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 funtional 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: $$ \widetilde{\mu_K}*\widetilde{\mu_L}=\widetilde{\mu_H} $$ ?

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: $$ \widetilde{\mu_K}*\widetilde{\mu_L}=\widetilde{\mu_H} $$ ?