Probability of generating group Let $G$ a finitely generated pro-$p$-group, with $p$ a prime;
if $\mu$ is the unique Haar measure on $G$ with $\mu(G)=1$, we can define $P_{G}(k)=\displaystyle \mu\left(G^{k}\setminus \bigcup_{M} M^{k}\right)$, where the union is over all the maximal open subgroups of $G$.
This is the probability of generating the group $G$ with $k$ elements.
If $\phi(G)$ is the frattini subgroup of $G$ we have also $P_{G/\phi(G)}(k)$, the probability of generating the finite group $G/\phi(G)$ with $k$ elements, that is the obvious one.
But why we have $P_{G}(k)=P_{G/\phi(G)}(k)$?.
(I think this is reasonable because say that $<g_{1},...,g_{k}>=G$ (topologically) is the same as $<g_{1},...,g_{k}>\phi(G)=G$, that is the same as $<\bar{g_{1}},...,\bar{g_{k}}>=G/\phi(G)$, but I would like a formal proof of the above equality ).
 A: Let start with an observation: if $G$ is a countable based group and if $X\subset G$, closed, then we have $X=\displaystyle \bigcap_{i=1,...,\infty} XN_{i}$, with $N_{1}\geq N_{2}\geq...$ normal open subgroup of $G$.
Now we have a formula: $\mu(X)=\displaystyle \inf_{N}\frac{|XN/N|}{|G/N|}$.
Proof: From the previous observations we have $\mu(X)=\mu(\displaystyle \bigcap_{i=1,...,\infty} XN_{i})=\displaystyle\inf_{i=1,...,\infty}\mu(XN_{i})\geq \inf_{N}\mu(XN)\geq\mu(X)$, because $X\subset XN$.
So we have $\displaystyle\mu(X)=\inf_{N}\mu(XN)$, but $XN=\displaystyle\bigcup_{j=1,...,m}x_{j}N$, with $m=|XN/N|$; by the properties of the Haar measure we obtain $\mu(X)=\displaystyle \inf_{N}\mu(XN)=\mu(X)=\inf_{N}\frac{|XN/N|}{|G/N|}$.
This formula tell us that we can reduce to consider abstract group in many situations, for example in the following:
Let $G$ a countable based group and if $K\subset G$, closed, and $\pi:G\longrightarrow G/K$. If $Y\subset G/K$, closed, then $\mu(\pi^{-1}(Y))=\mu(Y)$.
In fact, from the previous formula we can suppose $G$ finite.
And now if $X\subset G$, then $\mu(X)=\frac{|X|}{|G|}$.
Finally we have the following:
Let $G$ a pro-$p$-goup with $d(G)<\infty$. Then $P_{G}(k)=P_{G/\phi(G)}(k)$.
Proof: we consider for all the group $H$ the set $\phi_{k}(H)=\lbrace (h_{1},...,h_{k})\in H^{k} : \overline{<h_{1},...,h_{k}>}=H\rbrace$.
This is a closed subset in $H$. Moreover since $\overline{<g_{1},...,g_{k}>}=G \leftrightarrow <g_{1},...,g_{k}>\phi(G)=G \leftrightarrow <\bar{g}_{1},...,\bar{g}_{k}>=G/\phi(G)$ we get $\pi^{-1}(\phi_{k}(G/\phi(G))=\phi_{k}(G)$, with $\pi:G^{k}\longrightarrow(G/\phi(G))^{k}$, (in fact one can prove that for a finitely generated pro-$p$-goup $\phi(G)$ is closed). Finally since $P_{G}(k)=\mu(\phi_{k}(G))$ from the previous observations we get the conclusion.
