Frattini Subgroups and the Burnside Basis Theorem

If $M$ is maximal proper, then $\alpha (M)$ is proper and maximal: any subgroup strictly between $\alpha (M)$ and $G$ pulls back under ${\alpha }^{-1}$ to one strictly between $M$ and $G$. Thus $\alpha$ permutes the family of maximal proper subgroups.

An automorphism carries an intersection to the intersection of the images, so step 1.1 and [F1] give $\alpha (\mathrm{\Phi }(G))=\mathrm{\Phi }(G)$. This is characteristicity by [F2]. Every inner automorphism is an automorphism, so $\mathrm{\Phi }(G)$ is normal. The same argument covers $G=1$.

If $⟨S⟩=G$, then applying the quotient map gives $⟨\pi (S)⟩=G\mathrm{/}\mathrm{\Phi }(G)$.

Conversely, suppose $\pi (S)$ generates the quotient. Then $⟨S,\mathrm{\Phi }(G)⟩=G$. The subgroup $\mathrm{\Phi }(G)$ is finite, so list its elements and remove them one at a time from this generating set: [L1] says each is a nongenerator. After all have been removed, $⟨S⟩=G$.

Steps 1.1 and 1.2 prove both implications. When $G=1$, both the empty subset and its empty quotient image generate, so the boundary case also agrees.

If $a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}p)$, then $a-b=kp$ for an integer $k$. Since $x^p=e$ by [F1], the power laws in [L3] give $x^a=x^b(x^p{)}^k=x^b$. Thus $\bar{a}\cdot x:=x^a$ is independent of the representative.

The power laws in [L3] and commutativity give $(\bar{a}+\bar{b})\cdot x=(\bar{a}\cdot x)(\bar{b}\cdot x)$, $(\bar{a}\bar{b})\cdot x=\bar{a}\cdot (\bar{b}\cdot x)$, $\bar{a}\cdot (xy)=(\bar{a}\cdot x)(\bar{a}\cdot y)$, $\bar{1}\cdot x=x$, and $\bar{0}\cdot x=e$. Together with [L1] and [L2], these are the vector-space axioms.

The scalar structure uses the existing abelian group law and does not change its elements. In particular its additive group remains finite, abelian, and of exponent $p$, including the zero-dimensional trivial case.

The set of cardinalities of spanning subsets of $E$ is nonempty because $E$ spans itself, so [L3] gives its least member; choose a spanning subset $B$ of that size. It is inclusion-minimal, and if a nontrivial linear relation existed in $B$, one member with nonzero coefficient could be solved for using its inverse scalar, contradicting minimality. Thus $B$ is a basis by [F1].

Starting from $I$, adjoin an element outside its span while one exists; adjoining such an element preserves independence, and finiteness makes the process terminate at a spanning independent set. This extends $I$ to a basis. Applying the deletion argument of step 1.1 inside $S$ extracts a basis from every spanning set.

If $B$ is a basis, uniqueness of coordinates gives a bijection ${\mathbb{F}}_p^B\to E$, so [L1] gives $\mathrm{\mid }E\mathrm{\mid }=p^{\mathrm{\mid }B\mathrm{\mid }}$. For two bases $B,C$, the equality $p^{\mathrm{\mid }B\mathrm{\mid }}=p^{\mathrm{\mid }C\mathrm{\mid }}$ and uniqueness of the exponent of the prime $p$ in [L2] give $\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }C\mathrm{\mid }$.

By [L1], each maximal subgroup $M$ of $P$ is normal with $P\mathrm{/}M$ cyclic of order $p$. Hence every commutator and every $p$th power lies in every $M$. Their images are therefore trivial modulo the normal subgroup $\mathrm{\Phi }(P)$ from [L4] and [F1], so $P\mathrm{/}\mathrm{\Phi }(P)$ is abelian of exponent at most $p$, and thus elementary abelian, including the trivial quotient.

For the forward direction of the kernel criterion, suppose $P\mathrm{/}N$ is elementary abelian and let $x\mathrm{\notin }N$. The nonzero vector $xN$ extends by [L2] to a basis of $P\mathrm{/}N$. The span of the other basis vectors is a maximal proper subgroup not containing $xN$; by [L3], its preimage is a maximal subgroup of $P$ containing $N$ but not $x$. Thus $x\mathrm{\notin }\mathrm{\Phi }(P)$, and so $\mathrm{\Phi }(P)\le N$.

For the reverse direction, suppose $\mathrm{\Phi }(P)\le N$. Step 1.1 places every commutator and every $p$th power of $P$ inside $\mathrm{\Phi }(P)$ and hence inside $N$, so $P\mathrm{/}N$ is abelian and every element of it has $p$th power the identity. It is therefore elementary abelian, including the trivial quotient $N=P$. Together with step 1.2 this proves the iff.

Every automorphism sends $g^p$ to $\alpha (g{)}^p$, so it preserves the generating set in [F1] and hence $P^p$ is characteristic; conjugation by an element of $P$ is an automorphism, so $P^p$ is normal. By [L2], $P^{\mathrm{\prime }}$ is normal as well, so [L3] makes $K:=P^{\mathrm{\prime }}{P}^p$ a subgroup, and $gK{g}^{-1}=(g{P}^{\mathrm{\prime }}{g}^{-1})(g{P}^p{g}^{-1})=P^{\mathrm{\prime }}{P}^p=K$ for every $g\in P$ makes it normal.

The elementary abelian quotient $P\mathrm{/}\mathrm{\Phi }(P)$ is abelian and has exponent $p$ by [L1], so it kills every commutator and every $p$th power. Thus $K=P^{\mathrm{\prime }}{P}^p\le \mathrm{\Phi }(P)$.

The quotient $P\mathrm{/}K$ is abelian because it kills $P^{\mathrm{\prime }}$, and every element has $p$th power one because it kills $P^p$. It is therefore elementary abelian, so the kernel criterion in [L1] gives $\mathrm{\Phi }(P)\le K$. Together with step 2.1 this proves equality.

By [L2], $H^{\mathrm{\prime }}\le P^{\mathrm{\prime }}$. Every $p$th power of an element of $H$ is also a $p$th power of an element of $P$, so $H^p\le P^p$.

Multiplying the inclusions in step 1.1 gives $H^{\mathrm{\prime }}{H}^p\le P^{\mathrm{\prime }}{P}^p$, and [L1] identifies these products with $\mathrm{\Phi }(H)$ and $\mathrm{\Phi }(P)$.

By [L2], $\pi (P^{\mathrm{\prime }})=(P\mathrm{/}N{)}^{\mathrm{\prime }}$. Also $\pi (g^p)=\pi (g{)}^p$, so $\pi (P^p)=(P\mathrm{/}N{)}^p$.

Apply [L1] in $P\mathrm{/}N$ and use step 1.1: $\mathrm{\Phi }(P\mathrm{/}N)=\pi (P^{\mathrm{\prime }})\pi (P^p)=\pi (P^{\mathrm{\prime }}{P}^p)=\pi (\mathrm{\Phi }(P))$. Directly, ${\pi }^{-1}(\pi (\mathrm{\Phi }(P)))=\mathrm{\Phi }(P)N$, so the correspondence in [L3] gives $\pi (\mathrm{\Phi }(P))=\mathrm{\Phi }(P)N\mathrm{/}N$.

Componentwise commutators and powers give $(P\times Q{)}^{\mathrm{\prime }}=P^{\mathrm{\prime }}\times Q^{\mathrm{\prime }}$ and $(P\times Q{)}^p=P^p\times Q^p$.

By [L1] and step 1.1, $\mathrm{\Phi }(P\times Q)=(P^{\mathrm{\prime }}\times Q^{\mathrm{\prime }})(P^p\times Q^p)=P^{\mathrm{\prime }}{P}^p\times Q^{\mathrm{\prime }}{Q}^p=\mathrm{\Phi }(P)\times \mathrm{\Phi }(Q)$.

By [L1], $P^2$ is characteristic and hence normal. Every element of $P\mathrm{/}{P}^2$ has square one by [F1]. In any group of exponent at most two, $(xy{)}^2=e$ gives $xy=(xy{)}^{-1}=y^{-1}{x}^{-1}=yx$, so $P\mathrm{/}{P}^2$ is abelian.

By [L2], step 1.1 gives $P^{\mathrm{\prime }}\le P^2$. Substituting in [L1] yields $\mathrm{\Phi }(P)=P^{\mathrm{\prime }}{P}^2=P^2$, including $P=1$.

For the forward direction, if $\mathrm{\Phi }(P)=1$, then [L1] identifies $P$ with its elementary abelian Frattini quotient.

For the reverse direction, if $P$ is elementary abelian, apply the kernel criterion in [L1] with $N=1$ to obtain $\mathrm{\Phi }(P)\le 1$, hence $\mathrm{\Phi }(P)=1$.

For the forward direction, suppose $X$ is minimally generating. By [L1], $\pi (X)$ spans. If $x\in X$ mapped to zero, or if distinct $x,y\in X$ had the same image, deleting one of them would leave the same spanning image and [L1] would make a proper subset generate, contrary to [F1]. Thus $\pi {\mathrm{\mid }}_X$ is injective. Any proper subset of $\pi (X)$ that spanned would likewise pull back to a proper generating subset of $X$, so $\pi (X)$ spans and no proper subset of it does. By [L2] it contains a basis $B$; since $B$ spans, the failure of every proper subset to span forces $B=\pi (X)$, so $\pi (X)$ is itself a basis.

For the reverse direction, suppose $\pi {\mathrm{\mid }}_X$ is a bijection onto a basis. The basis spans, so [L1] gives $⟨X⟩=P$. Removing any $x\in X$ removes its distinct basis vector; the remaining basis vectors do not span, so [L1] says $X\setminus \{x\}$ does not generate. Thus $X$ is minimal by [F1].

Steps 1.1 and 1.2 prove both implications. For $P=1$, the quotient has the empty basis and the empty set is the minimal generating set, so the same statement applies.

By [L1], every minimal generating set is in bijection with a quotient basis. All such bases have size $d(P)$ by [F1] and [L2].

If $Y$ generates $P$, its quotient image spans. Choose by [L2] a basis contained in that finite image and, for each basis vector, retain one element of $Y$ mapping to it. The resulting subset maps bijectively onto the basis, so [L1] makes it minimally generating; step 1.1 gives its size.

By [L3] the quotient $P\mathrm{/}\mathrm{\Phi }(P)$ is a finite elementary abelian $p$-group, so [L2] applies to it. The coset $x\mathrm{\Phi }(P)$ is nonzero, so its singleton is independent in $P\mathrm{/}\mathrm{\Phi }(P)$.

Extend that singleton by [L2] to a finite basis. For every other basis vector choose one lift in $P$ and adjoin it to $x$. The quotient map restricts to a bijection from this lifted set onto the basis, so [L1] makes it a minimal generating set containing $x$. The assertion is vacuous for $P=1$.

For the reverse direction, if $d(P)=1$, choose a one-element quotient basis and lift it. By [L1] the lift is a one-element generating set, so $P$ is cyclic by [F2].

For the forward direction, if nontrivial $P$ is cyclic with generator $g$, then $\{g\}$ is minimally generating because the empty set generates only the trivial subgroup. By [L1], its quotient image is a one-element basis, so $d(P)=1$ by [F1].

If $P=1$, its Frattini quotient has the empty basis, hence $d(P)=0$ by [F1]; this is why nontriviality is needed in the biconditional.

If $M$ is maximal in $P$, [F1] and [L3] make $M\mathrm{/}\mathrm{\Phi }(P)$ maximal proper in $E$. Choose a basis of this subgroup and extend it by [L2] to a basis of $E$. Maximality permits exactly one added basis vector, since two would create a proper intermediate span. Thus $M\mathrm{/}\mathrm{\Phi }(P)$ has codimension one.

The coordinate of the omitted basis vector defines a nonzero linear homomorphism $E\to \mathbb{Z}\mathrm{/}p$ whose kernel is $M\mathrm{/}\mathrm{\Phi }(P)$. Conversely, let $\lambda :E\to \mathbb{Z}\mathrm{/}p$ be linear and nonzero and choose $v\in E$ with $\lambda (v)=1$. Every $x\in E$ splits as $x=(x-\lambda (x)v)+\lambda (x)v$ with the first summand in $\ker \lambda$, and $v\mathrm{\notin }\ker \lambda$, so a basis of $\ker \lambda$ together with $v$ spans $E$ and is independent; it is therefore a basis of $E$ by [L2], and $\ker \lambda$ has codimension one. Any subgroup strictly between $\ker \lambda$ and $E$ would contain some $y$ with $\lambda (y)\mathrm{\ne }0$ and hence a scalar multiple of $y$ equal to $v$ modulo $\ker \lambda$, so it would be all of $E$; thus $\ker \lambda$ is maximal proper and [L3] makes its inverse image maximal in $P$.

The quotient and inverse-image maps in [L3] are inverse, so steps 1.1 and 2.1 give the stated classification. The trivial group has neither maximal subgroups nor nonzero linear homomorphisms.

By [L1], $\alpha (\mathrm{\Phi }(P))=\mathrm{\Phi }(P)$, so $\pi \circ \alpha$ kills $\mathrm{\Phi }(P)$. The universal property [L4] gives a quotient automorphism $\bar{\alpha }$ with $\bar{\alpha }(x\mathrm{\Phi }(P))=\alpha (x)\mathrm{\Phi }(P)$; applying the same construction to ${\alpha }^{-1}$ supplies its inverse.

For $\bar{a}\in {\mathbb{F}}_p$, $\bar{\alpha }(\bar{a}\cdot x\mathrm{\Phi }(P))=\bar{\alpha }(x^a\mathrm{\Phi }(P))=\alpha (x^a)\mathrm{\Phi }(P)=\alpha (x{)}^a\mathrm{\Phi }(P)=\bar{a}\cdot \bar{\alpha }(x\mathrm{\Phi }(P))$, so $\bar{\alpha }$ is linear by [L2].

The induced map of the identity is the identity, and uniqueness in [L4] gives $\overline{\alpha \beta }=\bar{\alpha }\bar{\beta }$. Thus $\alpha \mapsto \bar{\alpha }$ preserves the group law in [L3] and defines ${\rho }_P$.

Suppose, for contradiction, that $A\mathrm{\ne }1$. Choose a prime $q$ dividing $\mathrm{\mid }A\mathrm{\mid }$; [L2] gives $\alpha \in A$ of order $q$. Since $p\nmid \mathrm{\mid }A\mathrm{\mid }$, one has $q\mathrm{\ne }p$.

Triviality of the quotient action in [L4] means that $\alpha$ preserves each coset of $\mathrm{\Phi }(P)$. Each coset has $\mathrm{\mid }\mathrm{\Phi }(P)\mathrm{\mid }$, a power of $p$ by [L5], elements. The cyclic $q$-group $⟨\alpha ⟩$ acts on that coset, and $q\mathrm{\ne }p$ makes [L3] provide an $\alpha$-fixed representative in every coset.

Starting from the finite generating set $P$, delete redundant elements until a minimal generating set remains; [L1] sends it bijectively onto a basis of $P\mathrm{/}\mathrm{\Phi }(P)$. Using step 2.1, choose a fixed representative of each of these finitely many basis cosets. By [L1] those representatives generate $P$. Since $\alpha$ fixes every generator, it fixes every element of $P$, so $\alpha$ is the identity, contradicting its prime order.