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Consider a locally compact group with Haar measure $m$, and recall that the commutator if the subgroup generated by all elements of the form $[a,b]:=a^{-1}b^{-1}ab$.

I'd like to show that the continuous modular homomorphism $\phi:G\to \mathbb{R}^+$ defined by the equality $(R_g)_\ast (m)=\phi(g)m$ (where $R_g$ is the right translation by $g$ and $(R_g)_\ast(m)$ is the measure defined by $(R_g)_\ast(m)(B)=m(Bg)$) is trivial on the commutator $[G,G]$, that is $[G,G]\subset \ker \phi$. This amounts to showing

$$m(B[g,h])=m(B)$$ for $g,h\in G$ and $B\subset G$ measurable. My original goal was to show that if $G$ is perfect (i.e. $[G,G]=G$), then $G$ is unimodular (i.e. $m$ is right invariant), which is obviously implied by the previous claim.

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As $\phi$ is a homomorphism into an Abelian group, it must be trivial in the commutator.

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  • $\begingroup$ Ah, I feel like an idiot. Thanks! $\endgroup$ – Reveillark Jul 27 '17 at 17:05
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Unimodular means the image of $G$ under the modular character is the trivial subgroup. $G$ is a product of things like $xyx^{-1}y^{-1}$.

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