If $G/[G,G]$ is compact, then $G$ is unimodular.

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How to prove Proposition 2.29 in page no. 52 of the book " A Course in Abstract Harmonic Analysis (2nd edition)" by G.B. Folland which states that:

If $G/[G,G]$ is compact, then $G$ is unimodular.

Note: A proof is provided but I am unable to grasp it.

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This is me rewriting Folland's argument.

The unimodular map is $\Delta: G \to (0,\infty)_\times$. Notice that $[G,G] \subset \text{ker}(\Delta)$ because $\Delta([x,y]) = [\Delta(x),\Delta(y)] = 1$. Thus this induces a map $\tilde \Delta:G/[G,G] \to (0,\infty)_\times$ such that $\tilde\Delta([G,G]x) = \Delta(x)$. Since $G/[G,G]$ is compact, the range of $\tilde\Delta$ is compact. And the only compact subgroup of $(0,\infty)_\times$ is $\{1\}$.