Let $G$ be a Lie Group and $H$ a closed normal subgroup of $G$. Then $G/H$ is also a Lie group. Denote $\pi: G\to G/H$ as the usual projection of Lie groups $\pi(g) = gH$.
Now, consider the short exact sequence on the complex chains $$\{1\} \rightarrow C_n(H) \xrightarrow[]{i_\#} C_n(G) \xrightarrow[]{\pi_\#} C_n(G/H) \rightarrow \{1\},\quad \quad (*)$$ where, if $\sigma: \Delta^n \to H$, then $$i_{\#}(\sigma) = i\circ \sigma = \sigma,\ i\text{ is the inclusion map,} $$ and $i_{\#}$ is extended in a linear fashion. In the same way, if $\omega: \Delta^n \to G$, then $$ \pi_\# (\omega) = \pi \circ \omega, $$ and once again, $\pi_\#$ is extended in a linear fashion, it is easy to check that $(*)$ is an exact sequence.
Moreover, it is clear that $\partial\circ i_\# = i_\# \circ\partial$, and \begin{align*} \pi_\# (\partial\omega) &= \pi_\# \left(\sum_{i=0}^{n}\left.\omega\right|_{[e_0,...\hat{e_i},...,e_n]}\right)\\ &=\left(\sum_{i=0}^{n}\left.\pi_{\#}\omega\right|_{[e_0,...\hat{e_i},...,e_n]}\right)\\ &=\left(\sum_{i=0}^{n}\left.\pi\circ\omega\right|_{[e_0,...\hat{e_i},...,e_n]}\right)\\ &=\partial (\pi\circ \omega)\\ &=\partial \pi_\# ( \omega)\\ \end{align*} implying $\partial \circ \pi_{\#} = \pi_{\#}\circ \partial$. So we can induce a long exact sequence in homology
$$...\to H_{n+1}(G/H)\to H_{n}(H)\to H_{n}(G)\to H_{n}(G/H) \to H_{n-1}(H)\to ... $$
It is right the above long exact sequence? I am asking it here because I did not find such a sequence anywhere and differential topology is not my specialty, so I would like some confirmation.