1

Assume that a profinite group $G$ is generated by $X$ as abstract group. Here $X$ is assumed to be a closed subset of $G$ such that $X=X^{-1}$ and $1\in X$. For each $n$ let $X_n$ be the set of all finite products $x_1^{\pm 1}\cdots x_{n}^{\pm 1}$ where $x_1,\cdots, x_n\in X$. It is clear that $$G=\bigcup_{n=1}^nX_n$$

How to prove that each $X_n$ is closed in $G$?

I think that it should be simple but I have not got an answer yet.

Thankssssss.

9999
  • 959
  • So you're saying $,\langle X\rangle=G,$ (as abstract groups) , yet you require from $,X,$ to closed (in the profinite topology, I pressume...?) and contain the unit...why? Such is the exercise or you think that it must be so? – DonAntonio Mar 07 '13 at 16:04
  • Well, I assume here that $X$ is an symmetric set which contains the unit element. Oh, maybe I should edit the expression of elements in $X_n$. – 9999 Mar 07 '13 at 16:24

1 Answers1

4

Note that $X_n$, for $n > 1$, is the product of $X$, which is closed by assumption, and $X_{n-1}$, which we may assume to be closed by induction. (Here as usual the product of two subsets $A$ and $B$ of a group is the set $\{ a b : a \in A, b \in B\}$.)

Now in a topological group the product of two closed subsets need not be closed. A nice example is the sum of the subgroups $\mathbf{Z}$ and $\sqrt{2} \, \mathbf{Z}$ of $\mathbf{R}$.

But if one of the two subsets is compact, then the product is closed.

Since a profinite group is compact, its closed subsets are compact, and we are done.

Community
  • 1
Andreas Caranti
  • 68,873