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



> 
> 2. Suppose $H$ is compact, does the following equality hold:
> $$
>  \widetilde{\mu\_K}\*\widetilde{\mu\_L}=\widetilde{\mu\_H}
>  $$
> ?
> 
> 
> 



Haar measures of compact subgroups
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.


