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I am looking specifically at whether the measure is translation invariant. The group in question is the group $M$ of $n\times n$ unipotent upper triangular matrices, i.e. matrices like this one $$ \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 5 & 6 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 1\end{bmatrix}$$ with ones on the diagonal. The normal subgroup I am looking at is the group $N$ of unipotent upper triangular matrices which are zero within one element of the diagonal; e.g. this matrix $$ \begin{bmatrix} 1 & 0 & 2 & 3 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$ This group of matrices is isomorphic to to the group of $(n-1) \times (n-1)$ unipotent upper triangular matrices. Translation is defined componentwise, ignoring the diagonal, so translating the first matrix by the second would result in $$ \begin{bmatrix} 1 & 2 & 5 & 7 \\ 0 & 1 & 5 & 10 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 1\end{bmatrix}$$

Now, if we look at the quotient space of $M/N$, members of which can be represented by matrices like this one:$$ \begin{bmatrix} 1 & 2 & * & * \\ 0 & 1 & 5 & * \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 1\end{bmatrix},$$ The quotient space is isomorphic to $\mathbb{R}^{n-1}$. We can define translation in $M/N$ as left-multiplication by another member of $M/N$, so the left Haar measure on $M/N$ is definitely translation-invariant.

Now, what I want to show is that the left Haar measure on $M$ itself is translation-invariant (meaning it is the Lebesgue measure, considering each matrix as a vector in $\mathbb{R}^{n(n-1)/2}$). Is it true that if

a.) the Haar measure is translation-invariant over the normal subgroup $N \trianglelefteq M$, and

b.) it is translation-invariant on sets of cosets in $M/N$, then

c.) it is translation-invariant on $M$?

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Define $\Delta_G(y) = \frac{\mu_G(Ay)}{\mu_G(A)}$ to be the modular function on $G$. Then, we have the following theorem:

Let $G$ be a locally compact group and $H$ be a closed subgroup. Then, there exists a left-invariant Radon measure on the quotient $G/H$ if and only if $\Delta_G|_H = \Delta_H$. Further, there is a unique choice of measure $\mu$ that satisfies the following version of Fubini's Theorem: $$\int_G f(g) dg = \int_{G/H}\int_H f(xh) dh d\mu(x)$$

A proof is in Chapter 1 of Deitmar's Principles of Harmonic Analysis, under the name Quotient Integral Formula.

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    $\begingroup$ Likewise, if $G$ locally compact, $H\leq G$ closed, $K\leq G$ compact, $G=HK$, then we can arrange Haar measures such that $\displaystyle \int_G f\ dg = \int_H \int_K f(hk)\ dk\ dh$. Proof is in the same place. $\endgroup$
    – Araske
    Dec 5 '17 at 0:13

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