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?


  1. Suppose $H$ is compact, does the following equality hold: $$ \widetilde{\mu_K}*\widetilde{\mu_L}=\widetilde{\mu_H} $$ ?
  • 2
    $\begingroup$ 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 ?? $\endgroup$ Apr 10 at 10:14
  • $\begingroup$ @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. $\endgroup$ Apr 10 at 10:27

1 Answer 1


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.

  • $\begingroup$ Stefaan, I don't understand, where is $H$ in this reasoning? $\endgroup$ Apr 10 at 12:51
  • $\begingroup$ I have updated the answer to make this more clear. $\endgroup$ Apr 10 at 13:06
  • $\begingroup$ 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... :} $\endgroup$ Apr 10 at 13:28
  • $\begingroup$ Stefaan, OK, this answers the first question, but your proof of the second one looks complicated. I need some time to analyse this. $\endgroup$ Apr 10 at 13:33
  • $\begingroup$ 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. $\endgroup$ Apr 10 at 14:01

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