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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 $C_p\times C_p$.

In finite-group character theory, an elementary group is a group that is p-elementary group for some prime $p$. Such groups are solvable and provide the subgroup tests in Brauer's characterization of characters. This does not mean elementary abelian group.

The group of maps $x\mapsto Ax+b$ of a finite-dimensional vector space, with $A$ invertible. Identifying the underlying points with a regular elementary abelian group identifies its normalizer in the symmetric group with this semidirect product.

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.

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.

A nontrivial minimal normal subgroup $N$ of a finite solvable group is abelian: its commutator subgroup is characteristic, and minimality plus solvability excludes $N^{\prime }=N$. A nontrivial Sylow subgroup of this abelian group is characteristic, so minimality makes $N$ a p-group. The characteristic subgroup of pth powers is proper, hence trivial by minimality. Therefore $N$ is an elementary abelian group.

Let $\mathrm{Irr}(G)$ denote the irreducible characters of finite-dimensional complex group representations of $G$. A generalized character, or virtual character, is an integral linear combination of these irreducible characters. A class function is a function constant on conjugacy classes. For class functions use the character inner productA p-elementary group is a direct product of groups $P\times C$, where $P$ is a p-group and $C$ is a cyclic group of order coprime to $p$. An elementary subgroup is a subgroup of this form for some prime $p$. Either factor may be trivial. This notion is different from an elementary abelian group.

Brauer's characterization of characters says that a complex class function $f$ on $G$ is a generalized character exactly when its restriction to every elementary subgroup is a generalized character there. Equivalently,The character orthogonality relations make these inner products the coefficients of the restrictions in their irreducible bases. The theorem detects the integral character lattice. An ordinary character of a representation additionally requires every global irreducible coefficient $⟨f,\chi {⟩}_G$ to be nonnegative; integrality alone characterizes virtual characters, not necessarily actual representations.

Since the two-transitive group action is primitive, the orbits of any nontrivial normal subgroup form only the single block $\Omega$. Thus $N$ is transitive. If $N$ is regular, identify $\Omega$ with $N$ by $x\mapsto x\alpha$. Then $G=N{G}_{\alpha }$, $N\cap G_{\alpha }=1$, and conjugation by $G_{\alpha }$ gives automorphisms of $N$. The two-transitive group action makes these automorphisms transitive on $N\setminus \{1\}$. All nonidentity elements therefore have the same order, which must be a prime $p$: a suitable power of an element of composite order would have a smaller nontrivial order. Cauchy's theorem for finite groups now shows $|N|=p^d$.

The center of a group of prime-power order is nontrivial and is a characteristic subgroup. The transitive conjugation action on nonidentity elements therefore forces $Z(N)=N$. Hence $N$ is an elementary abelian group, which we identify with the additive vector space ${\mathbb{F}}_p^d$. Conjugation by $G_{\alpha }$ is linear on this vector space. It is faithful: an element fixing $\alpha$ and centralizing $N$ fixes every $x\alpha$, hence is the identity permutation. ConsequentlyHere the general affine group acts by translations followed by invertible linear maps.

For each $j\ge 0$, the elementary abelian group $P_j/P_{j+1}$ is spanned by the images of $a_i^{p^j}$. We construct exponents $e_i$ so that $g$ equals the ordered product $a_1^{e_1}\cdots a_n^{e_n}$ to successively greater precision. Suppose this product agrees with $g$ modulo $P_j$; initially this is automatic for $j=0$. Its error modulo $P_{j+1}$ can be expressed using the images of $a_i^{p^j}$. Add the corresponding digits $p^j{t}_i$ to the exponents $e_i$. Since $[P_j,G]\subseteq P_{j+1}$, moving each new factor $a_i^{p^j{t}_i}$ past the other factors changes nothing modulo $P_{j+1}$. The new ordered product therefore agrees with $g$ modulo $P_{j+1}$. Termination of the lower p-series provesNo uniqueness of this factorization is asserted for a redundant generating tuple.

The printed converse is false without an additional condition. For any odd $p$, take the Heisenberg group $G={\mathrm{UT}}_3({\mathbb{F}}_p)$ withEvery element is uniquely $x^a{y}^b{z}^c$, with $a,b,c\in {\mathbb{F}}_p$, so $G$ is a product of three cyclic subgroups. For $A$ strictly upper triangular, $A^3=0$ andin characteristic $p$, including $p=3$. Therefore $G^p=1$, whereas $[G,G]=⟨z⟩\ne 1$. This finite p-group is not powerful, disproving the literal request in the authoritative PDF.

A precise nearby correction is the minimal cyclic factorization criterion for powerful p-groups: require a product of $d(G)$ cyclic subgroups, where $d(G)$ is the minimal generator number. To prove it, put $d=d(G)$ and $H=G/G^p$. The quotient has exponent $p$ and still has minimal generator number $d$, because $G^p\subseteq \Phi (G)$. Each cyclic factor has image of order at most $p$, so $|H|\le p^d$. But $|H/\Phi (H)|=p^d$ by the Burnside basis theorem. Consequently $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. A product of exactly $d(G)$ cyclic subgroups is powerful. A direct product of cyclic groups is also powerful, since it is abelian; neither stronger hypothesis appears in the printed converse.

Take $u\in L\setminus K$, whose irreducible polynomial is $t^2+ct+d$. Since the characteristic is not two, $x=2u+c$ satisfies $x^2=c^2-4d=a\in K$ and $L=K(x)$. Irreducibility makes $a$ nonzero and nonsquare. If another quadratic extension is $K(y)$ with $y^2=b$, an isomorphism sends $x$ to $s+ty$, with $t\ne 0$. Squaring gives $a=s^2+t^{2}b+2sty$. Linear independence of $1,y$ forces $s=0$, so $a=t^{2}b$ and $b/a$ is a square. Conversely, if $b/a=q^2$, sending $x$ to $y/q$ gives an isomorphism. If $b$ itself is a square, the target is just $K$ and cannot be isomorphic; the square-ratio condition also fails.

The multiquadratic field $F$ is the splitting field of ${\prod }_i(t^2-a_i)$, a product of separable quadratic polynomials: each has distinct roots because $a_i\ne 0$ and the characteristic is not two. Even if factors repeat, this splitting field is a separable normal extension, and therefore Galois. Each automorphism sends $x_i$ to $\pm x_i$, and these signs determine it, embedding its Galois group into $(\mathbb{Z}/2{)}^n$. Every subgroup of this elementary abelian group is $(\mathbb{Z}/2{)}^m$ for some $m\le n$.

For a quadratic intermediate field, its fixing subgroup $H$ has index two and is the kernel of a nontrivial one-dimensional character $\chi :G\to \{\pm 1\}$. The coordinate sign characters span the dual of $G$: a linear functional on the subspace $G\subseteq {\mathbb{F}}_2^n$ extends to the ambient space by extending a basis. Hence $\chi ={\prod }_{i\in I}{\chi }_i$ for some subset $I$. Set $y={\prod }_{i\in I}{x}_i$. Then $\sigma (y)=\chi (\sigma )y$, so $y$ is fixed by $H$ but not by all of $G$, and $y^2={\prod }_{i\in I}{a}_i\in K$. Thus $K(y)$ is quadratic and equals the fixed field of $H$. This constructs all quadratic subfields of a multiquadratic extension:

For odd $p$, a subgroup $N$ is powerfully embedded in $G$ if $[N,G]\le N^p$, where $N^p=⟨n^p:n\in N⟩$. The condition implies normality since $n^g=n[n,g]\in N$. The group $G$ is a powerful p-group if $[G,G]\le G^p$, equivalently if $G$ itself is powerfully embedded. For a finite p-group, the Burnside basis theorem gives $d(Q)={\dim }_{{\mathbb{F}}_p}Q/\Phi (Q)$, and $\Phi (Q)=Q^p[Q,Q]$. Thus for powerful $Q$, $d(Q)=\dim Q/Q^p$.

We prove $d(H)\le d(G)$ by induction on $|G|$, with the trivial group immediate. The supplied embedding result applied to $N=G$ makes $G^p$ powerfully embedded, and hence powerful. It is a proper subgroup when $G\ne 1$, because $G^p=\Phi (G)$ is proper in a nontrivial finite p-group. SetAll the displayed quotients are elementary abelian. The supplied onto homomorphism $\theta :V\to W$, $x{G}^p\mapsto x^p{G}^{p^2}$, is therefore linear. Write $v=\dim V=d(G)$, $w=\dim W=d(G^p)$, $r=\dim A$, and $t=\dim \theta (A)$.

Since $H/K$ embeds in $V$, $\Phi (H)\le K$. AlsoHere $(G^p{)}^p=G^{p^2}$ is the iterated-power convention, valid for powerful groups. Hence the natural map $K/\Phi (K)\to W$ is defined. The image under this map of the subspace $\Phi (H)/\Phi (K)$ contains $\theta (A)$, because for every $h\in H$ its power $h^p$ lies in $\Phi (H)\cap K$. It follows thatThe exact sequence $1\to K/\Phi (H)\to H/\Phi (H)\to H/K\to 1$ consists of elementary abelian groups, and $\dim H/K=r$. By induction inside $G^p$, $d(K)\le d(G^p)=w$. ThereforeThis proves $d(H)\le d(G)$, including subgroups $H$ that are not themselves powerful.

The rank of a profinite group is $\mathrm{rk}(P)=\sup \{d(K):K\le P\text{ closed}\}$, where $d(K)$ is its least topological generator number. If $P$ has finite rank then every closed subgroup, including the open subgroup $H$, has no larger rank. Conversely suppose $H$ has rank $r<\infty$. Its normal core $C={\bigcap }_{g\in P}{H}^g$ is an open normal subgroup: there are finitely many conjugates because $H$ has finite index. It has rank at most $r$.

For every closed $K\le P$, the subgroup $K\cap C$ is normal in $K$ and has at most $r$ topological generators. The quotient $K/(K\cap C)$ embeds in the finite group $P/C$. A finite group of order at most $M=[P:C]$ has at most $\lfloor {\log }_{2}M\rfloor$ generators: in a minimal sequential generating list every strict enlargement at least doubles the subgroup order. Lift such quotient generators and adjoin generators of $K\cap C$. Their closed generated subgroup contains the group homomorphism kernel and maps onto the quotient, so equals $K$. ConsequentlyThus an open subgroup has finite rank exactly when the whole profinite group does.

Let a finite group $G$ act on a set $X$. For $x\in X$, define its stabilizer subgroup $G_x=\{g:g\cdot x=x\}$ and its group orbit $Gx=\{g\cdot x:g\in G\}$. The mapis well-defined, because multiplying $g$ on the right by an element fixing $x$ does not change its image. It is onto by the definition of the orbit. If $g\cdot x=h\cdot x$, then $h^{-1}g\in G_x$, so $g{G}_x=h{G}_x$; hence it is one-to-one. Every coset has $|G_x|$ elements and the cosets partition $G$. This proves the orbit-stabilizer theorem:For a group element $x$, its cyclic subgroup $H=⟨x⟩$ has $\mathrm{ord}(x)$ elements. Apply the theorem to the action of $G$ on its left cosets $G/H$, whose stabilizer at $H$ is $H$. It follows thatThis is the needed case of Lagrange's theorem, obtained directly from the action.

To prove Cauchy's theorem for finite groups, let $p$ be a prime divisor of $|G|$ and considerThe first $p-1$ entries are arbitrary and determine the last, so $|\mathcal{X}|=|G{|}^{p-1}$ is divisible by $p$. Cyclically rotate the entries. This preserves the product constraint even if $G$ is nonabelian, sinceThus a cyclic group of order $p$ acts on $\mathcal{X}$. By the orbit-stabilizer theorem, its orbits have size one or $p$. Its fixed tuples are exactly $(g,\dots ,g)$ with $g^p=e$. If $m$ is their number, counting the remaining size-$p$ orbits gives $m\equiv |\mathcal{X}|\equiv 0(\mathrm{mod}p)$. The identity provides one fixed tuple, so $m\ge p$ and a nonidentity $g$ with $g^p=e$ exists. The order of this $g$ divides $p$: division of $p$ by its order and minimality of that order prove this directly. Since $p$ is prime and $g\ne e$, the order is exactly $p$. This is the cyclic-tuple proof of Cauchy theorem.

Now suppose every nonidentity element has order two. Then every element equals its inverse, and for any $a,b\in G$,Thus group of exponent two is abelian applies. Starting from the identity subgroup, choose an element outside the subgroup already generated. Because the group is abelian and that element squares to the identity, adjoining it doubles the subgroup: it is the disjoint union of the old subgroup and its translate by the new element. Repeating until the finite group is exhausted shows $|G|=2^r$. Conversely the direct product of groups $C_2^r$ has order $2^r$ and every nonidentity element has order two. ThereforeThe case $r=0$ is the trivial group and satisfies the condition vacuously. These finite examples are elementary abelian groups of exponent two.

An infinite example is the group of all binary sequences, with coordinatewise addition modulo two:Every sequence added to itself is zero, so every nonzero sequence has order exactly two. The sequences having a single $1$ in coordinate $j$ are all distinct, proving that the group is infinite.

For a profinite group, its profinite Frattini subgroup isUse $G$ for an empty intersection. This is a closed normal subgroup, invariant under continuous group automorphisms, since these automorphisms permute the maximal open subgroups. Let $H=\overline{⟨X⟩}$. If $H=G$, its image certainly topologically generates $G/\Phi (G)$. Conversely, suppose $H\ne G$. For $g\notin H$, the open normal subgroup basis of a profinite group gives $N{⊴}_{o}G$ with $g\notin HN$: choose a sufficiently small neighbourhood $gN$ disjoint from the closed subgroup $H$. The proper subgroup $HN/N$ of the finite group $G/N$ lies in a maximal proper subgroup. Its preimage $M$ is a maximal proper open subgroup of $G$, contains $H$, and contains $\Phi (G)$. Since $H\Phi (G)$ is compact and therefore closed, density of the image of $X$ in $G/\Phi (G)$ would force $H\Phi (G)=G$, contradicting $H\Phi (G)\le M<G$. Generation can therefore be tested modulo the profinite Frattini subgroup.

For a finite p-group, every maximal proper subgroup is normal and has index $p$. One proof uses the normalizer condition for finite p-groups: maximality forces its normalizer to be the whole group, and its simple $p$-group quotient has order $p$. Therefore every such maximal subgroup contains powers and group commutators. Conversely, the quotient by the algebraic $G^p[G,G]$ is an elementary abelian p-group; its nonzero elements are separated by linear maps to ${\mathbb{F}}_p$. Their kernels give maximal subgroups. This proves the finite Frattini subgroup formula.

Now take a pro-p group and put $D=\overline{G^p[G,G]}$, retaining the closure of the algebraically generated subgroups. A maximal open subgroup is the preimage of a maximal subgroup of a finite $p$-group quotient, so it contains $D$. If $g\notin D$, some finite quotient of the profinite group $G/D$ detects $gD$. That quotient is elementary abelian, and a linear functional nonzero at $gD$ gives a continuous group homomorphism $G\to {\mathbb{F}}_p$ whose kernel is a maximal open subgroup missing $g$. HenceThe closure bar is present in the original PDF; the converted TeX fraction is an extraction error.

For the finite-index assertion, first pass from the arbitrary abstract subgroup $K$ to its normal core of a subgroup $C$. The action on the finitely many cosets of $K$ shows $[G:C]<\infty$, without assuming that $K$ or $C$ is closed. Write $m=[G:C]=p^{r}q$ with $(p,q)=1$, and let $X_m=\{h^m:h\in G\}$. By Lagrange theorem, $X_m\subseteq C$. For $N{⊴}_{o}G$, the quotient $G/N$ is a finite p-group of order, say, $p^s$. Choose an integer $a$ with $aq\equiv 1(\mathrm{mod}{p}^s)$. Thenso $g^{p^r}\in X_{m}N$ for every such $N$. The set $X_m$ is the continuous image of the compact space $G$, hence is closed. Consequently $g^{p^r}\in {\bigcap }_N{X}_{m}N=X_m\subseteq C$. Every element of the abstract finite quotient $G/C$ thus has order dividing $p^r$. By Cauchy theorem for groups, no prime other than $p$ divides its order. Therefore $[G:C]$, and its divisor $[G:K]$, are powers of $p$. This proof does not assume continuity of the finite quotient map.

To prove openness under finite generation, first establish commutator collection with fixed generators in nilpotent groups. If a nilpotent group $P$ is generated by $a_1,\dots ,a_d$, every element of $P\prime$ isInduct on the nilpotency class $c$. The assertion is trivial for $c=1$. Apply the induction hypothesis in $P/{\gamma }_c(P)$; the remaining error lies in the central subgroup ${\gamma }_c(P)=[{\gamma }_{c-1}(P),P]$. Since commutators in this last layer are central, collecting against words in the $a_i$ expresses the error as ${\prod }_i[y_i,a_i]$ with $y_i\in {\gamma }_{c-1}(P)$. Also $[P\prime ,{\gamma }_{c-1}(P)]\le {\gamma }_{c+1}(P)=1$. Thus replacing each $x_i$ by $x_i{y}_i$ multiplies the desired product by precisely that central error, proving the formula.

Let $a_1,\dots ,a_d$ now topologically generate $G$. Every finite continuous quotient is a nilpotent group, so the formula holds there with these fixed generators. For $z\in \overline{[G,G]}$, the sets of tuples $(x_1,\dots ,x_d)\in G^d$ solving $z\equiv {\prod }_i[x_i,a_i](\mathrm{mod}N)$ are nonempty closed subsets of a compact space. They have the finite intersection property, by replacing finitely many $N$ by their intersection. A common tuple solves the equality in $G$. It follows that the algebraic commutator subgroup is the compact image of $G^d$ under this product map, and is closed.

Its quotient $A=G/[G,G]$ is a topologically finitely generated abelian pro-p group. The continuous map ${\mathbb{Z}}_p^d\to A$ taking $({\lambda }_i)$ to ${\prod }_i{\bar{a}}_i^{{\lambda }_i}$ is onto, because its compact image contains a dense generating subgroup. Therefore $A/pA$ is generated as an elementary abelian group by at most $d$ elements, and $pA$ is compact. The preimage $G^p[G,G]$ of $pA$ is closed of finite index, hence open: its complement is a finite union of closed cosets. The finite-index result and the maximal-subgroup property of finite p-groups show that any abstract index-$p$ subgroup of $G$ is normal. It contains this open subgroup, so is itself open.

Finally, a finite-index normal subgroup $C$ has finite $p$-group quotient. If $[G:C]>1$, choose a normal index-$p$ subgroup in that quotient and call its preimage $H$. The preceding argument makes $H$ open. An open subgroup of a finitely generated pro-p group is finitely generated: intersect a finitely generated dense abstract subgroup with $H$, use the Schreier lemma, and observe that the intersection is dense in the open subgroup $H$. Induction on $[G:C]$ now makes $C$ open in $H$, hence in $G$. Returning to the normal core of a subgroup, $K$ is a union of open cosets of $C$. Every finite-index subgroup of a finitely generated pro-p group is open.

A topological group module $A$ is a topological abelian group with a jointly continuous action of $G$ by group automorphisms. Its continuous cochains with topological coefficients arewith pointwise addition. The inhomogeneous differential isFor $n=0$, this means $(da)(g)=ga-a$. Continuity of multiplication and of the action ensures that $d$ preserves continuous cochains. Expanding $d(dc)$ gives cancellation in pairs: the two orders of multiplying neighbouring arguments agree by associativity, and the two ways of applying the action agree by its action law. Thus $d^2=0$. The cocycles are $Z^n=\ker (d:C^n\to C^{n+1})$ and the coboundaries are $B^n=\mathrm{im}(d:C^{n-1}\to C^n)$, with $B^0=0$. DefineFor general topological coefficients this is an algebraic quotient of the continuous cochains; its optional quotient topology need not be Hausdorff. For discrete coefficients it is the usual continuous cohomology of a profinite group. In degree zero it is the subgroup of invariant elements.

Give the discrete prime field ${\mathbb{F}}_p$ its trivial $G$-action. The one-cocycle condition becomes $c(gh)=c(g)+c(h)$ and every one-coboundary is zero. HenceEvery such group homomorphism kills powers and group commutators, hence their closure $\Phi (G)$, and every continuous functional on the quotient pulls back. If $G$ is topologically finitely generated, $V=G/\Phi (G)$ is a finite elementary abelian group. To verify finiteness without assuming it, the closure of a subgroup generated by $r$ elements of this abelian exponent-$p$ quotient has at most $p^r$ elements; a finite subgroup is already closed. Thus $V$ is a finite-dimensional vector space. Question 2 says that a generating family of $G$ is minimal precisely when its images give a basis of $V$. The dual space has the same finite dimension. This proves

There is a genuine missing hypothesis if the printed assertion is read for arbitrary pro-p groups and $d(G)$ means the least cardinality of any dense generating family. Here is an explicit infinite-rank failure of the first-cohomology generator formula. Let $I=\{0,1{\}}^{\mathbb{N}}$ and $G={\prod }_{r\in I}{\mathbb{F}}_p$. Enumerate the finite binary strings as $s_1,s_2,\dots$ and define $g_n(r)=1$ if $s_n$ is a prefix of $r$, and $0$ otherwise. For finitely many distinct sequences $r_1,\dots ,r_k$, choose long enough prefixes to distinguish each individually. The corresponding $g_n$ project onto the coordinate unit vectors of ${\mathbb{F}}_p^k$, so $⟨g_n:n\ge 1⟩$ surjects onto every finite coordinate projection and is dense in the product topology. No finite family can generate densely, since a finitely generated abelian exponent-$p$ subgroup is finite and closed. Thus $d(G)={\aleph }_0$. On the other hand, continuity of a linear functional to ${\mathbb{F}}_p$ means its kernel contains a basic neighbourhood, so it depends on finitely many coordinates. The coordinate projections are consequently a basis of the continuous dual space, givingThe unrestricted infinite-rank formula is false. A convention requiring a generating basis tending to the identity is different from the printed unrestricted topological generator count; the finite-rank identity proved above has no such ambiguity.

For finite $X$, a free pro-p group $F_p(X)$ is characterized by the universal property that every map $X\to P$ to a pro-p group extends uniquely to a continuous group homomorphism $F_p(X)\to P$. It is the pro-p completion of the abstract free group on $X$: each map to a finite $p$-group factors through one of the defining finite quotients, and passing to the inverse limit gives the extension to any pro-p group; uniqueness follows from density. This unrestricted universal property can also define the completion on an arbitrary set $X$. When using an infinite basis required to tend to the identity, one instead imposes the universal property only for maps tending to the identity. Only the finite-rank version is needed for the following argument.

A finite pro-p presentation is an isomorphismThere are finitely many generators and finitely many relators, and the denominator is their closed normal closure. The relators are elements of the free pro-p group; they may be limits of ordinary words. That distinction is important when writing conjugation relations below.

Now suppose $K{⊴}_{o}G$ has a finite pro-p presentation with generators $k_1,\dots ,k_a$ and relators $R_1,\dots ,R_t$. The quotient $Q=G/K$ is a finite p-group. Choose a finite finite pro-p presentation $Q=⟨{\bar{y}}_1,\dots ,{\bar{y}}_b\mid S_1,\dots ,S_c⟩$. Such a presentation exists, for example by using its entire finite multiplication table. Lift the quotient generators to $y_j\in G$. For every $i,j$, choose elements $W_{ij},V_{ij}$ of the free pro-p group on the $x_i$ that map respectively to $y_j^{-1}{k}_i{y}_j$ and $y_j{k}_i{y}_j^{-1}$ in $K$. Choose $U_{\ell }(x)$ mapping to $S_{\ell }(y)$, which lies in $K$. Form the finite finite pro-p presentationAll listed relations hold in $G$, giving a continuous surjective group homomorphism $P\to G$: its compact image contains $K$ and maps onto $Q$.

Let $X$ be the closed subgroup of $P$ topologically generated by the $x_i$. Both conjugation directions were included, so each $y_j$ normalizes $X$. The $x_i$ also normalize it, and its closed normalizer contains the generators of $P$. Thus $X⊴P$. The relations $R_h$ give a continuous surjection $K\to X$. Its composite with $X\to K$ induced by $P\to G$ fixes every $k_i$, and hence is the identity. Therefore the surjection $K\to X$ is injective as well, so $X\cong K$. On quotienting by $X$, the conjugation and kernel relations disappear, and the remaining presentation is exactly $⟨y_1,\dots ,y_b\mid S_1,\dots ,S_c{⟩}_{\mathrm{pro}\text{-}p}=Q$. Thus $P/X\to G/K$ is an isomorphism. The kernel of $P\to G$ lies in $X$ and meets $X$ trivially, so is trivial. Consequently $P\cong G$, proving

For a finite group, define the Frattini subgroup bythe intersection is $G$ when there are no maximal subgroups, which for a finite group happens only for $G=1$. A second definition is the set of non-generators: an element $x$ such that $⟨S,x⟩=G$ always implies $⟨S⟩=G$.

If $x\in \Phi (G)$ and $H=⟨S⟩<G$, put $H$ in a maximal subgroup $M$. Then $x\in M$, so $⟨S,x⟩\le M<G$. Conversely, if $x\notin \Phi (G)$, some maximal subgroup $M$ omits it; maximality gives $⟨M,x⟩=G$, although $M\ne G$. This proves equivalence.

Every group automorphism permutes maximal subgroups, so $\Phi (G)$ is a characteristic subgroup, in particular a normal subgroup. Alsoa proper $H$ would lie in a maximal subgroup containing both factors.

Set $F=\Phi (G)$ and take a Sylow subgroup $P$ of $F$. For $g\in G$, normality of $F$ makes $P^g$ a Sylow subgroup of $F$. Conjugating inside $F$ therefore gives the Frattini argument factorization $G=F{N}_G(P)$. The preceding generating property forces $N_G(P)=G$. Thus every Sylow subgroup of $F$ is normal, and part (i) shows that $F$ is a nilpotent group.

Now suppose $G$ is a finite p-group. The normalizer condition makes every maximal subgroup $M$ normal: $M<N_G(M)$ and maximality force $N_G(M)=G$. The quotient $G/M$ has no proper nontrivial subgroup, hence is cyclic of prime order; its p-power order makes that prime $p$. Consequently every such $M$ containswhere the two factors are the commutator subgroup and the subgroup generated by pth powers. Thus $D\le \Phi (G)$.

The quotient $G/D$ is an elementary abelian group, hence a vector space over ${\mathbb{F}}_p$. If $x\notin D$, extend the nonzero vector $xD$ to a basis and choose a linear functional nonzero on it. Its kernel pulls back to a subgroup of index $p$, a maximal subgroup omitting $x$. Therefore $x\notin \Phi (G)$, provingThe two factors in the last formula are multiplied; the superscripted expression in the TeX aid is not the formula printed in the PDF.

Choose a complement $W$ to the fixed $k$-dimensional vector subspace $U$. With column vectors and the decomposition $V=U\oplus W$, its parabolic stabilizer of a subspace isThe arbitrary upper-right block forms a normal elementary abelian group. The block-diagonal matrices form a complement, acting on it by $B\mapsto AB{D}^{-1}$. ThusWith row vectors the off-diagonal block is transposed; the subgroup description is the same after the convention change.

The rank of a transitive permutation group is the number of orbits of a stabilizer subgroup, not the rank of a matrix. For $Y\in X_k$, the invariant of its $P_k$-orbit is $r=\dim (U\cap Y)$. This invariant is complete. Choose a basis of $U\cap Y$, extend it to a basis of $U$, then add a basis of a complement of $U\cap Y$ in $Y$, and extend the resulting independent vectors to a basis of $V$. For $Y^{\prime }$ with the same intersection dimension $r$, do the same. The map between these adapted bases preserves $U$ and sends $Y$ to $Y^{\prime }$.

Each value $r=0,1,\dots ,k$ occurs: in a basis $e_1,\dots ,e_n$ with $U=⟨e_1,\dots ,e_k⟩$, takeThe hypothesis $2k\le n$ guarantees all these vectors exist. Hence

If a nonidentity power $x^j$ belongs to $H$, reduce $j$ modulo $p$ so that $1\le j<p$. There is an integer $a$ with $aj\equiv 1(\mathrm{mod}p)$, and therefore $(x^j{)}^a=x\in H$, a contradiction. ThusNow suppose that $G$ is abelian and finite. Start with $H_0=\{1\}$, and whenever $H_k\ne G$, choose $x_{k+1}\notin H_k$. The multiplication mapis a group homomorphism because all elements commute. It is surjective by construction and injective because the intersection of its two factors is trivial: $hx=h^{\prime }{x}^{\prime }$ implies $h^{\prime -1}h=x^{\prime }{x}^{-1}$ lies in that intersection. Hence $H_{k+1}\cong H_k\times C_p$, and its order is $p$ times the previous order. Strict growth and finiteness ensure that the construction ends at $G$. Thereforewhere $n$ is the number of factors, possibly zero. This is an elementary abelian group, obtained by an explicit direct product of groups construction.

Lagrange's theorem for finite groups states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$. For $x\in G$, two powers among $1,x,\dots ,x^{|G|}$ coincide, giving $x^m=1$ for some positive $m$. The least such exponent $d$ is the order of a group element, and the cyclic subgroup generated by $x$ consists of the $d$ distinct powers $1,x,\dots ,x^{d-1}$. Apply Lagrange's theorem for finite groups to obtain $d\mid |G|$.

Cauchy's theorem for finite groups states that if a prime $p$ divides the order of a finite group, the group contains an element of order $p$.

The classification of a group of order eight isHere $D_8$ denotes the dihedral group of order eight, not a group of order sixteen. To prove exhaustiveness, every element order is $1,2,4$ or $8$. An element of order eight generates the whole group, giving $C_8$.

If no element has order four or eight, every nonidentity element is an involution. Since $(ab{)}^{-1}=ab$ while $a^{-1}=a$ and $b^{-1}=b$, we have $ab=ba$. Choose an involution $a$, another $b$ outside $⟨a⟩$, and then $c$ outside the four-element subgroup $⟨a,b⟩$. The factors commute and each added factor has trivial intersection with the preceding subgroup. The internal direct product theorem gives $G\cong C_2^3$, an elementary abelian group.

Otherwise choose $a$ of order four and set $A=⟨a⟩$. It has index two, so is normal. Choose $b\notin A$. The two cosets show that $a,b$ generate $G$ and that $b^2\in A$. There is no element of order eight, so $b^2$ is either $1$ or $a^2$.

If $G$ is abelian and $b^2=1$, the internal direct product theorem yields $G=A\times ⟨b⟩\cong C_4\times C_2$. If instead $b^2=a^2$, let $c=a^{-1}b$; commutativity gives $c^2=1$ and $c\notin A$, producing the same direct product.

If $G$ is nonabelian, conjugation by $b$ is a nonidentity group automorphism of $A$: otherwise $a,b$ would commute and hence so would all of $G$. Thus $ba{b}^{-1}=a^{-1}$. With $b^2=1$ these are the relations of $D_8$; with $b^2=a^2$ they are the relations of the quaternion group $Q_8$. In either presentation every word reduces to $a^i{b}^j$, $0\le i<4$, $0\le j<2$. The standard eight-element model maps onto $G$, and equal orders make that group homomorphism an isomorphism.

Finally group isomorphisms preserve element orders. The following numbers of elements distinguish all five possibilities:

Thus the list is both exhaustive and pairwise non-isomorphic.

Regard bit strings as the elementary abelian group $G=({\mathbb{Z}}_2{)}^n$ under bitwise exclusive or. The promise saysThus $f$ is constant exactly on the cosets of $H$ and distinct between them. Determining the hidden subgroup determines its nonzero element $p$, so this is Simon's problem as an instance of the hidden subgroup problem.