Finite group theory

Finite group theory studies groups with finitely many elements.

These are all the groups of group order $70$ up to group isomorphism, with $D_n$ denoting a dihedral group of order $n$. The Sylow theorems make the Sylow subgroups of orders five and seven unique: the possible counts divide fourteen and ten, respectively, and only one has the required residue. These normal subgroups of coprime order commute, giving a normal subgroup $N\cong C_{35}$. Cauchy's theorem for finite groups provides an element of order two outside $N$, so $G=N⋊C_2$.

The automorphism of a finite cyclic group defining this semidirect product is a unit $a$ modulo $35$ with $a^2\equiv 1$. The Chinese remainder theorem makes its restrictions modulo five and seven independently $+1$ or $-1$. The resulting exponents are $1,34,6,29$, giving the four displayed groups in order. Their center of a group orders are $70,1,5,7$, so no two are isomorphic. Every case exists as the specified semidirect product.

Up to group isomorphism, the finite groups with at most six elements are the trivial group, the cyclic groups $C_2,C_3,C_4,C_5,C_6$, the Klein four-group $C_2\times C_2$, and the symmetric group $S_3$.

A group of prime cardinality is a cyclic group, since Lagrange's theorem forces any nonidentity element to generate it. For four elements, either an element has order of a group element four, or every nonidentity element is an involution. In the latter case the group of exponent two is abelian identity shows that two distinct nonidentity elements $a,b$ generate $\{e,a,b,ab\}\cong C_2\times C_2$.

For a group of order six, either an element generates it or some element $a$ has order of a group element three: otherwise all nonidentity elements would be involutions, producing a four-element subgroup in contradiction with Lagrange's theorem. Pairing each element with its inverse element gives an involution $b$ outside $⟨a⟩$. The subgroup $⟨a⟩$ has index of a subgroup two and is a normal subgroup, so $ba{b}^{-1}=a$ or $a^{-1}$. The commuting case makes $ab$ have order of a group element six, whereas the second case gives the multiplication rules of $S_3$ on the six distinct elements $a^i,a^{i}b$, $0\le i<3$.

All these groups embed in $S_4$ except $C_5$ and $C_6$. The former is excluded by Lagrange's theorem; the latter is excluded because the disjoint cycle decomposition of a permutation of four letters cannot have least common multiple of cycle lengths equal to six.

Every group of order six is isomorphic to the cyclic group $C_6$ or the symmetric group on three points $S_3$. Lagrange's theorem restricts element orders to $1,2,3,6$. If there is an element of order six, it generates the group. If every nonidentity element had order two, the group of exponent two is abelian result would produce a subgroup $\{e,a,b,ab\}$ of order four, contradicting Lagrange's theorem. Thus the noncyclic case has an element $x$ of order three.

There cannot be two distinct subgroups of order three: their intersection would be trivial and their nine pairwise products would be distinct elements of a six-element group. Hence exactly two elements have order three, and the other three nonidentity elements have order two. The subgroup $⟨x⟩$ has index two and is normal. Choose an involution $h$ outside it. Conjugation sends $x$ to $x$ or $x^{-1}$. The first case would make $xh$ have order six, so $hx{h}^{-1}=x^{-1}$. The generators therefore obey $x^3=h^2=e$, $hx{h}^{-1}=x^{-1}$, the relations realized by $(123)$ and $(12)$ in $S_3$.

The largest nilpotent normal subgroup of a finite group $G$. If $O_p(G)$ denotes its p-core, distinct p-cores commute, and $F(G)={\prod }_p{O}_p(G)$. Every normal nilpotent group has Sylow subgroups which are characteristic subgroups, so it is contained in this product.

In a finite soluble group, $C_G(F(G))\le F(G)$. Put $C=C_G(F)$ and $D=C\cap F=Z(F)$. If $C>D$, choose an elementary abelian minimal normal subgroup of $G/D$ inside $C/D$, and let $K$ be its preimage. Then $K^{\prime }\le D\le Z(K)$, so $K$ is a nilpotent normal subgroup, contradicting $K\nleqq F$.

For a finite group, $F(G)$ centralizes its group socle. A nonabelian minimal normal subgroup meets $F(G)$ trivially: otherwise it is nilpotent, and its nontrivial characteristic center makes it abelian. Their group commutators therefore vanish. An abelian minimal normal subgroup $M$ is an elementary abelian p-group; $M\le O_p(G)$ and $M\cap Z(O_p(G))\ne 1$ by the fixed-point counting of a finite p-group on $M$. Minimal normality makes $M\le Z(O_p(G))$, and other p-cores commute with $M$.

A normal subgroup which is a nilpotent group. In a finite group its unique Sylow subgroups are characteristic subgroups and hence normal in the ambient group. Products of nilpotent normal subgroups are nilpotent groups: their normal Sylow subgroups of different prime characteristics commute.

The largest normal p-subgroup of a finite group. Products of normal p-subgroups are again normal p-subgroups, since $|AB|=|A||B|/|A\cap B|$. Thus the subgroup generated by all such subgroups is itself a p-subgroup. It is a characteristic subgroup.

A group of order ten is either a cyclic group or the dihedral group of the pentagon. Lagrange's theorem and the fact that a group of exponent two has power-of-two order force an element of order five or ten. In the latter case the group is cyclic. In the former case its order-five cyclic subgroup is normal by index two; an outside involution acts on it either trivially or by inversion. The trivial action gives $C_5\times C_2\cong C_{10}$, and inversion gives $D_{10}$.

A group between a nonabelian simple group and its automorphism group. Finite two-transitive groups are either of this type or affine two-transitive groups, a structural theorem rather than an order-divisibility test.

Upper-unitriangular three-by-three matrices over the prime finite field form the finite analogue of the real Heisenberg group. In coordinates their multiplication is $(a,b,c)(a^{\prime },b^{\prime },c^{\prime })=(a+a^{\prime },b+b^{\prime },c+c^{\prime }+a{b}^{\prime })$. The order is $p^3$, and $(a,b,c{)}^k=(ka,kb,kc+\left(\genfrac{}{}{0ex}{}{k}{2}\right)ab)$. For odd $p$, every nonidentity element has order $p$, but the group is nonabelian. For $p=2$ the exponent statement changes, so odd characteristic is essential for the nonisomorphic Gassmann equivalent regular subgroups construction.

Use the coordinate conventionIn this convention the Heisenberg group over a prime field has the productThe subset $H=\{A(t,b,t):t,b\in {\mathbb{F}}_p\}$ is an Abelian subgroup of order $p^2$: the middle coordinate in a product is $b+b^{\prime }+t{t}^{\prime }$, which is symmetric in the factors. Also $A(t,b,t{)}^{-1}=A(-t,t^2-b,-t)$ lies in $H$. The map $A(a,b,c)\mapsto a-c$ is a surjective group homomorphism to the additive group of ${\mathbb{F}}_p$, with kernel of a group homomorphism $H$. Hence $H$ is a normal subgroup of subgroup index $p$, and its left cosets are exactly the level sets of $a-c$.

For odd $p$, an explicit group isomorphism from $H$ to the additive group ${\mathbb{F}}_p^2$ isThe term $t{t}^{\prime }$ in matrix multiplication cancels the cross term in $(t+t^{\prime }{)}^2/2$. Thus $H$ is an elementary abelian group, isomorphic to $C_p\times C_p$.

A regular tetrahedron has a proper rotational symmetry group $T$ of order 12, isomorphic to the alternating group $A_4$ acting on its four vertices. Its full spatial symmetry group $T_d$ has order 24 and is isomorphic to the symmetric group $S_4$; the additional operations reverse orientation. Three-fold vertex/face-axis rotations and half-turns about opposite-edge midpoints generate the proper group. The distinction matters when a field symmetry uses only proper rotations, but a scalar density also admits reflections.

The orientation-preserving symmetries of a regular tetrahedron form the alternating group on its four vertices. There are twelve elements: the identity, eight rotational symmetries of order three, and three half-turns.

The full symmetry group of a regular tetrahedron is the symmetric group $S_4$ acting on its vertices. Its subgroups with at most six elements have eight types up to conjugation, with the following representative subgroups and counts:A transposition is a reflection through the plane containing the two fixed vertices and the midpoint of the exchanged edge. A double transposition is a half-turn about the axis joining the midpoints of the two exchanged edges. A three-cycle is a one-third-turn about a vertex-to-opposite-face axis. A four-cycle is a quarter-turn composed with a reflection perpendicular to the opposite-edge midpoint axis; it is an improper orthogonal transformation.

The unique Klein four-group containing only double transpositions consists of the three opposite-edge half-turns and the identity element. The other three Klein four-groups consist of two perpendicular mirror reflections, their half-turn product, and the identity element. Each $S_3$ is a stabilizer subgroup of a vertex, including the three rotations and three mirror reflections preserving that vertex.

For completeness, a Klein four-group containing $(12)$ must contain the commuting involution $(34)$ or $(12)(34)$, yielding the displayed subgroup; one containing no transpositions must contain all three double transpositions. An $S_3$ subgroup has a unique cyclic subgroup of cardinality three, whose unique fixed vertex must therefore be preserved by the entire $S_3$. These observations and the disjoint cycle decomposition exhaust the types allowed by groups of order at most six. The counts sum to $25$.

The identity and three half-turns form a normal Klein four-group $V_4$ inside the proper rotational tetrahedral symmetry group $A_4$. Conjugation preserves double-transposition cycle type, establishing normality. The quotient group has order three and is therefore cyclic. If $r$ is any 3-cycle, every element is uniquely $r^{j}k$ with $j\in \{0,1,2\}$ and $k\in V_4$. The formula $r^{j}k\mapsto e^{2\pi ij/3}$ defines a surjective group homomorphism with kernel $V_4$. Normality allows the factors from $V_4$ to be moved past powers of $r$, making exponents add modulo three and proving the homomorphism property directly.

The proper rotational tetrahedral symmetry group is $A_4$, acting by even permutations of the vertices. Its conjugacy classes consist of the identity, the three half-turns about opposite-edge midpoint axes, four rotations through $+2\pi /3$ about outward-pointing vertex axes, and their four inverses. A 3-cycle has a three-element centralizer, so its class has size $12/3=4$. Conjugating it to its inverse would require an odd permutation on its three moved vertices, so the two orientations form different classes in $A_4$ even though they lie in one class in $S_4$.

Up to group isomorphism, the groups of order eight are $C_8$, $C_4\times C_2$, $C_2^3$, $D_8$ and $Q_8$, with $D_8$ here denoting the eight-element dihedral group. An element of order eight gives the first case. If every nonidentity element has order two the group is an elementary abelian group. Otherwise, after excluding elements of order eight, a cyclic subgroup of order four has index two; an element outside it either commutes with its generator or inverts it. In the latter case its square is either the identity or the central involution, giving the dihedral group or quaternion group.

Every finite group of odd order is a solvable group. In particular, a nonabelian simple group has even order and therefore has an involution by Cauchy's theorem for finite groups. This is a deep theorem, not an elementary consequence of the Sylow theorems.

Every maximal proper subgroup normalizes one of its minimal normal subgroups. As $A_5$ has no proper nonabelian simple subgroup, that subgroup is elementary abelian. Its possible types are $C_2,C_2^2,C_3,C_5$. Their normalizers have orders $4,12,6,10$ respectively. The order-four normalizer lies in an order-twelve point stabilizer; the other three types are maximal and each forms a single conjugacy class. There are five, six and ten maximal subgroups of orders twelve, ten and six, respectively.

Except in degree six, the $3$-cycles are characterized among order-three elements by the largest centralizer. Their cyclic subgroups correspond to three-element supports. Two different such subgroups commute exactly for disjoint supports; with one common point their generator products have order five, and with two common points those products have order two or three. Thus an automorphism induces an automorphism of $J(n,3)$. The maximal cliques of a Johnson graph reconstruct a permutation of the points. After undoing that permutation, an automorphism fixes every cyclic $3$-subgroup. Product orders force its choices of generator inversion to be consistent on the connected graph, and inversion of every $3$-cycle would reverse a noncommuting product. Hence it fixes all $3$-cycles, which generate $A_n$.

A connected finite hypergraph of three-element supports, with a three-cycle on each support, generates the full alternating group. To merge a new triple meeting an old alternating support in two letters, conjugate over unordered pairs of old letters. For one-letter overlap compare $(abc)$ and $(a^{\prime }bc)$: their quotient is $(ab{a}^{\prime })$, reducing to two two-letter-overlap steps.

The alternating conjugacy class splitting criterion gives seven conjugacy classes in $A_6$. Their cycle types and sizes are $1^6:1$, $2^2{1}^2:45$, $31^3:40$, $3^2:40$, $42:90$, and two $51$ classes of size 72. Representatives of the split classes are a 5-cycle and its square. Their sizes sum to 360. A normal subgroup must be a union of these classes containing the identity, so these explicit sizes can prove simplicity using Lagrange's theorem.

The permutation group of a three-element set has six elements: the identity, three transpositions, and two three-cycles. A three-cycle and any transposition generate it, with relations $x^3=h^2=e$, $hx{h}^{-1}=x^{-1}$. Its alternating group is the normal cyclic subgroup of order three. It is the unique nonabelian isomorphism type of a group of order six.

The conjugacy classes of the symmetric group on three letters have sizes $1,3,2$: the identity, the three transpositions and the two 3-cycles. A normal subgroup is a union of whole conjugacy classes and contains the identity. Its candidate orders are therefore $1,4,3,6$. Lagrange's theorem excludes four, leaving precisely $\{e\}$, $A_3$ and $S_3$. The order-three subgroup is the cyclic group of the two 3-cycles and the identity; it is also the kernel of a group homomorphism of the sign map.

The first isomorphism theorem gives the corresponding quotients $S_3$, $C_2$ and the trivial group. Thus these are exactly the possible images of surjective group homomorphisms from $S_3$. Each occurs, by the identity map, the sign map and the constant identity map, respectively.

An automorphism preserving the transposition class is inner: commuting relations recover the vertex stars of the complete graph, with a small-case adjustment in degrees three and four. An involution class has the same size as the transposition class only in degree six, when triple transpositions also have size fifteen. Hence outer automorphism groups of symmetric groups are trivial outside degree six and have size at most two in degree six.

// Target: combinatorics.bigb

A permutation of order two is a product of $k$ disjoint transpositions. Its conjugacy class has the displayed size. For $k\ge 2$, equality with the transposition class requires $(n-2)!/(n-2k)!=2^{k-1}k!$. The left side grows strictly with $n$. At $n=2k$, its ratio to the right side is one only for $k=3$; at $k=2$ it jumps from below to above one between $n=4$ and $n=5$, and for $k\ge 4$ it is already greater than one. Thus degree six is the only possible exceptional degree for an group automorphism not preserving transpositions.

A subgroup of the symmetric group of a set, acting faithfully by permutations. A finite permutation group specified by generators can be enumerated by repeatedly multiplying the elements already found by those generators until no new element appears. Its orbits and conjugacy classes can then be computed directly.

The historical study of permutations of letters or quantities and their compositions, closely connected with the development of group theory and the theory of polynomial equations.

A permutation group on a set is oligomorphic when its diagonal action on ordered $n$-tuples has finitely many orbits for every positive integer $n$. For a countable model of a complete theory, an oligomorphic automorphism group makes every finite-arity type space finite: realized types are dense, and only finitely many are possible because types are constant on orbits.

For $n\ge 2$ and $1\le k<n$, an $n$-cycle $c=(12\dots n)$ and $t=(11+k)$ generate $S_n$ exactly when $\gcd (n,k)=1$. Their conjugate transpositions connect labels differing by $k$ modulo $n$. This graph is connected exactly at coprime separation, so transpositions on a connected graph generate the symmetric group. Otherwise residue classes modulo $\gcd (n,k)$ form a nontrivial block system preserved by both generators.

Associate a transposition to each edge of a finite connected graph. For a simple path $v_0,\dots ,v_{\ell }$, let $t_j=(v_{j-1}{v}_j)$. Then $t_1\cdots t_{\ell -1}{t}_{\ell }{t}_{\ell -1}\cdots t_1=(v_0{v}_{\ell })$, using rightmost-first composition. Thus edge transpositions generate every transposition, and hence the full symmetric group.

The parity of a permutation records whether it is a product of an even or odd number of transpositions. It is well-defined and equals the sign of the corresponding permutation matrix.

The sign homomorphism sends an even permutation to $1$ and an odd permutation to $-1$; its kernel is the alternating group.

A symmetric group is the group of all bijections from a set to itself, with function composition as its operation.

A permutation cycle $(a_1{a}_2\dots a_k)$ sends $a_i$ to $a_{i+1}$, sends $a_k$ to $a_1$, and fixes every other element. Its order is $k$.

For $n>1$, every permutation of order of a permutation $n$ in $S_n$ is one full cycle if and only if $n$ is a prime power. If $n=p^r$, all lengths in its disjoint cycle decomposition divide $p^r$, and their least common multiple is their maximum. One cycle therefore has length $n$ and uses every symbol. If $n$ has more than one prime divisor, write $n=ab$ with $a,b>1$ coprime by separating one full prime-power factor from the others. Since $a+b\le ab=n$, disjoint cycles of lengths $a$ and $b$ fit inside $S_n$. Their product has order of a permutation $\mathrm{lcm}(a,b)=n$, but is not a full cycle.

A disjoint cycle decomposition writes a permutation as a product of disjoint permutation cycles. Repeatedly apply the permutation to a point until it returns to that point; these successive points form one cycle. Repeat on points not yet used. Finiteness and injectivity ensure that each process returns to its starting point. This partitions the underlying finite set into cycles. The decomposition is unique up to reordering the cycles, cyclically rotating the entries within each cycle, and optionally omitting fixed-point one-cycles. Include the one-cycles when specifying cycle type.

The cycle type of a permutation is the multiset of lengths in its decomposition into disjoint permutation cycles, including one-cycles for fixed points. Two permutations in the same symmetric group are conjugate exactly when they have the same cycle type.

Conjugation preserves order of a group element, because $(gx{g}^{-1}{)}^k=g{x}^k{g}^{-1}$, and preserves the sign of a permutation, because sign is a group homomorphism into $\{1,-1\}$. These invariants do not determine cycle type. In $S_6$, $(123)$ and $(123)(456)$ both have order three and positive sign, but have three and zero fixed points respectively. Conjugation relabels cycles, so these permutations are not conjugate.

Squaring an odd-length permutation cycle retains its length, whereas an even-length cycle splits into two cycles of half its length. Thus if any even cycle is present, the number of cycles strictly increases, so the cycle type changes. Since conjugate permutations in a symmetric group have the same cycle type, a permutation is conjugate to its square precisely when all its cycle lengths are odd, equivalently when its order of a group element is odd.

Left multiplication of a permutation by a transposition $(ab)$ swaps the two target labels $a,b$. If they occur in different cycles, $(a{x}_1\dots x_p)(b{y}_1\dots y_q)$ becomes the single cycle $(a{x}_1\dots x_{p}b{y}_1\dots y_q)$. If they occur in the same cycle, that formula reverses and splits the cycle in two. All other cycles are unchanged, so the total number of cycles, including fixed points, decreases or increases by exactly one. Empty intermediate lists yield one-cycles and require no exceptional case.

A three-cycle is a permutation cycle of length three. On any chosen set of three letters, the two possible three-cycles are inverses of one another.

Permutation cycles with disjoint supports commute. The order of their product is the least common multiple of their lengths.

The support of a permutation is the set of elements it does not fix. A permutation has finite support when this set is finite.

The sign homomorphism $\mathrm{sgn}:S_n\to \{\pm 1\}$ is the determinant of the corresponding permutation matrix. A transposition has sign $-1$, so this also proves that the parity of any transposition decomposition is well-defined.

An odd permutation has sign $-1$, equivalently it is a product of an odd number of transpositions.

An even permutation has sign $+1$, equivalently it is a product of an even number of transpositions.

The quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$ is a nonabelian Dedekind group.

In the quaternion group, $-1=i^2=j^2=k^2$ commutes with each generator because any element commutes with its powers. Hence $N=\{1,-1\}$ is central and is a normal subgroup. The reverse products satisfy $ji=-k$, $kj=-i$, $ik=-j$, so none of $\pm i,\pm j,\pm k$ is central. Therefore $Z(Q_8)=N$.

The four cosets are $N,iN,jN,kN$. Each nonidentity coset squares to $N$, and $(iN)(jN)=kN$. Thus the quotient group is the Klein four-group. Explicitly, mapping $\pm 1$ to its identity and mapping $\pm i,\pm j,\pm k$ to its three different nonidentity elements is a surjective group homomorphism with group kernel $N$. The first isomorphism theorem gives the title result.

The quaternion group has conjugacy classes $\{1\}$, $\{-1\}$, $\{\pm i\}$, $\{\pm j\}$, $\{\pm k\}$, with sizes $1,1,2,2,2$. Over the complex numbers, the number of classes and the sum of squares of irreducible representation dimensions force four one-dimensional representations and one two-dimensional representation. For $ϵ,\eta \in \{\pm 1\}$, the linear character rows on these classes are $(1,1,ϵ,\eta ,ϵ\eta )$. Orthogonality then gives the remaining row $(2,-2,0,0,0)$. A representation with character $(4,-4,0,0,0)$ contains two copies of the two-dimensional irrep and none of the linear irreps, by the character inner product. The choice of scalar field matters: the corresponding quaternionic four-dimensional real representation is irreducible over the reals.

The least size of a finite set on which the quaternion group acts faithfully is eight. Every nontrivial subgroup of $Q_8$ contains its central involution $-1$. In an action on fewer than eight points, the orbit-stabilizer theorem makes every stabilizer subgroup nontrivial, so $-1$ fixes every point. This rules out a faithful group action, even with several group orbits. The regular action on eight points is faithful by Cayley theorem.

The dicyclic group of order $4n$ has generatorsEvery element has a unique form $a^k$ or $a^{k}b$, with $0\le k<2n$.

This dicyclic group has classes $1$, $a^3$, $\{a,a^5\}$, $\{a^2,a^4\}$ and the two parity classes of $b{a}^j$, of sizes $1,1,2,2,3,3$. Its quotient by $⟨a^2⟩$ is cyclic of order four, supplying four linear characters with $a\mapsto (-1{)}^j$, $b\mapsto i^j$. Its two remaining irreducible representations have $a\mapsto \mathrm{diag}({\zeta }^k,{\zeta }^{-k})$ and $b\mapsto \left(\begin{array}{cc}0& (-1{)}^k\\ 1& 0\end{array}\right)$ for $k=1,2$, where $\zeta =e^{i\pi /3}$. The traces are $2\cos (km\pi /3)$ on $a^m$ and zero on $b{a}^m$. Distinct diagonal eigenlines are exchanged by $b$, proving irreducibility; the degree-square sum $4+4+4=12$ proves completeness.

Four one-dimensional characters and $n-1$ two-dimensional characters exhaust a dicyclic group of order $4n$. In the two-dimensional rows, the values at $a^r$ are $2\cos (\pi jr/n)$ and those on the two noncyclic cosets are zero. The one-dimensional rows depend on whether $n$ is odd or even.

A finite p-group is a finite group whose order is a power of one prime number $p$.

Induct on the order of the finite p-group. Its center of a group is nontrivial unless the group itself is trivial. Quotient by the center to obtain a smaller finite p-group and lift a central series supplied by induction. Prepending the center gives a central series for the original group: the first term is central, and the remaining commutator-containment conditions are exactly the lifted quotient conditions. Hence every finite p-group is a nilpotent group. Since central factors are Abelian, every nilpotent group is also a soluble group.

A p-subgroup is a subgroup which is a finite p-group, for a specified prime $p$. Every p-subgroup lies in a Sylow subgroup.

A p-subgroup is p-radical if it equals the p-core of its normalizer. Every defect group of a block is p-radical by defect groups are centralizer-conjugate Sylow intersections. Being an intersection of arbitrary Sylow subgroups alone is weaker than the centralizer-conjugate statement used in that argument.

The displayed descending filtration combines powers and group commutators. In a pro-p group, take the closure of the generated product at each step. For a powerful p-group these layers are $P_i(G)=G^{p^{i-1}}$, with iterated closed power subgroups in the infinite case. The successive quotients are elementary abelian p-groups.

A finite p-group is powerful if $[G,G]\le G^p$ for odd $p$, or $[G,G]\le G^4$ for $p=2$. Here powers denote generated subgroups. The same definition applies to pro-p groups using closed power and commutator subgroups. Its Frattini subgroup is $G^p$. For odd $p$, successive power layers admit linear correction of roots, which makes every element of $G^p$ an actual $p$th power.

For an odd prime number $p$, a finite p-group that is a product of $d(G)$ cyclic subgroups is a powerful p-group. Indeed, $H=G/G^p$ has exponent of a finite group $p$ and the same minimal generator number $d$. Its cyclic factorization gives $|H|\le p^d$, while its Frattini quotient has order $p^d$. Thus $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. The condition on the number of factors is essential: the nonabelian Heisenberg group over ${\mathbb{F}}_p$ has exponent $p$ and is a product of three cyclic subgroups, although its minimal generator number is two.

For odd $p$, a subgroup $H$ of a finite powerful p-group $G$ satisfies $d(H)\le d(G)$. Here $d$ is the dimension of the Frattini quotient. Set $K=H\cap G^p$, $V=G/G^p$, $W=G^p/G^{p^2}$, and $A=H{G}^p/G^p$. The onto power map $\theta :V\to W$ sends $A$ into the image of $H^p$ in $K/\Phi (K)$. Since $\Phi (K)\le G^{p^2}$, this image has dimension at least $\dim \theta (A)$. Also $\Phi (K)\le \Phi (H)\le K$. Induction on $|G|$, applied to the powerful subgroup $G^p$, gives $d(K)\le \dim W$. ConsequentlyThe last inequality is rank-nullity theorem.

For odd $p$, set $P_0=G$ and $P_{i+1}=P_i^p$. Powerful embedding gives $[P_i,G]\le P_{i+1}$. Modulo $P_{i+2}$, commutators with $P_i$ are central and have exponent dividing $p$, so $(xy{)}^p\equiv x^p{y}^p$ when $y\in P_i$. Also the power map $P_i/P_{i+1}\to P_{i+1}/P_{i+2}$ is a surjective homomorphism. First find a root modulo $P_2$, then multiply it by successive corrections in $P_i$. The process terminates for a finite p-group, and converges for a pro-p group with separated power filtration. This proves that the generated power subgroup consists of actual powers.

A normal subgroup $N$ is powerfully embedded in a finite p-group or pro-p group $G$ if $[N,G]\le N^p$ for odd $p$, or $[N,G]\le N^4$ for $p=2$; use closed generated subgroups in the pro-p group case. The condition implies that $N$ is a powerful p-group. For odd $p$, both $[N,G]$ and $N^p$ remain powerfully embedded.

The nontrivial center of a finite p-group theorem implies that the center of a group of order $p^2$ has order $p$ or $p^2$. If it has order $p$, the quotient by the center has prime order and is cyclic; a cyclic quotient by the center forces the whole group to be abelian, contradicting the assumed size of its center. Thus the group is abelian. An element of order $p^2$ makes it cyclic. Otherwise choose $a\ne 1$ and $b\notin ⟨a⟩$; their order-$p$ cyclic subgroups intersect trivially and generate the direct product of groups $C_p\times C_p$. The two possibilities are nonisomorphic because only the cyclic one has an element of order $p^2$.

Every proper subgroup $H$ of a finite p-group is properly contained in its normalizer. Let $H$ act on the left cosets $G/H$ by left multiplication. The fixed cosets are those represented by elements of $N_G(H)$, so there are $[N_G(H):H]$ of them. All other orbits of a group action have sizes divisible by $p$. Since $[G:H]$ is divisible by $p$ and at least one coset is fixed, the fixed-coset count is a positive multiple of $p$. This also proves $[N_G(H):H]\ge p$.

A nontrivial normal subgroup $N$ of a finite p-group meets the center of a group nontrivially. The conjugation action of $G$ on $N$ has singleton orbits exactly at $N\cap Z(G)$. The orbit-stabilizer theorem makes every other orbit size divisible by $p$, so $|N\cap Z(G)|\equiv |N|\equiv 0(\mathrm{mod}p)$. Since the intersection contains the identity, its size is at least $p$.

Every nontrivial finite p-group has a nontrivial center. The conjugation action partitions the group into conjugacy classes whose noncentral sizes are positive powers of $p$; the class equation then implies $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$.

The Frattini subgroup of a finite p-group is the intersection of its maximal subgroups. It equals $G^p[G,G]$.

Let $\Phi =\Phi (G)$ in a finite group. If $H⊴G$ contains $\Phi$ and $H/\Phi$ is a nilpotent group, every Sylow subgroup $P$ of $H$ has $P\Phi ⊴G$. The Frattini argument gives $G=(P\Phi )N_G(P)=\Phi N_G(P)$, and the defining non-generator property of $\Phi$ forces $N_G(P)=G$. Hence $H$ is a nilpotent group.

An element that can always be omitted from a generating family. These are exactly the elements of the Frattini subgroup: a proper generated subgroup is contained in a maximal subgroup, while an element outside a maximal subgroup generates the group together with it.

The Frattini quotient of a finite p-group is an elementary abelian p-group. Its dimension over ${\mathbb{F}}_p$ is the minimum number of generators of $G$; in particular, it has dimension at least two when $G$ is noncyclic.

A subset generates a finite p-group precisely when its images span the Frattini quotient. Indeed $⟨S⟩\Phi (G)=G$ implies $⟨S⟩=G$, since otherwise a maximal subgroup contains both factors. Minimal generating sets correspond to vector-space bases in the quotient.

A nontrivial simple group has no normal subgroups other than the identity subgroup and itself.

For a finite simple group with a subgroup of index $n\ge 3$, the coset action is nontrivial and its kernel is a normal subgroup, hence trivial. Its sign homomorphism must also be trivial: otherwise simplicity would give an isomorphism to $C_2$, which cannot act transitively on at least three points. Thus the image lies in the alternating group. In particular Lagrange's theorem gives $|G|\mid n!/2$. This can exclude simplicity with very little counting: a group of order $112$ has a subgroup of index seven by Sylow theorems, but $112$ does not divide $7!/2$.

For a nontrivial simple soluble group, the commutator subgroup is either the identity or the whole group. It cannot be the whole group because the derived series would never terminate. The group is therefore Abelian. Every nonidentity element then generates a normal cyclic subgroup, which must be the whole group. An infinite cyclic group has proper nontrivial subgroups; a finite cyclic group is simple exactly when its order is prime. Thus the soluble simple groups are precisely the cyclic groups of prime order.

A nontrivial finite group whose only normal subgroups are the trivial group and the whole group. A cyclic group of prime number order gives an abelian example; $A_5$ gives a nonabelian alternating group example. Finite versions of classical groups can produce further examples after taking suitable subgroups and quotients; small parameters require separate checks.

For a finite simple group of this order, $e=0$ gives a normal Sylow $5$-subgroup, and $e=1$ gives a nontrivial sign homomorphism from the regular action, because an involution swaps fifteen pairs. Thus $e\ge 2$. The Sylow $2$-count divides $15$; it cannot be one or three. If it is five, a Sylow subgroup's normalizer has index five. If it is fifteen, the strengthened Sylow congruence from intersections produces two Sylow subgroups meeting in index two. Their common subgroup is normal in both, so its normalizer contains both and has index three or five. Simplicity excludes index three. The resulting faithful degree-five action embeds $G$ in $A_5$ and forces order $60$, proving the conclusion.

For a faithful primitive group action, suppose a point stabilizer has an abelian normal subgroup whose conjugates generate the whole group. Every nontrivial normal subgroup then contains the derived subgroup. Indeed the normal subgroup is transitive, and all those conjugates have the same image in the quotient, making the quotient abelian. If the whole group is nontrivial and perfect, it is simple.

A normal subgroup of $A_6$ has order dividing 360 and consists of the identity plus whole conjugacy classes of the alternating group on six letters. If it omits the odd-sized class of size 45, its order is an odd divisor and no nontrivial class sum fits. If it includes that class, the possible class sums not exceeding 180 are $46,86,118,126,136,158,176$, none dividing 360. Every proper subgroup has order at most 180; therefore $A_6$ is a simple group.

The alternating group $A_n$ is simple for every $n\ge 5$. In particular it has no proper subgroup of index two, since every index-two subgroup is normal.

A finite nonabelian simple group is a finite group that is a nonabelian group and a simple group. Every such group is a perfect group because its commutator subgroup is a nontrivial normal subgroup.

Let $p$ be the smallest prime divisor of the order of a finite nonabelian simple group. A cyclic Sylow subgroup would give a normal p-complement, so if $p^3$ does not divide the order, the Sylow subgroup must be $C_p^2$. Conjugation embeds its normalizer modulo centralizer into $G{L}_2(p)$, and this quotient has order prime to $p$. For odd $p$, every other prime divisor of $|G{L}_2(p)|=p(p-1{)}^2(p+1)$ is smaller than $p$, again forcing trivial conjugation and a contradiction to the Burnside transfer theorem. For $p=2$, nontrivial conjugation must have order three, giving $12\mid |G|$. The groups $PS{L}_2(5^{2m+1})$ give infinitely many examples with exactly two factors of two in their orders.