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Let G be a locally compact topological group with Haar measure $d_G$, H be a compact subgroup of G with normalized Haar measure $d_H$ and N be the smallest normal subgroup of G containing of H with Haar measure $d_N$.

First question: Is there any relation between the Haar measures $d_G,d_H$ and $d_N$?

Second question: Can we write $d_G|_H=d_H$?

Third question: Can we write $d_H(H)=d_N(N)$?

Thank you for the help.

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    $\begingroup$ Frequently $H$ would have measure zero in $G$, so the restriction of $d_G$ to $H$ would yield the zero measure. For instance, $SO(n)$ as a subgroup of $GL(n)$. I think the construction you're looking for is not restriction but disintegration. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2019 at 1:48
  • $\begingroup$ For the third question, assuming $N$ is compact, how do you normalize $d_N$? If $d_N$ is normalized Haar measure then the statement $d_H(H)=d_N(N)$ is trivial since they are both 1 by definition. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2019 at 1:59
  • $\begingroup$ It could also happen that $N$ is not compact. For instance, take $G = GL(n)$ and $H = O(n)$. Then $d_N(N)=\infty \ne d_H(H)$. $\endgroup$
    – Nate Eldredge
    Commented Mar 31, 2019 at 2:14
  • $\begingroup$ Tank you very much for your useful guidance. $\endgroup$
    – B.Gillan
    Commented Nov 12, 2019 at 2:05

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