Pro P Groups and the P Adic Integers

If $G$ is pro-$p$, then [F1] gives a presentation with each $P_i$ a finite $p$-group. By [L1] this already makes $G$ profinite, and [L2] supplies an open normal basis whose quotients are the coordinate images ${\pi }_i(G)\le P_i$, hence finite $p$-groups.

Conversely, suppose $G$ is profinite and let $\mathcal{N}$ be an identity-neighbourhood basis of open normal subgroups with each $G\mathrm{/}N$ a finite $p$-group. Replace it by the basis of finite intersections of its members. The new basis has the same properties, because a quotient by $N_1\cap \cdots \cap N_r$ embeds in the product of the finite $p$-groups $G\mathrm{/}{N}_j$, and it is directed by reverse inclusion. The quotient maps define a continuous homomorphism . Because the basis separates points, $\eta$ is injective. A basic cylinder in the target prescribes finitely many compatible cosets, and their common refinement in $\mathcal{N}$ has a representative in $G$, so $\eta$ is surjective. Its coordinate maps are exactly the quotient maps by open normal subgroups from the refined basis, so $\eta$ is a homeomorphism. The target is an inverse limit of finite $p$-groups, hence [F1] makes $G$ pro-$p$.

Steps 1.1 and 1.2 prove both implications. The trivial group is included: it is the inverse limit of the constant system on $1$, and its only finite quotient is the trivial $p$-group.

Because reduction modulo $p^n$ is a group homomorphism, compatibility of the tuples in [F1] gives $(x_{n+1}+y_{n+1})<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p^n=x_n+y_n$ and $(-x_{n+1})<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p^n=-x_n$ for every $n$. Thus $x+y$ and $-x$ are again compatible tuples, so they lie in ${\mathbb{Z}}_p$.

Associativity, commutativity, the zero element, and additive inverses all hold coordinatewise because they hold in every finite quotient $\mathbb{Z}\mathrm{/}{p}^n\mathbb{Z}$. The inverse-limit topology is the subspace topology from the product of the discrete coordinate groups, so coordinatewise addition and negation are continuous. Hence ${\mathbb{Z}}_p$ is a topological abelian group.

Let $U_n(x):=\{y=(y_r)\in {\mathbb{Z}}_p:y_r=x_r\text{ for }1\le r\le n\}$. By [F1], this is exactly the metric ball $U_n(x)=\{y\in {\mathbb{Z}}_p:d_p(x,y)\le p^{-n}\},$ because the first $n$ residue coordinates agree if and only if the largest initial block of equal coordinates has length at least $n$.

By [L1], the same set $U_n(x)$ is the trace on ${\mathbb{Z}}_p$ of the cylinder in the product space that fixes the first $n$ coordinates, so every basic metric ball is inverse-limit open. Conversely, let a basic inverse-limit cylinder $C$ contain $x$ and restrict the finite set of coordinates $F$. Taking $n\ge r$ for every $r\in F$, compatibility in [L2] shows that $U_n(x)\subseteq C$. Thus the sets $U_n(x)$ refine every inverse-limit neighbourhood of $x$.

The sets $U_n(x)$ therefore form a neighbourhood basis for both topologies at every point $x$. So the inverse-limit topology and the metric topology coincide.

By construction, ${\mathbb{Z}}_p$ is an inverse limit of the finite discrete groups $\mathbb{Z}\mathrm{/}{p}^n\mathbb{Z}$. Therefore [L1] gives that ${\mathbb{Z}}_p$ is Hausdorff and totally disconnected, and also compact under the extra Choice hypothesis named in the Statement. The group structure from [L3] is already compatible with this topology.

Fix $n\ge 1$. Since the sequence is Cauchy and [L2] identifies the metric balls with the cylinder neighbourhoods, there exists $M_n$ such that for all $m,r\ge M_n$ the first $n$ coordinates of $x^{(m)}$ and $x^{(r)}$ agree. Let $a_n$ be that eventual common $n$-th coordinate. The compatibility of the $x^{(m)}$ forces the tuple $a=(a_n{)}_{n\ge 1}$ itself to be compatible, hence an element of ${\mathbb{Z}}_p$.

For each $n$ and every $m\ge M_n$, the first $n$ coordinates of $x^{(m)}$ and $a$ agree, so step 1.2 gives $d_p(x^{(m)},a)\le p^{-n}$. Given $\epsilon >0$, choose $n$ with $p^{-n}<\epsilon$; then every $m\ge M_n$ satisfies $d_p(x^{(m)},a)<\epsilon$. Thus $x^{(m)}\to a$ in the $p$-adic metric. Every Cauchy sequence therefore converges, so ${\mathbb{Z}}_p$ is complete.

Steps 1.1 and 2.1 prove the stated properties.

If an integer $m$ maps to $0$, then [F1] says $m<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p^n=0$ for every $n\ge 1$. Thus every power $p^n$ divides $m$, which is possible only for $m=0$. So the canonical map is injective.

A basic neighbourhood in ${\mathbb{Z}}_p$ fixes some residue class modulo $p^n$. Let $x=(x_r)\in {\mathbb{Z}}_p$ and let $n\ge 1$. Choose an integer $m$ representing the coordinate $x_n\in \mathbb{Z}\mathrm{/}{p}^n\mathbb{Z}$. Then [F1] gives that the image of $m$ has $n$-th coordinate $x_n$, and [L1] implies that this image agrees with $x$ in every earlier coordinate as well. So every basic neighbourhood of $x$ meets the embedded copy of $\mathbb{Z}$, which is therefore dense.

Step 1.1 proves injectivity and step 1.2 proves density. The zero element is treated in both arguments without any extra case split.

Write $m=p^{a}u$ with $a\ge 0$ and $p\nmid u$. Multiplication by $u$ is an automorphism of each cyclic group $\mathbb{Z}\mathrm{/}{p}^n\mathbb{Z}$, so from $mx=0$ it follows that $p^{a}x=0$. Assume for contradiction that $x\mathrm{\ne }0$, and choose the least index $t\ge 1$ with $x_t\mathrm{\ne }0$. By [F1], the earlier coordinates vanish and the tuple is compatible, so $x_t$ is represented by an integer divisible by $p^{t-1}$; because $x_t\mathrm{\ne }0$ in $\mathbb{Z}\mathrm{/}{p}^t\mathbb{Z}$, that representative is not divisible by $p^t$. Thus $x_t$ has exact $p$-adic divisibility $p^{t-1}$.

Compatibility propagates that exact divisibility to the coordinate $x_{t+a}\in \mathbb{Z}\mathrm{/}{p}^{t+a}\mathbb{Z}$, because reducing $x_{t+a}$ modulo $p^t$ gives the nonzero class $x_t$. Hence $p^a{x}_{t+a}$ is divisible by $p^{t+a-1}$ but not by $p^{t+a}$, so it is nonzero in $\mathbb{Z}\mathrm{/}{p}^{t+a}\mathbb{Z}$. This contradicts $p^{a}x=0$. Therefore $x=0$, and ${\mathbb{Z}}_p$ is torsion-free.

The subgroup generated by $1$ consists exactly of the integer multiples $n\cdot 1$, so it is the canonical image of $\mathbb{Z}$ in ${\mathbb{Z}}_p$. Therefore [L1] says that its closure is all of ${\mathbb{Z}}_p$.

For every binary sequence $ϵ=({ϵ}_i{)}_{i\ge 0}\in \{0,1{\}}^{\mathbb{N}}$, define a compatible tuple $x(ϵ)$ by $x(ϵ{)}_n:={\sum }_{i=0}^{n-1}{ϵ}_i{p}^i(\mathrm{m}\mathrm{o}\mathrm{d}{p}^n)\mathrm{.}$ If two binary sequences first differ at index $j$, their $(j+1)$-st coordinates differ because their difference is divisible by $p^j$ but not by $p^{j+1}$. Hence $ϵ\mapsto x(ϵ)$ is injective.

Cantor's diagonal argument shows that $\{0,1{\}}^{\mathbb{N}}$ is uncountable, so step 1.2 makes ${\mathbb{Z}}_p$ uncountable. Every abstract cyclic group is the image of the countable group $\mathbb{Z}$ under $n\mapsto n\gamma$ for one generator $\gamma$, and is therefore countable. Hence the additive group of ${\mathbb{Z}}_p$ cannot be cyclic.

Step 1.1 proves topological generation by $1$, while steps 1.2 and 2.1 show that no element can generate the additive group abstractly.

Every subgroup of the additive group $\mathbb{Z}$ has the form $d\mathbb{Z}$. The quotient $\mathbb{Z}\mathrm{/}d\mathbb{Z}$ is a finite $p$-group exactly when $d=p^n$ for some $n\ge 0$, because every finite quotient of a cyclic group is cyclic. Thus the inverse system in [F1] is exactly the system of quotients $\mathbb{Z}\mathrm{/}{p}^n\mathbb{Z}$ with the usual reduction maps.

The compatible-tuple inverse limit of the system from step 1.1 is precisely ${\mathbb{Z}}_p$, and [L1] is the resulting canonical cone map from $\mathbb{Z}$ to that inverse limit. By [L2], this is the universal pro-$p$ completion map. Therefore the pro-$p$ completion of $\mathbb{Z}$ is ${\mathbb{Z}}_p$.

For each positive integer $n={\prod }_p{p}^{v_p(n)}$, the Chinese remainder theorem gives $\mathbb{Z}\mathrm{/}n\mathbb{Z}\cong {\prod }_{p\mid n}\mathbb{Z}\mathrm{/}{p}^{v_p(n)}\mathbb{Z}$. These decompositions are compatible with the reduction maps as $n$ varies by divisibility.

Passing to the inverse limit over all $n$ therefore separates the finite quotients prime by prime: . The right-hand inverse limit is ${\mathbb{Z}}_p$ by [L1], and [L2] identifies the induced map as the unique compatible morphism. Hence $\widehat{\mathbb{Z}}\cong {\prod }_p{\mathbb{Z}}_p$.

Choose $x\in G\setminus H$. By [L1], write $G$ as an inverse limit of finite groups and view $H$ as a closed subset of that product. Since $x\mathrm{\notin }H$, some cylinder neighbourhood of $x$ misses $H$; equivalently, there is an open normal subgroup $N$ of $G$ such that $xN\cap H=\mathrm{\varnothing }$.

The subgroup $HN$ is open because it is a union of $N$-cosets, and it is closed because it is a finite union of closed cosets. If $HN=G$, then $x=hn$ for some $h\in H$ and $n\in N$, so $h=x{n}^{-1}\in xN\cap H$, contradicting step 1.1. Thus $HN$ is a proper closed subgroup containing $H$. By maximality of $H$, one must have $HN=H$, hence $N\le H$. Therefore $H$ contains an open neighbourhood of the identity and is itself open.

By [F1] and [L1], $\mathrm{\Phi }(G)$ is the intersection of the maximal open subgroups of $G$. If $M_i$ is a maximal subgroup of some finite quotient $G_i$, then ${\pi }_i^{-1}(M_i)$ is a maximal open subgroup of $G$. Conversely, every maximal open subgroup $U$ of $G$ contains the kernel of some coordinate projection, so $U={\pi }_i^{-1}(M_i)$ for a maximal subgroup $M_i<G_i$.

Therefore an element $x\in G$ lies in $\mathrm{\Phi }(G)$ exactly when ${\pi }_i(x)$ lies in every maximal subgroup of every finite quotient $G_i$, that is, exactly when ${\pi }_i(x)\in \mathrm{\Phi }(G_i)$ for every $i$. The coordinatewise condition defines the inverse limit of the subgroups $\mathrm{\Phi }(G_i)$, so .

Let $K:=\overline{[G,G]G^p}$. For every open normal subgroup $N⊴G$, the finite quotient $G\mathrm{/}N$ is a finite $p$-group, and the image of $K$ in $G\mathrm{/}N$ is $\overline{[G,G]G^p}N\mathrm{/}N=[G\mathrm{/}N,G\mathrm{/}N](G\mathrm{/}N{)}^p=\mathrm{\Phi }(G\mathrm{/}N)$ by [L2] and [F1].

By [L1], $\mathrm{\Phi }(G)$ is the subgroup of all elements whose image in every finite quotient $G\mathrm{/}N$ lies in $\mathrm{\Phi }(G\mathrm{/}N)$. Step 1.1 shows that $K$ has exactly the same image in every such quotient. Closed subgroups of a profinite group are determined by their images in all finite quotients, so $K=\mathrm{\Phi }(G)$.

Suppose $S$ topologically generates $G$. Then for every open normal subgroup $N$, the image of $S$ generates the finite $p$-group $G\mathrm{/}N$. By [L3], its image therefore generates $(G\mathrm{/}N)\mathrm{/}\mathrm{\Phi }(G\mathrm{/}N)$, and by [L2] that is the same as spanning the Frattini quotient over ${\mathbb{F}}_p$. Passing over all finite quotients shows that the image of $S$ spans $G\mathrm{/}\mathrm{\Phi }(G)$.

Conversely, suppose the image of $S$ spans $G\mathrm{/}\mathrm{\Phi }(G)$. Let $N$ be any open normal subgroup. The image of $S$ in $(G\mathrm{/}N)\mathrm{/}\mathrm{\Phi }(G\mathrm{/}N)$ then spans by functoriality of the Frattini quotient and [L1], so [L2] says that the image of $S$ generates $G\mathrm{/}N$. Hence the closure of the subgroup generated by $S$ surjects onto every finite quotient $G\mathrm{/}N$, and therefore equals $G$. So $S$ topologically generates $G$.

Steps 1.1 and 2.1 prove the equivalence. The empty set fits the statement when $G=1$, because then $G\mathrm{/}\mathrm{\Phi }(G)=0$.

If $f$ is surjective, then every quotient map induced by $f$, including $G\mathrm{/}\mathrm{\Phi }(G)\to H\mathrm{/}\mathrm{\Phi }(H)$, is surjective.

Conversely, suppose $G\mathrm{/}\mathrm{\Phi }(G)\to H\mathrm{/}\mathrm{\Phi }(H)$ is surjective, and let $K:=f(G)\le H$. The image of $K$ in $H\mathrm{/}\mathrm{\Phi }(H)$ is all of $H\mathrm{/}\mathrm{\Phi }(H)$ by hypothesis, so [L1] says that $K$ topologically generates $H$. But $K$ is compact as the continuous image of the profinite group $G$, hence closed in the Hausdorff group $H$. A closed subgroup whose closure is all of $H$ must equal $H$, so $f$ is surjective.

Steps 1.1 and 1.2 prove both implications. In the trivial-group boundary case, both maps are automatically surjective.

By [L1], ${\mathbb{Z}}_p$ remembers only the finite quotients of $\mathbb{Z}$ whose order is a power of $p$.

By [L2], the full profinite completion also has the $q$-primary factor ${\mathbb{Z}}_q$ for every prime $q\mathrm{\ne }p$. Therefore ${\mathbb{Z}}_p$ omits the prime-to-$p$ information and cannot be the whole profinite completion.

So the stated claim is false.

The element $1\in {\mathbb{Z}}_p$ is nonzero.

If the additive group of ${\mathbb{Z}}_p$ were torsion, some positive multiple of $1$ would be $0$. That contradicts [L1], since [L1] says that no nonzero element of ${\mathbb{Z}}_p$ has finite order.

Therefore the statement is false. Finite coordinate groups do not force torsion in the inverse limit.

By [L1], the underlying set of ${\mathbb{Z}}_p$ is uncountable, whereas the subgroup generated by any one element is the countable image of $\mathbb{Z}$. Thus no element generates ${\mathbb{Z}}_p$ as an abstract group.

In particular, step 1.1 rules out every infinite cyclic realization, while [L2] rules out every nontrivial finite cyclic realization. Hence the claim is false.

Therefore ${\mathbb{Z}}_p$ is topologically generated by one but not cyclic as an abstract group.

The theorem [L1] applies to the finitely generated pro-$p$ group fixed in the Statement, and the notation $G^p$ is exactly that of [F2].

Therefore $\mathrm{\Phi }(G)=\overline{[G,G]G^p}$ as stated. The closure cannot be discarded merely by reading the abstract-group notation $[G,G]G^p$.

The profinite group $\widehat{\mathbb{Z}}$ has nontrivial continuous quotients ${\mathbb{Z}}_q$ for every prime $q$ by [L1].

If $\widehat{\mathbb{Z}}$ were pro-$p$ for some fixed prime $p$, then every finite quotient of $\widehat{\mathbb{Z}}$ would be a $p$-group by [F1]. For each prime $q\mathrm{\ne }p$, the projection to the $q$-adic factor followed by reduction modulo $q$ gives a quotient $\widehat{\mathbb{Z}}\twoheadrightarrow {\mathbb{Z}}_q\twoheadrightarrow {\mathbb{Z}}_q\mathrm{/}q{\mathbb{Z}}_q\cong \mathbb{Z}\mathrm{/}q\mathbb{Z},$ which is a nontrivial $q$-group. This contradiction shows that not every profinite group is pro-$p$.

Therefore the statement is false.