Isometrically embedding $L^1(H)$ into $L^1(G)$

Question

Let $G$ be a locally compact group and $H$ a closed subgroup of $G$. Denote $\mu_G$ as Haar measure on $G$ and $\mu_H$ as Haar measure on $H$. Is it possible to embed $L^1(H)$ into $L^1(G)$? By Radon-Nikodym both can be identified as the absolutely continuous measures, $M_a(H)$ and $M_a(G)$ respectively, so I believe that it would suffice to show that $\mu_H$ is absolutely continuous with respect $\mu_G$. This holds in the case when $H$ is open because then $\mu_H$ is just the restriction of $\mu_G$ on $H$, but it is unclear to me what happens when $H$ is not closed. As a more general question, what can one say is the relationship between $\mu_G$ and $\mu_H$?