# Compact subgroups are contained in open compact subgroups in locally profinite groups

Let $G$ be a totally disconnected, Hausdorff, locally compact group. In the wikipedia page about these groups there is a claim that any compact subgroup of $G$ is contained in some compact open subgroup. Is this true? If so, why?

It seems the following.

By [Pon, Theorem 16], the group $G$ has a base at the unit consisting of its open compact subgroups. In particular, the group $G$ contains a compact open subgroup $K$. Now let $C$ be a compact subgroup of the group $G$. Put $N=\bigcap \{x^{-1}Kx:x\in C\}$. It is easy to check that $N$ is a closed subgroup of the group $K$ (and hence a compact) and $CN$ is a group. The set $CN$ is compact as a product of two compact sets. We claim that $N$ is an open subset of the group $G$. Indeed, let $x\in C$ be an arbitrary element. Since $x^{-1}ex\in K$, and $K$ is open, there exist open neighborhoods $V_x\ni e$ and $W_x\ni x$ such that $W_x^{-1}V_xW_x\subset K$. The family $\{W_x:x\in C\}$ is an open cover of a compact space $C$. Therefore there exists a finite subset $F$ of the group $C$ such that $C\subset\bigcup\{W_x:x\in F\}$. Put $V=\bigcap\{V_x:x\in F\}$. Then $x^{-1}Vx\subset K$ for each element $x\in C$. Thus $V\subset N$. Hence for each element $y\in N$ we have $Vy\subset N$. Therefore, the set $N$ is open. Similarly we can show that the set $CN$ is open. Therefore $CN$ is an open compact subgroup of the group $G$, which contains the group $C$.

References

[Pon] Lev S. Pontrjagin, Continuous groups, 2nd ed., M., (1954) (in Russian).

• Thanks for posting this proof. Let me ask for some details: why is $CN$ a subgroup? Consider two elements $ck^x$ and $dh^y$, with $c,d,x,y\in C$, $h,k\in K$ and $k^x,h^y\in N$. Then, there exists $k'\in K$ such that $k^x=(k')^{yd^{-1}}$, so that $ck^x dh^y=c(k')^{yd^{-1}}dh^y=(cd)(k')^y$. Still, why is this $(k')^y$ belonging to $N$? – Simone May 12 '15 at 8:43
• @Simone It seems the following. I think that a key observation here is that $y^{-1}Ny\subset N$ for each $y\in C$. Indeed, let $z\in N$ be an arbitrary element. Then for any element $x\in C$ we have $x^{-1}y^{-1}zyx=(yx)^{-1}z(yx)\in N$ because $z\in N$ and $yx\in C$. Thus, $y^{-1}zy\in N$. – Alex Ravsky May 12 '15 at 10:13
• I do not see why the proof in your comment works, it seems a circular argument: to show that $y^{-1}Ny\subseteq N$ for a given $y\in C$, you use that $(yx)^{-1}N(yx)\subseteq N$ for all $x\in C$. But for $x=1$ this was the statement you wanted to prove... – Simone May 12 '15 at 14:26
• @Simone Oops, it was a misprint, sorry. It should be $(yx)^{−1}z(yx)\in K$ instead. – Alex Ravsky May 12 '15 at 22:43
• Now I see, thanks! I think your argument perfectly works now. There is a misprint in your answer: on the 5th line you claim that $C$ is open, I think you mean either $N$ or $CN$, any of the two would work. – Simone May 13 '15 at 7:36