and $P$ is Sylow in $N_G(P)$, the quotient has order prime to $p$. Any remaining prime factor would divide $p-1$ and hence be smaller than $p$, but would also divide $|G|$, an impossibility. Thus $N_G(P)=C_G(P)$ and $P\subseteq Z(N_G(P))$. The Burnside transfer theorem proves that a cyclic Sylow subgroup at the least prime divisor gives a normal complement.

Apply this to a finite nonabelian simple group. If $p^3\nmid |G|$, its Sylow $p$-subgroup has order $p$ or $p^2$. A cyclic Sylow subgroup would give a normal complement; simplicity would then force either a trivial complement and a $p$-group, or an improper complement, neither possible for a nonabelian simple group. Groups of order $p^2$ are Abelian: their nontrivial group center, supplied by the class equation, is either the whole group or has cyclic quotient, which also forces the group Abelian. Thus the only remaining Sylow possibility is $P\cong C_p^2$.

The conjugation quotient $N_G(P)/C_G(P)$ now embeds in $G{L}_2(p)$ and is still prime to $p$. Its possible prime divisors come fromIf $p$ is odd, every prime divisor of $p-1$ is smaller than $p$, and every prime divisor of the even number $p+1$ is at most $(p+1)/2<p$. Since $p$ was the least prime divisor of $|G|$, the quotient must be trivial. Then the Burnside transfer theorem again contradicts simplicity. If $p=2$, the only nontrivial odd divisor of $|G{L}_2(2)|=6$ is three. Trivial conjugation would again give a normal complement, so three must divide $|G|$; the Sylow subgroup has order four. We have proved the least-prime divisibility constraint for a finite simple group:

For infinitely many instances of the latter alternative only, takeThese are simple, as permitted in the question (and also covered by the next solution). For $q=5^{2m+1}$, the group order is $q(q^2-1)/2$. Since $q\equiv 5(\mathrm{mod}8)$, $q-1$ is divisible by four but not eight and $q+1$ by two but not four. Hence the order is divisible by four but not eight. Also $q\not\equiv 0(\mathrm{mod}3)$, so $3\mid q^2-1$. Their smallest prime divisor is two andTheir orders strictly increase with $m$, so they give infinitely many distinct examples.

Write $x_{ij}(t)=I+t{E}_{ij}$ for $i\ne j$ and $t\in {\mathbb{F}}_q$. When $t\ne 0$ these are transvections, and their inverses are $x_{ij}(-t)$. Row addition by such an elementary transvection matrix preserves the determinant. It also realizes signed interchanges and compensating diagonal scalings: in any two coordinate positions,Thus elimination can reduce any invertible matrix to diagonal form using only row additions and signed interchanges, all products of transvections. For a diagonal matrix with determinant one, $\mathrm{diag}(d_1,\dots ,d_n)$, use the displayed two-coordinate diagonal matrices in positions $(i,n)$ with parameter $d_i$, for $1\le i<n$. Their product has diagonal entries $d_1,\dots ,d_{n-1},{\prod }_{i<n}{d}_i^{-1}=d_n$. Hence the residual diagonal matrix is also a product of transvections, and reversing elimination provesThis proves transvections generate the special linear group, including the small fields.

For perfectness, use the convention $[x,y]=xy{x}^{-1}{y}^{-1}$ in the following calculations. If $n\ge 3$ and $i,j,k$ are distinct, direct matrix multiplication givesTaking $b=1$ puts every elementary generator in the commutator subgroup. Thus $S{L}_n(q)$ is perfect. When $n=2$ and $q>3$, choose $s\in {\mathbb{F}}_q^{\times }$ with $s^2\ne 1$; such an element exists because a degree-two polynomial has at most two roots and $q-1>2$. For $D=\mathrm{diag}(s,s^{-1})$,Both coefficients are nonzero, so varying $t$ supplies every upper and lower elementary generator as a commutator. ThereforeThe exceptions really fail: $S{L}_2(2)\cong S_3$ has a nontrivial sign quotient, and $S{L}_2(3)$ has quotient $PS{L}_2(3)\cong A_4$, which has the nontrivial Abelian quotient $A_4/V_4\cong C_3$. For the latter identification, the faithful projective action embeds the order-twelve group $PS{L}_2(3)$ as an index-two subgroup of $S_4$, necessarily $A_4$.

The Iwasawa simplicity lemma states that, in a faithful primitive group action, if a point stabilizer has an Abelian normal subgroup whose conjugates generate the whole group, then every nontrivial normal subgroup contains the commutator subgroup. In particular a nontrivial perfect group with this property is simple. The lemma is applied here, as allowed, without assuming the desired simplicity theorem.

Let $S{L}_n(q)$ act on the one-dimensional subspaces of ${\mathbb{F}}_q^n$. This action is two-transitive: a linear map taking two specified distinct lines to two others can be made determinant one by rescaling one of the target basis vectors, without changing either target line. Its kernel consists exactly of scalar matrices. Indeed fixing all lines makes a matrix diagonal in any fixed basis, and fixing the lines through $e_i+e_j$ makes all its diagonal entries equal. The scalar matrices in $S{L}_n(q)$ are $aI$ with $a^n=1$. They also form its group center, since a central matrix commutes with every elementary generator, forcing all off-diagonal entries to vanish and all diagonal entries to agree. Their number is $\gcd (n,q-1)$. Consequently the quotient $PS{L}_n(q)$ acts faithfully and primitively.

Fix a line $L=⟨v⟩$ and considerHere $vf$ is the rank-at-most-one map $z\mapsto f(z)v$. These matrices have determinant one, and $(vf)(vh)=0$, so $T_L$ is Abelian under addition of the functionals. If $gL=L$, then $gv=av$ and $g(I+vf)g^{-1}=I+av(f{g}^{-1})\in T_L$. Thus $T_L$ is normal in the line stabilizer. Its image in $PS{L}_n(q)$ is a nontrivial Abelian normal subgroup of the point stabilizer. Conjugates of these groups include every elementary transvection, hence generate $PS{L}_n(q)$. Perfectness passes to a quotient. The Iwasawa simplicity lemma therefore givesIn those exceptions the projective groups are $S_3$ and $A_4$, as above.

For $q=4$, counting columns gives $|S{L}_2(4)|=4(4^2-1)=60$. Its scalar group center is trivial, so $PS{L}_2(4)$ has order sixty. Its faithful action on the five points of the projective line embeds it in $S_5$. Because it is perfect, its sign homomorphism is trivial and its image lies in $A_5$. Equal orders then giveFor $q=5$, use an explicit subgroup to obtain a five-point action. In $S{L}_2(5)$ setAll three have determinant one. Matrix multiplication gives $i^2=j^2=-I$, $ij=-ji$, and $r^3=I$. The eight distinct matrices $\{\pm I,\pm i,\pm j,\pm ij\}$ form a quaternion group. Moreover,Thus $r$ normalizes that quaternion group and has order three. Their generated subgroup has order $24$ and contains the scalar group center $\{\pm I\}$. Its image in $PS{L}_2(5)$ has order twelve, hence index five since $|PS{L}_2(5)|=5(25-1)/2=60$. The resulting coset action is nontrivial and, by the simplicity already proved, faithful. Perfectness again puts its image in $A_5$, and equal orders implyThis is the quaternion construction of an index-five subgroup of PSL2 over F5; it avoids assuming a classification of groups of order sixty.

For the last comparison, the order of a general linear group over a finite field comes from choosing independent columns:The determinant map onto ${\mathbb{F}}_q^{\times }$ has kernel $S{L}_n(q)$, and the scalar quotient divides by $\gcd (n,q-1)$. HenceThe highest power of two dividing either order is $64$. Their Sylow $2$-subgroups are respectively the upper unitriangular groups $U_4(2)$ of order $2^6$ and the image of $U_3(4)$ of order $4^3$. The latter image is isomorphic to $U_3(4)$ because its intersection with the odd-order scalar group center is trivial.

The center of an upper unitriangular group can be computed directly here. Commuting $X\in U_4(2)$ with $I+E_{12}$ forces $X_{23}=X_{24}=0$; commuting with $I+E_{23}$ forces $X_{12}=X_{34}=0$; commuting with $I+E_{34}$ forces $X_{13}=0$. Only $X_{14}$ can remain, and those matrices commute with the whole group. Thus $Z(U_4(2))=\{I+t{E}_{14}:t\in {\mathbb{F}}_2\}$ has order two. In $U_3(4)$, commutation with $I+E_{12}$ and $I+E_{23}$ similarly forces $X_{23}=X_{12}=0$, leaving the arbitrary entry $X_{13}\in {\mathbb{F}}_4$. Its group center has order four. An isomorphism of the whole groups would send a Sylow subgroup to a Sylow subgroup and preserve its group center order, which is impossible. Therefore