Is the group algebra separable?

Let $G$ be a locally compact Hausdorff group. Is the group algebra $L^1(G)$ separable? Would you please help me or introduce references that can help me. Thanks

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No, because any discrete group is locally compact and admits a Haar measure that assigns the measure 1 to any element. $L^1$ on a non-countable set with such measure isn’t separable.
BTW I doubt that we necessarily have separability even for compact groups. Compactness is not a very strong condition to infer “nice” properties of function spaces. There are spaces that are compact and Hausdorff albeit very thick, in the sense there are damn much functions on them (although I am not sure it is the case for $L^1$).
No, even if the group is compact, Abelian and separable. For instance, the Cantor group $\{0,1\}^{\omega_1}$, where $\omega_1$ is the first uncountable cardinal, is a good counter-example. This group is obviously compact and Abelian. It is also separable by the Hewitt–Marczewski–Pondiczery theorem. However $L_1(\{0,1\}^{\omega_1})$ is non-separable.
Indeed, if it were separable, the group C*-algebra $C^*_r(\{0,1\}^{\omega_1})$ would be separable as it contains $L_1(\{0,1\}^{\omega_1})$ as a dense subspace. This is not true because $$\widehat{\{0,1\}^{\omega_1} } = \bigoplus_{\alpha<\omega_1} \{0,1\},$$ so you have uncountably pair-wise orthogonal non-zero projections in $C^*_r(\{0,1\}^{\omega_1})$.