# Haar measures of compact subgroups

Let $$G$$ be a locally compact group, $$K$$ a compact subgroup in $$G$$, and $$\mu_K$$ the normalized Haar measure on $$K$$: $$\mu_K(K)=1.$$ Let us denote by $$\widetilde{\mu_K}$$ the measure on $$G$$ defined as the functional on $${\mathcal C}(G)$$ by the formula $$\widetilde{\mu_K}(u)=\int_K u(t) \ \mu_K(dt), \qquad u\in{\mathcal C}(G)$$ Suppose now that $$K$$ and $$L$$ are two normal compact subgroups in $$G$$, and $$H$$ is the closed normal subgroup in $$G$$ generated by $$K$$ and $$L$$: $$H=\operatorname{Gr}(K\cup L)$$ I have two questions:

1. Is it true that $$H$$ is always compact?

and

1. Suppose $$H$$ is compact, does the following equality hold: $$\widetilde{\mu_K}*\widetilde{\mu_L}=\widetilde{\mu_H}$$ ?
• Can you do 1. in case of discrete groups? If two normal subgroups $K,L$ are finite, then the group generated by $K \cup L$ is finite ?? Apr 10 at 10:14
• @GeraldEdgar you know my imagination only suggests that if $G$ is generated by two elements $a$ and $b$ with the identities $a^2=1$ and $b^2=1$, then we can take $K=\{1,a\}$ and $L=\{1,b\}$, and the answer will be "no" if we remove the requirement that $K$ and $L$ are normal. It seems to me the answer anyway is "no", but my intuition is not good here. Apr 10 at 10:27

The answer to both questions seems yes to me.

First, only assume that $$K$$ is a normal subgroup of $$G$$. Then, $$(\text{Ad } g)_{g \in L}$$ defines a continuous action of $$L$$ by automorphisms of $$K$$ and we can define the semidirect product group $$K \rtimes L$$, which is just the set $$K \times L$$ with product $$(x,g) \cdot (y,h) = (x (gyg^{-1}),gh) \; .$$ Then $$K \times L$$ is a compact group. Because the Haar measure $$\mu_K$$ of $$K$$ is invariant under every continuous automorphism of $$K$$, we get that $$\mu_K \times \mu_L$$ is the Haar measure of $$K \times L$$.

The product map $$\theta : K \times L \to G : \theta(x,g) = xg$$ is a continuous group homomorphism. The image $$\theta(K \times L)$$ is compact and equals the group considered in the question. The Haar measure on this group is given by $$\theta_*(\mu_K \times \mu_L)$$, which is equal to $$\widetilde{\mu_K} * \widetilde{\mu_L}$$.

Note that $$\theta(K \times L) = K \cdot L$$ equals the subgroup of $$G$$ generated by $$K$$ and $$L$$.

Finally, if both $$K$$ and $$L$$ are normal in $$G$$, then for every $$g \in G$$, we have that $$g \cdot (K \cdot L) \cdot g^{-1} = (gKg^{-1}) \cdot (gLg^{-1}) = K \cdot L$$ so that $$K \cdot L$$ is normal in $$G$$. This means that $$K \cdot L$$ equals the group $$H$$ in the question.

• Stefaan, I don't understand, where is $H$ in this reasoning? Apr 10 at 12:51
• I have updated the answer to make this more clear. Apr 10 at 13:06
• Ah, yes, if $K$ and $L$ are normal subgroups, then $K\cdot L$ is a normal subgroup... I was wondering, why I couldn't prove non-compactness... :} Apr 10 at 13:28
• Stefaan, OK, this answers the first question, but your proof of the second one looks complicated. I need some time to analyse this. Apr 10 at 13:33
• Stefaan, yes, actually, this obsevration, that $H=K\cdot L$ implies everything immediately: first, for each $x\in K$ and $y\in L$ we have $$(\widetilde{\mu_K}*\widetilde{\mu_L})*\delta^{x\cdot y}=\widetilde{\mu_K}*\widetilde{\mu_L}*\delta^x*\delta^y=\widetilde{\mu_K}*\delta^x*\widetilde{\mu_L}*\delta^y=\widetilde{\mu_K}*\widetilde{\mu_L}$$ - and this means that $\widetilde{\mu_K}*\widetilde{\mu_L}$ is the Haar measure on $K\cdot L$, and second, $$(\widetilde{\mu_K}*\widetilde{\mu_L})(1)=1$$ - and this means that it is normalized. Apr 10 at 14:01