# Haar Measure: Unimodular Locally Compact Groups

I have the following problem: "Let $G$ be a locally compact group, all of whose normal subgroups are contained in $Z(G)$. Prove that $G$ is unimodular."

My attempt at attacking the problem was to consider the homomorphism $\Delta:G \rightarrow \mathbb{R}^{\times}_+$ defined by \begin{equation*} \int_G f(yx)d_rx = \Delta(y)\int_Gf(x)d_rx \end{equation*} where $d_rx$ is a right Haar measure on $G$. Then $G$ will be unimodular if and only if $\Delta = 1$. However, I am having some difficulty showing that this holds. Any advice is appreciated.

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I'm guessing proper normal subgroups is meant, to keep the problem from being trivial? – Harry Altman Oct 15 '12 at 16:55
I don't know the answer, but note that $\ker(\Delta)$ is a normal subgroup of $G$ (you need to show it's $G$), and $\int_G f(yxy^{-1}) \, d_r x = \Delta(y) \int_G f(x) \, d_r x$. Maybe that could help. – Joel Cohen Oct 15 '12 at 17:09
Harry Altman: You are correct of course. – James Miller Oct 15 '12 at 17:16
Joel Cohen: A similar argument shows that $Z(G) \leq \ker \Delta$ so that certainly helps. – James Miller Oct 15 '12 at 17:17