Finite p-group ($|G|=p^n$)

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 $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$.

For a finite group $G$, the Frattini subgroup $\Phi (G)$ is the intersection of all maximal subgroups. Its elements are exactly the non-generators: an element in every maximal subgroup cannot enlarge a proper generated subgroup to $G$, while an element outside a maximal subgroup generates $G$ together with it.

For a finite p-group,where $G^p$ is generated by all $p$th powers and $[G,G]$ is the commutator subgroup. Every maximal subgroup has index $p$ and is normal, so its quotient group is cyclic of order $p$ and it contains $G^p[G,G]$. Conversely $G/G^p[G,G]$ is an elementary abelian p-group; each nonzero vector is excluded by some hyperplane, whose preimage is a maximal subgroup. This proves equality.

Thus $\Phi (C_{p^2})$ is its order-$p$ subgroup, whereas $\Phi (C_p\times C_p)=1$. The Frattini quotient records the independent generators in the Burnside basis theorem.

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.