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For the fixed degree-two irreducible polynomial $f$, a matrix with minimal polynomial a power of $f$ has only one primary component. Its conjugacy classes correspond to integer partitions $\lambda$ of $m$. By the primary matrix centralizer formula, its centralizer has exactly the same order $c_{\lambda }(q^2)$ as the centralizer of the corresponding unipotent matrix in $G{L}_m(q^2)$.

Sum the conjugacy class sizes:The same sum times $|G{L}_m(q^2)|$ is the number of unipotent matrices of $G{L}_m(q^2)$, which 5(c) shows is $(q^2{)}^{m^2-m}$. Hence counting matrices with a fixed primary polynomial givesIn particular the exponent $a$ of the minimal polynomial is not fixed in advance: summing all integer partitions includes every possible largest block size. The usual statement takes $m\ge 1$; dimension zero can instead be included with its identity convention.

Take finite-dimensional modules over $k$ and finite $R$-free lattices over the complete valuation ring, the usual integral modular representation theory convention. If a completion has several local factors, use the factor supporting the modular block. A block of a group algebra is an indecomposable two-sided ideal $B=RGb$, where $b$ is a primitive central idempotent. The identity decomposes as the sum of orthogonal block idempotents, and $M={⨁}_{b}bM$ for every module. Hence an indecomposable module belongs to $B$ exactly when $bM=M$.

A defect group of a block in characteristic $p$ is a maximal p-subgroup $D$ with ${\mathrm{Br}}_D(b)\ne 0$. Equivalently, it is minimal with $b\in {\mathrm{Tr}}_D^G((kG{)}^D)$; the trace criterion for defect groups and conjugacy are proved below. For a complete discrete valuation ring, use the corresponding reduced modular block; modular reduction of block idempotents preserves its defect groups. The equivalent bimodule description says that the module vertices of $B$, with $(x,y)$ acting by $a\mapsto xa{y}^{-1}$, are conjugates of $\Delta D=\{(d,d):d\in D\}$. This basic characterization follows from the relative-projectivity trace criterion for the bimodule action.

Fix a Sylow subgroup $P$ containing $D$. The regular group algebra is the permutation module $R[G]$ for $G\times G$, and multiplication by $b$ is its block projection. Restriction to a subgroup containing a chosen module vertex retains a summand with that module vertex: a module source at that module vertex is a summand of the module restriction, and the Mackey restriction formula and minimality prevent all the restricted summands from having smaller module vertices. Thus $B{\downarrow }_{P\times P}$ has an indecomposable summand with module vertex $\Delta D$.

For a finite p-group $Q$, the transitive permutation module $k[Q/L]$ is indecomposable: the group algebra of a p-group in characteristic p is local, and this module has a one-dimensional top, incompatible with a sum of two nonzero modules. Its module vertex is $L$, because it is induced from the trivial $L$-module and its Brauer quotient of a module at $L$ has the nonzero fixed coset $L$. The fixed-basis rule for this quotient follows by removing nonsingleton orbit sums, which are proper-subgroup traces. It excludes a smaller module vertex. Reduction gives the same statements for permutation lattices. Therefore Krull-Schmidt decomposition identifies the chosen block summand with an entire transitive $P\times P$ orbit in $G$.

Choose a point $c$ in that orbit whose stabilizer is exactly $\Delta D$. Being fixed by every $(d,d)$ gives $c\in C_G(D)$. Its full stabilizer isEquality with $\Delta D$ gives $P\cap c^{-1}Pc=D$. Conjugating, and using that $c$ centralizes $D$, provesThis is defect groups are centralizer-conjugate Sylow intersections.

If $Q◃G$ is a p-subgroup, then $QP$ is a p-subgroup for every Sylow $P$, so $QP=P$. It lies in both Sylow factors above. Thus every normal p-subgroup of $G$ is contained in $D$.

Finally put $N=N_G(D)$, choose $S\in {\mathrm{Syl}}_p(N)$, and extend it to a Sylow $P$ of $G$. Then $D\le S$ and $P\cap N=S$. Apply the proved intersection result to this $P$. Its element $c$ belongs to $C_G(D)\le N$, so $cP{c}^{-1}\cap N=cS{c}^{-1}$ is Sylow in $N$ too. Hence $D=S\cap cS{c}^{-1}$. The p-core $O_p(N)$ is contained in both factors, whereas normality of $D$ in $N$ gives the opposite inclusion. Therefore $D=O_p(N_G(D))$, so $D$ is a p-radical subgroup. The centralizer condition on the conjugating element is important for this deduction.

The Brauer morphism retains precisely the fixed group-basis elements under conjugation by $D$:It is unital and surjective because $k{C}_G(D)\subseteq (kG{)}^D$. For multiplicativity, fix $x\in C_G(D)$. Its coefficient in a product is ${\sum }_{gh=x}{a}_g{b}_h$. Conjugation by $D$ permutes these pairs and the summand is constant on each orbit. Nonsingleton orbits have p-divisible size and contribute zero. Singleton pairs have both entries in $C_G(D)$, so they give exactly the coefficient in ${\mathrm{Br}}_D(a){\mathrm{Br}}_D(b)$. This proves the ring-homomorphism assertion.

Orbit sums form a basis of $(kG{)}^D$. A nonfixed orbit sum of $g$ is ${\mathrm{Tr}}_{C_D(g)}^D(g)$, a proper-subgroup relative trace. Conversely any proper-subgroup trace has zero coefficient on a fixed basis element, since that coefficient is multiplied by $[D:Q]=0$ in $k$. ThusThere is no assertion that this whole kernel is nilpotent.

Brauer first main theorem bijects the modular blocks of $kG$ with defect $D$ and those of $k{N}_G(D)$ with defect $D$, matching their nonzero Brauer images. Here is a proof with its auxiliary ingredients.

Write $N=N_G(D)$, $C=C_G(D)$ and defineThese are ideals in the indicated commutative algebras, since an invariant multiplier passes through a relative trace. The Brauer morphism and relative trace identityfollows by letting $D$ act on $G/D$. Fixed cosets have representatives in $N$; projection onto $C$ gives equal contributions on each other orbit, whose p-divisible size makes the contribution zero. Thus ${\mathrm{Br}}_D$ surjects $I_D(G)$ onto $J_D$.

To prove the trace criterion for defect groups, choose a p-subgroup $E$ of least order with a given block idempotent $b$ in $I_E(G)$. Such a choice exists because a Sylow $P$ gives $b={\mathrm{Tr}}_P^G(b/[G:P])$. Write $b={\mathrm{Tr}}_E^G(a)$ and replace $a$ by $ba$. If ${\mathrm{Br}}_E(b)=0$, the kernel formula and trace transitivity give $b\in {\sum }_{Q<E}{I}_Q(G)$. But $bZ(kG)$ is an Artinian local ring, and a finite sum of proper ideals cannot contain its identity. Multiplication by $b$ therefore forces $b\in I_Q(G)$ for some smaller $Q$, a contradiction. Hence ${\mathrm{Br}}_E(b)\ne 0$.

If $F$ is not conjugate into $E$, its action on $G/E$ has no fixed coset and the same orbit argument gives ${\mathrm{Br}}_F(I_E(G))=0$. Every subgroup with nonzero Brauer image of $b$ is therefore conjugate into $E$. Its maximal such subgroups are exactly the conjugates of $E$, proving the definition's conjugacy assertion andAmong block idempotents in this ideal, the Brauer image at $D$ survives precisely when their defect is $D$.

We also use primitive idempotents under a surjection of commutative Artinian ideals. Decompose the ambient algebras into local factors. The component of an ideal in a local factor is either the whole factor, possessing its identity, or a proper ideal in its nilpotent radical, possessing no nonzero idempotent. A nonzero quotient of a local algebra is local. A surjection of these ideals consequently bijects its surviving primitive idempotents with those of the target. Entire factors may be killed; the whole kernel need not be nilpotent.

Apply this result to $I_D(G)\to J_D$. Its surviving primitive idempotents are exactly the modular blocks of $G$ with defect $D$. Apply it also to $I_D(N)\to J_D$, where the target is identical since $N_N(D)=N$ and $C_N(D)=C$. Its surviving primitive idempotents are exactly the modular blocks of $N$ with defect $D$. Matching images in $J_D$ proves Brauer first main theorem, in both directions and with uniqueness.

For the last construction, the intended hypothesis is that $D$ is a defect group of the source modular block $b$. Put $L=N_H(D)$ and use the theorem inside $H$ to obtain the corresponding modular block $c$ of $kL$. The centralizer condition gives $C_H(D)=C_G(D)$, so ${\mathrm{Br}}_D$ maps $Z(kG)$ into $Z(kL)$. Let ${\omega }_c:Z(kL)\to F_c$ be the central character of a block, retaining its residue field $F_c$ rather than assuming it equals $k$. The unital compositesends exactly one global block idempotent to one, because their orthogonal idempotent images sum to one in a field. Define $b^G$ to be this unique modular block. This is block induction, independent of the chosen $H$-conjugate of $D$. It needs no containment of $N_G(D)$ in $H$, and its defect may be larger than $D$. With that containment it agrees with the first-main-theorem correspondence.

The printed PDF does not require $D$ to be a defect group of $b$. Its literal conditions do not always define the usual induced modular block. For example, over an algebraically closed field of characteristic three, take $G=A_5$, $H=A_4$ as a point stabilizer, and $D=⟨(123)⟩$. Then $D{C}_G(D)=D\le H$. The ordinary degree-three irreducible of $A_4$ gives a defect-zero modular block $b$ on reduction. The two distinct ordinary degree-three irreducibles of $A_5$ give distinct defect-zero modular blocks $B_{+}$ and $B_{-}$. Both restrict to that same character of $A_4$, with values $3,-1,0$ on its identity, double transpositions and three-cycles.

The standard defect-zero lifting theorem says that these reductions are simple projective modules and their block algebras are full matrix algebras. Thus both module restrictions $B_{+}{\downarrow }_{H\times H}$ and $B_{-}{\downarrow }_{H\times H}$ contain the same modular block bimodule $b$; indeed each module restriction is isomorphic to it here. The actual defect of $b$ is 1, whose centralizer is $G$, not a subgroup of $H$. The unrelated subgroup $D=C_3$ cannot select between the two modular blocks. This is block induction requires the block's defect centralizer; the qualified construction above answers the intended request.

The conjugate group elements relation is $x\sim y$ when $y=gx{g}^{-1}$ for some $g\in G$. It is reflexive by taking the identity, symmetric because $x=g^{-1}yg$, and transitive because $y=gx{g}^{-1}$ and $z=hy{h}^{-1}$ imply $z=(hg)x(hg{)}^{-1}$. Hence it is an equivalence relation.

For a finite group, the centralizer $C_G(x)=\{g:gx=xg\}$ is a subgroup. The map $g\mapsto gx{g}^{-1}$ has equal values at $g,h$ exactly when $h^{-1}g\in C_G(x)$. Its fibers are therefore the left cosets of $C_G(x)$, givingBy Lagrange's theorem, this index divides $|G|$. This is also the orbit-stabilizer theorem for the conjugation action, with the centralizer as stabilizer.

Work with finite groups, as required for these Sylow subgroup counts. Let $P$ act by conjugation on the set of Sylow $p$-subgroups. The stabilizer of $Q$ is $P\cap N_G(Q)$. Since $Q$ is the normal Sylow subgroup of its own normalizer, every $p$-subgroup of that normalizer lies in $Q$. Consequently the stabilizer is exactly $P\cap Q$, and the orbit-stabilizer theorem gives orbit size $[P:P\cap Q]$. The orbit of $P$ has size one. Every other orbit size is a power of $p$ at least $p^a$, hence is divisible by $p^a$. Adding orbit sizes proves the strengthened Sylow congruence from intersections:Now suppose $G$ is simple of order $2^e\cdot 15$. If $e=0$, its Sylow $5$-subgroup is normal because its number divides three and is one modulo five. If $e=1$, an involution in the regular permutation action of $G$ swaps fifteen pairs, so its sign is negative. The sign of a permutation would give a nontrivial group homomorphism $G\to C_2$ with a proper nontrivial normal kernel. Both cases contradict simplicity. Thus $e\ge 2$.

The Sylow theorems make $n_2$ a divisor of fifteen. It is not one, since a nontrivial Sylow subgroup would then be normal. It is not three, since the conjugation action on three Sylow subgroups would give a faithful group homomorphism $G\hookrightarrow S_3$, impossible by order. If $n_2=5$, take a Sylow subgroup $P$ itself: its normalizer has subgroup index five. If $n_2=15$, not all distinct Sylow subgroups can intersect $P$ in subgroup index at least four, since that would give $15\equiv 1(\mathrm{mod}4)$. Choose $Q\ne P$ with $[P:P\cap Q]=2$. Put $R=P\cap Q$, of order $2^{e-1}$. It has subgroup index two in both $P,Q$, so is normal in each. Hence $N_G(R)$ contains both distinct Sylow subgroups and has order greater than $2^e$. It is proper: otherwise $R$ would be a nontrivial proper normal subgroup of $G$. Its order is therefore $2^e\cdot 3$ or $2^e\cdot 5$. The latter would give an index-three subgroup and again an impossible faithful action on three cosets. Thus $[G:N_G(R)]=5$.

In either case we have the requested subgroup of order $2^e$ or $2^{e-1}$ with index-five normalizer. The action on the five cosets of that normalizer is nontrivial and faithful by simplicity. Its image lies in $A_5$, since composing with sign cannot give a nontrivial map from this simple group of order greater than two to $C_2$. Thus $2^e\cdot 15$ divides $60$, forcing $e=2$ and equality of orders. This proves the simple groups with order a power of two times fifteen conclusion:For $S{L}_2(5)$, choose the first column: there are $5^2-1=24$ nonzero choices. For each, determinant one is a nonzero linear equation in the second column and has five solutions. Therefore $|S{L}_2(5)|=120$. A central matrix commuting with $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ has zero lower-left entry and equal diagonal entries; commuting also with $\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ makes the upper-right entry zero. It is scalar. Determinant one then givesTo establish the projective special linear group over the field with five elements isomorphism without assuming its simplicity, compute its classes. For determinant-one matrices, the characteristic polynomial is $X^2-tX+1$ with trace $t$. The possibilities over ${\mathbb{F}}_5$ exhaust the group:For trace zero the eigenvalues are $2,3$, so the determinant-one centralizer is the split diagonal torus of order four. For traces $\pm 1$, the discriminant is the nonsquare $2$: the centralizer is the norm-one subgroup of ${\mathbb{F}}_{25}^{\times }$, of order six. In both semisimple cases the determinant map on the full linear centralizer is surjective onto ${\mathbb{F}}_5^{\times }$; consequently each trace type is a single special-linear class. For the noncentral traces $\pm 2$, a Jordan block has full linear centralizer $\left\{\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right):a\ne 0\right\}$ with determinant $a^2$. The determinant image consists of the two squares, so each general-linear class splits into two special-linear classes; the determinant-one centralizer has order $2\cdot 5=10$. These sizes sum to $120$.

Quotienting by $Z$ identifies a matrix with its negative. The trace-zero class gives one projective class of size fifteen; the two traces $\pm 1$ give one class of size twenty; the four noncentral unipotent classes give two classes of size twelve. Their respective projective orders are $2,3,5,5$, by the characteristic polynomials. A normal subgroup is a union of classes containing the identity. The nontrivial proper possible union sizes are $13,16,21,25,28,33,36,40,45,48$, none of which divides $60$. Thus the quotient is simple. Applying the order-$60$ result above givesFinally, if $A^2=I$ over an odd-characteristic field, its minimal polynomial divides the square-free polynomial $(X-1)(X+1)$, so it is diagonalizable with eigenvalues $\pm 1$. In dimension two, determinant one requires both eigenvalues equal. Hence $A=I$ or $A=-I$. This proves the unique involution in SL2 over an odd field assertion: the only element of order two in $S{L}_2(5)$ is $-I$. If an index-two subgroup $H$ existed, it would be normal and have order sixty. By Cauchy theorem it would contain an involution, hence contain $Z$. Then $H/Z$ would be a normal subgroup of order thirty in the simple quotient $PS{L}_2(5)$, a contradiction. Thus $S{L}_2(5)$ has no index-two subgroup.

First prove Simplicity of alternating groups. The $3$-cycles generate $A_n$: express an even permutation as an even number of transpositions and pair them. A pair sharing a point is a $3$-cycle, an identical pair cancels, and a disjoint pair can be written $(ab)(cd)=(acb)(acd)$. All $3$-cycles are conjugate in $A_n$ when $n\ge 5$: a conjugator in $S_n$ can have its parity corrected by a transposition on two points outside the target triple, without changing its effect on that triple.

Let $1\ne N◃A_n$ and choose $1\ne \sigma \in N$. Some $3$-cycle $\tau$ does not commute with $\sigma$. Indeed, commuting with every $3$-cycle would preserve every triple setwise; intersecting all triples containing a given point forces that point to be fixed, so would make $\sigma =1$. The nonidentity group commutator $\sigma \tau {\sigma }^{-1}{\tau }^{-1}\in N$ is supported on at most six points. Its possible nontrivial even cycle types are a $3$-cycle, a double transposition, a $5$-cycle, two disjoint $3$-cycles, or a $4$-cycle times a transposition. Each yields a $3$-cycle in $N$ by explicit group commutators:

For $(ab)(cd)$, use $\tau =(cde)$ with a fifth point $e$; conjugation reverses $\tau$, so the group commutator is ${\tau }^{-2}=\tau$. For $(abcde)$, use $(abc)$: the group commutator is $(bcd)(abc{)}^{-1}=(adb)$. The same calculation works for $(abcd)(ef)$. For $(abc)(def)$, use $(abd)$; its conjugate is $(bce)$, and the product $(bce)(abd{)}^{-1}$ is a $5$-cycle, reducing to the preceding case. Thus $N$ contains a $3$-cycle, hence the whole conjugacy class of $3$-cycles, and hence all of $A_n$. ThereforeFor the exceptional automorphism, a duad is an unordered pair of six points, a syntheme is a partition into three duads, and a pentad is a set of five synthemes covering every duad once. There are fifteen synthemes. To count pentads, fix $12|34|56$, where bars separate duads. A syntheme disjoint from it in duads is one of eight; the union of the two matchings is a six-cycle, so the stabilizer of the fixed syntheme is transitive on these eight choices. For the choice $13|25|46$, the remaining compatible synthemes are $14|26|35$, $15|24|36$, $16|23|45$ and $16|24|35$. The last shares a duad with each of the first three and cannot belong to a completion; the first three form the unique completion. Thus each of the eight choices has a unique completion through the fixed syntheme. Each completion contains four of them, so exactly two pentads contain a fixed syntheme, and there are $15\cdot 2/5=6$ pentads in total.

Here is the complete list, which also defines concrete labels for the induced action:Relabelling points gives a group homomorphism $\Phi :S_6\to S_6$ on these pentads. Direct substitution in the table givesThe restriction to $A_6$ is nontrivial, hence faithful by the simplicity just proved. Therefore the full kernel has order at most two. An order-two normal subgroup would be central, whereas $S_6$ has trivial centre: an element centralizing every transposition fixes every unordered pair and consequently every point. The kernel is trivial, and equal orders make $\Phi$ an automorphism. It preserves $A_6$, because that is the unique index-two subgroup of $S_6$: any nontrivial group homomorphism to $C_2$ sends every conjugate transposition to the nonidentity element, and is therefore the sign map. Its restriction is the pentad construction of the exceptional alternating-group automorphism. It maps a $3$-cycle to two disjoint $3$-cycles, while conjugation in $S_6$ preserves cycle type. Hence this automorphism of $A_6$ is not induced by any element of $S_6$.

For the other degrees, recover the points from $3$-cycles. An element of order three has type $3^k{1}^{n-3k}$. Its centralizer in $S_n$ has order $3^{k}k!(n-3k)!$. That centralizer contains an odd permutation: either swap two fixed points, or, when $k\ge 2$, swap two three-cycles by three transpositions. ThusFor $n=5$, only $k=1$ is possible. For $n\ge 7$, the $k=1$ value is strictly largest. For $k=2$ equality first occurs at $n=6$, and the ratio increases strictly with $n$. For $k\ge 3$, already at $n=3k$ one has $(3k-3)!>3^{k-1}k!$: it holds at $k=3$, and the inductive ratio is $(3k)(3k-1)(3k-2)>3(k+1)$. Increasing $n$ again increases the ratio. Consequently every automorphism preserves the set of $3$-cycles when $n\ne 6$.

Each cyclic subgroup generated by a $3$-cycle corresponds to its three-point support. Two distinct such subgroups commute exactly when the supports are disjoint. If they meet in one point, products of their nonidentity generators have order five. If they meet in two points, those products have order two or three. These intrinsic tests recover adjacency in the Johnson graph $J(n,3)$, so the automorphism acts on its vertices.

We can reconstruct the underlying points without an additional group-theoretic bound. The maximal cliques of a Johnson graph in $J(n,3)$ are the $n-2$ triples containing a fixed pair, and the four triples contained in a fixed four-set. To verify the classification, take adjacent triples $D\cup \{a\}$ and $D\cup \{b\}$ with $|D|=2$. Any common adjacent triple either contains $D$ or lies in $D\cup \{a,b\}$. A triple of the first kind outside that four-set is not adjacent to a triple of the second kind omitting a point of $D$, so a clique cannot mix the two alternatives. Extending a clique therefore gives exactly one of the two listed families.

For $n\ge 5$, $n\ne 6$, their sizes distinguish the pair-containing cliques. Their intersections reconstruct the pair graph $J(n,2)$: two such cliques intersect precisely when their defining pairs meet in one point. In this graph the maximal cliques are stars of $n-1$ pairs through a point and triangles of three pairs within a triple. These sizes are different for $n\ge 5$, so its automorphisms permute the stars and hence the points. Every pair is the intersection of its two point stars, and every triple is determined by its three pairs. Thus the original graph action comes from some $\pi \in S_n$.

Undo conjugation by $\pi$. The remaining group automorphism $\beta$ fixes each cyclic $3$-subgroup, so it fixes or inverts each generator. For adjacent supports, inverting exactly one generator changes the order of their product between two and three; inverting both preserves that order. Hence the two inversion choices must agree. The triple-support graph is connected, since one can replace differing points one at a time, so the choice is uniform. Inverting every $3$-cycle is impossible: take $c=(123)$, $d=(142)$, whose product is the $3$-cycle $(143)$. Then $\beta (cd)=(cd{)}^{-1}=d^{-1}{c}^{-1}$, whereas the group homomorphism property would give $c^{-1}{d}^{-1}$; these are unequal because $c,d$ do not commute. Therefore $\beta$ fixes every $3$-cycle and is the identity. Conversely, conjugation by $S_n$ acts faithfully on $A_n$, since centralizing all $3$-cycles fixes every triple and every point. The automorphisms of alternating groups from triple supports are exactly

Choose a minimal nontrivial normal subgroup $L$ of $K$ itself. Every conjugate $L^g$ for $g\in G$ is also a minimal normal subgroup of $K$, since $K$ is normal in $G$. Distinct such subgroups intersect trivially: their intersection is normal in $K$ and minimality forces either equality or trivial intersection. Moreover $[L_i,L_j]\subseteq L_i\cap L_j=1$, so distinct ones commute.

Choose a maximal family of these conjugates whose product $M=L_1\times \cdots \times L_r$ is direct. For any further conjugate $L^g$, its intersection with $M$ is normal in $K$ and is either all of $L^g$ or trivial. In the latter case it commutes with all the existing factors and could be adjoined to the direct product of groups, contrary to maximality. Thus every conjugate is contained in $M$. The product of all conjugates is consequently $M$, which is nontrivial and normal in $G$. Minimality of $K$ as a $G$-normal subgroup gives $M=K$.

Any normal subgroup of a factor $L_i$ is normal in $K$: the other direct factors commute with it, and $L_i$ normalizes it. Minimality of $L_i$ as a $K$-normal subgroup makes it simple. All factors are isomorphic because they are conjugate to $L$. This proves the direct-product structure of a finite minimal normal subgroup:An abelian simple group is cyclic of prime order, so the abelian case is elementary abelian. Now let $H$ be a maximal proper subgroup of a nonabelian simple group $X$. It is nontrivial, since $X$ has nontrivial proper cyclic subgroups. Choose a minimal normal subgroup $K$ of $H$. Then $H\le N_X(K)$. This normalizer cannot be $X$, since that would make $1<K\le H<X$ normal in $X$. Maximality givesFor each of the two groups in this question, the allowed absence of proper nonabelian simple subgroups makes the factors of such a $K$ cyclic of prime order. Hence it suffices to examine normalizers of nontrivial elementary abelian subgroups.

In $A_5$, the possible primes are two, three and five. The Sylow subgroups for the last two primes are cyclic of orders three and five. Every involution is a double transposition; for $(12)(34)$ the centralizer in $A_5$ isthe Klein four-group fixing point five. Thus an elementary abelian $2$-subgroup is cyclic of order two or a Klein four-group. The normalizer of a cyclic order-two subgroup equals its centralizer and has order four, so is contained in a point stabilizer of order twelve. A Klein four-group's common fixed point is intrinsic, and the full point stabilizer $A_4$ normalizes it; its normalizer is therefore exactly that stabilizer.

An order-three subgroup has support a triple. Its normalizer preserves the triple and the remaining pair, and is $(S_3\times S_2)\cap A_5$, of order six. An order-five subgroup acts regularly on all five points. Label the points by ${\mathbb{F}}_5$, with the subgroup acting by translations. Its normalizer in $S_5$ is the twenty affine maps $x\mapsto ax+b$. The translations are even, whereas multiplication by a generator of ${\mathbb{F}}_5^{\times }$ is a four-cycle and is odd. Exactly half these maps are even, so the normalizer in $A_5$ has order ten.

The only candidates for maximal subgroups are thus the normalizers of orders twelve, ten and six. They are all maximal: none can lie in a larger candidate, except potentially an order-six subgroup in an order-twelve $A_4$. But an index-two subgroup of $A_4$ would contain every order-three element; there are eight such elements, plus the identity, already more than six. Hence that containment is impossible. Point stabilizers are conjugate; order-three and order-five subgroup normalizers each form one class by the Sylow theorems. Therefore the maximal subgroups of A5 form preciselyTheir numbers are five, six and ten respectively, since these proper maximal subgroups are self-normalizing in the simple group.

For $X=G{L}_3(2)$, choosing independent columns gives $|X|=(8-1)(8-2)(8-4)=168$. If $t$ is an involution, write $t=I+N$. Field characteristic two gives $N^2=0$, and $\mathrm{im}N\subseteq \ker N$ in dimension three forces its nonzero rank to be one. Thus $t=I+vf$, with nonzero vector $v$, nonzero linear functional $f$ and $f(v)=0$. This transvection fixes the plane $\ker f$ and has image line $⟨v⟩$. Every flag consisting of an incident line and plane occurs, giving $7\cdot 3=21$ involutions, all conjugate. Its centralizer, and hence the normalizer of its cyclic subgroup, is the flag stabilizer of order $168/21=8$. It lies in both the line and plane stabilizers of order $24$, so cannot be maximal.

For the elementary abelian subgroups in GL3 over F2, consider commuting transvections $I+vf$, $I+wh$. Their rank-one parts satisfyIf $v,w$ are independent, equality forces both sides zero; thus $f,h$ vanish on $⟨v,w⟩$ and are the same nonzero functional. If $v=w$, they have a common image line. Consequently the two types of order-four elementary abelian subgroup areNo extra transvection can commute with all elements of $K_L$: if its image were different, its functional would have to equal each of the two independent functionals occurring in $K_L$. Dually, the same holds for $K_W$. Thus these exhaust the elementary abelian $2$-subgroups of rank at least two. The normalizer of $K_L$ is the line stabilizer and the normalizer of $K_W$ is the plane stabilizer, each of order $168/7=24$. They give different conjugacy classes because their common fixed spaces have dimensions one and two. All line stabilizers are conjugate, as are all plane stabilizers.

For an order-three element, its minimal polynomial is $(X-1)(X^2+X+1)$: it has a fixed line and an invariant complementary irreducible plane. Its normalizer preserves both and is the full $G{L}_2(2)$ on that plane, of order six, since the order-three subgroup is normal in $G{L}_2(2)$. This normalizer lies in the corresponding line and plane stabilizers, so is not maximal.

An order-seven element is irreducible. Over the finite field ${\mathbb{F}}_2$ one has $X^7-1=(X-1)(X^3+X+1)(X^3+X^2+1)$, and both cubic factors are irreducible because neither has a root in ${\mathbb{F}}_2$. Its minimal polynomial divides this square-free product and cannot be just $X-1$, since the element is not the identity. It therefore contains a cubic factor; the three-dimensional space forces that irreducible cubic to be the whole minimal polynomial. Identify the space with ${\mathbb{F}}_8$ and the element with multiplication by a generator $\alpha$ of ${\mathbb{F}}_8^{\times }$. Its centralizer consists of the seven nonzero multiplication maps. A normalizer induces an ${\mathbb{F}}_2$-automorphism of this field, so it sends $\alpha$ to $\alpha$, ${\alpha }^2$ or ${\alpha }^4$. All three possibilities occur via the field automorphisms $x\mapsto x^{2^j}$. Therefore the normalizer of the Singer cycle has order $7\cdot 3=21$, and all such normalizers are conjugate by the Sylow theorems.

The maximal-normalizer reduction leaves exactly the two order-twenty-four stabilizer types and the order-twenty-one type. Each is maximal because it cannot be contained in any larger remaining candidate; equal-order subgroups cannot properly contain one another and $21$ does not divide $24$. Thus

For $H\le G$ of finite subgroup index, let $H^{\prime }=[H,H]$ and choose a right-coset transversal $T$ for the cosets $Ht$. For each $g\in G$ writeThe transfer homomorphism isThe product is taken in the abelianization, so its order is irrelevant. If another transversal has representative $a_{t}t$ for $Ht$, with $a_t\in H$, its factor is $a_{t}h(t,g)a_{t_g}^{-1}$. The map $t\mapsto t_g$ permutes $T$, so the extra factors cancel in $H/H^{\prime }$. This proves independence of the transversal. AlsoMultiplying over $t$ and again permuting the second indices proves $V(g_1{g}_2)=V(g_1)V(g_2)$. Thus it is indeed a group homomorphism.

The Burnside transfer theorem states that if a Sylow $p$-subgroup $P$ of a finite group satisfies $P\subseteq Z(N_G(P))$, then $G$ has a normal $p$-complement: a normal subgroup of order $|G|/|P|$. The hypothesis first makes $P$ abelian. It also makes fusion inside $P$ trivial. To prove this, suppose $u,v\in P$ and $u^g=v$, using $u^g=g^{-1}ug$. Both $P$ and $P^g$ lie in $C_G(v)$: the former because $P$ is abelian, and the latter because $v\in P^g$. They are Sylow subgroups of that centralizer. Choose $c\in C_G(v)$ with $(P^g{)}^c=P$. Then $gc\in N_G(P)$ and $u^{gc}=v^c=v$, while the central-normalizer hypothesis gives $u^{gc}=u$. Thus $u=v$.

Evaluate transfer to the abelian group $P$ on $u\in P$. In the action of $u$ on right cosets, a cycle of length $r$ contributeswith $t_r=t_0$. This is a conjugate of $u^r$ inside $P$ and hence equals $u^r$ by the fusion argument. Summing the cycle lengths therefore givesThe exponent is coprime to $p$, so this power map is an automorphism of the finite abelian $p$-group $P$. Transfer is consequently onto $P$, and its kernel is normal of subgroup index $|P|$ and of order prime to $p$. This proves the theorem. Notice also that a normal $p$-complement consists exactly of the elements of prime-to-$p$ order: they map trivially to the $p$-group quotient, while all elements of the complement have such order. It is therefore unique and a characteristic subgroup.

Now let $G\le A_p$ be primitive of degree $p=2q+1$, with $q$ prime. Every nontrivial normal subgroup of a faithful primitive group action is transitive: its orbits form a $G$-invariant block partition, and singleton orbits would make it trivial. Such a subgroup has order divisible by $p$. A Sylow $p$-subgroup $P$ has order $p$, since $p!$ contains only one factor $p$. The normalizer of a regular $p$-cycle in $S_p$ is the affine group $x\mapsto ax+b$ on ${\mathbb{F}}_p$, of order $p(p-1)$. Translations are even, whereas a generating multiplier gives a $(p-1)$-cycle and is odd. Its normalizer in $A_p$ therefore has order $p(p-1)/2=pq$.

Suppose $G$ is not simple and choose $1<N◃G$ proper. It is transitive, so we can choose $P\le N$. The Frattini argument gives $G=N{N}_G(P)$: for each $g$, the subgroup $P^g$ is a Sylow subgroup of $N$ and can be conjugated back to $P$ within $N$. Since $N_G(P)$ contains $P$ and lies in the order-$pq$ normalizer, it has order $p$ or $pq$. If its order were $p$, the Burnside theorem applied to $G$ would give a normal $p$-complement. A nontrivial such complement cannot be transitive because its order is prime to $p$, contradicting primitivity. It must be trivial, giving $G=P$, contrary to the choice of $N$. Thus $|N_G(P)|=pq$.

The intersection $N_N(P)$ again has order $p$ or $pq$. If it had order $pq$, the whole normalizer would lie in $N$ and the Frattini equality would give $G=N$. Hence $N_N(P)=P$. Applying Burnside transfer within $N$ gives its normal $p$-complement $K$. By the uniqueness argument, $K$ is a characteristic subgroup of $N$, and so normal in $G$. Primitivity again forces this prime-to-$p$ subgroup to be trivial. Thus $N=P$ is normal in $G$, and $G=N_G(P)$ has order $pq$. The even primitive groups of safe-prime degree conclusion is

For the centralizer of a subset, the identity commutes with every $h\in A$. If $g,k$ commute with each member of $A$, then for every such $h$,Also $gh=hg$ implies $g^{-1}h=h{g}^{-1}$ after multiplication by $g^{-1}$ on both sides. Thus identity, products and inverses lie in $C(A)$, proving that it is a subgroup.

For the conjugation action, identity acts trivially, andThese are the defining laws of a left group action. Its orbits of a group action are the conjugacy classes. The stabilizer of $h$ is exactly its centralizer, since $gh{g}^{-1}=h$ if and only if $gh=hg$. The orbit-stabilizer theorem therefore givesA class has one element exactly when that element commutes with all of $G$, so the singleton classes are precisely the elements of $C(G)$, the centre of a group. Partitioning $G$ into all its conjugacy classes now proves the class equationFor $|G|=p^r$, Lagrange's theorem for finite groups makes every class size a power of $p$. A nonsingleton class has size at least $p$ and is therefore divisible by $p$. Reducing the class equation modulo $p$ yields $|C(G)|\equiv 0(\mathrm{mod}p)$, because $r>0$. The identity lies in $C(G)$, so its order is positive and divisible by $p$. ConsequentlyThis proves the nontrivial center of a finite p-group directly from the conjugation classes.

Suppose $1\ne N⊲G$ is proper. The defining property makes $N$ cyclic of order $p$. Conjugation gives a homomorphismwhose kernel is the centralizer $C_G(N)$. The automorphism group has order $p-1$, so the kernel has finite index in $G$. It contains $N$, because $N$ is abelian. If this kernel were proper, it would be an order-$p$ subgroup, and its finite index would force $G$ itself to be finite. Therefore $C_G(N)=G$, so $N$ is central.

Choose $h\notin N$. Part (i) shows that $h$ has order $p$. Its cyclic subgroup meets $N$ trivially, since two order-$p$ subgroups with nontrivial intersection are equal. Since $N$ is central,This subgroup has order $p^2$ and is proper because $G$ is infinite, contradicting the assumed order of every proper nontrivial subgroup. Thus no such $N$ exists:Together with part (i), this proves that Tarski monster groups are two-generated and simple.