I am trying to understand the following puzzle, at least for me, in the context of one-parameter subgroups of the Special Euclidean group $\mathbb{SE}_{n}(\mathbb{R})$ in n dimensions.
Specifically, I consider $\mathbb{SE}_{n}$ as a subgroup of $\mathbb{GL}_{n+1}(\mathbb{R})$, that is $$ \mathbb{SE}_{n}(\mathbb{R})=\left\{\left(\matrix{R & v \\ 0 &1}\right) , R\in SO(n), v\in\mathbb{R}^{n}\right\}. $$ An element in the Lie algebra of $\mathbb{SE}_{n}(\mathbb{R})$ is then written as $$ H=\left(\matrix{\Omega & v \\ 0 &0}\right) $$ where $\Omega$ is a skew-symmetric $(n\times n)$ matrix with real entries. Then, the exponential of $H$ can be computed by direct application of the series expansion of the exponential, and it is found to be $$ \exp(H)=\left(\matrix{\mathrm{e}^{\Omega} & Bv \\ 0 &1}\right). $$ where $$ \mathbb{I}_{n} + \sum_{k\geq 1} \frac{\Omega^{k}}{(k+1)!} $$ $\mathbb{I}_{n}$ being the identity matrix in n dimensions.
At this point, we use the identity $$ B=\mathbb{I}_{n} + \sum_{k\geq 1} \frac{\Omega^{k}}{(k+1)!}=\int_{0}^{1}\mathrm{e}^{\lambda\Omega}\mathrm{d}\lambda $$ and the expression of $\mathrm{e}^{H}\in\mathbb{SE}_{n}(\mathbb{R})$ is complete.
Now, the puzzle begins. Suppose we want to compute the one-parameter subgroup generated by $H$. To do so, and by definition of one-parameter subgroup, we simply have to replace $H$ with $tH$ in the formula above, obtaining $$ \mathrm{e}^{tH}=\left(\matrix{\mathrm{e}^{t\Omega} & t\,B_{t}v \\ 0 &1}\right) $$ where $$ B_{t}=\int_{0}^{1}\mathrm{e}^{\lambda t\,\Omega}\mathrm{d}\lambda $$
Let us check the subgroup property, that is, the property $$ \mathrm{e}^{(t+s)H}=\mathrm{e}^{tH}\,\mathrm{sH}. $$ The LHS is immediately found to be $$ \mathrm{e}^{(t+s)H}=\left(\matrix{\mathrm{e}^{(t+s)\Omega} & (t+s)\,B_{t+s}v \\ 0 &1}\right) $$ with $$ B_{t+s}=\int_{0}^{1}\mathrm{e}^{\lambda (t+s)\,\Omega}\mathrm{d}\lambda . $$ On the other hand, performing the matrix multiplication, the RHS is found to be $$ \mathrm{e}^{tH}\,\mathrm{e}^{sH}=\left(\matrix{\mathrm{e}^{(t+s)\Omega} & t\,\mathrm{e}^{s H}\,B_{t}v + s B_{s}v\\ 0 &1}\right). $$
If the subgroup property holds, as it should be, then we must have $$ \left(\matrix{\mathrm{e}^{(t+s)\Omega} & (t+s)\,B_{t+s}v \\ 0 &1}\right)=\left(\matrix{\mathrm{e}^{(t+s)\Omega} & t\,\mathrm{e}^{s H}\,B_{t}v + s B_{s}v\\ 0 &1}\right) $$ which I am not able to prove. Furthermore, since $[tH,sH]=0$, we must also have $$ \mathrm{e}^{sH}\,\mathrm{e}^{tH}=\mathrm{e}^{tH}\,\mathrm{e}^{sH} $$ which means $$ \left(\matrix{\mathrm{e}^{(t+s)\Omega} & t\,\mathrm{e}^{s H}\,B_{t}v + s B_{s}v\\ 0 &1}\right)=\left(\matrix{\mathrm{e}^{(t+s)\Omega} & s\,\mathrm{e}^{t H}\,B_{s}v + t B_{t}v\\ 0 &1}\right), $$ and, again, I am not able to prove this statement.
What am I missing?
Try this: \begin{align} \text{LHS}&=\\ te^{s\Omega}\int_0^1e^{\lambda t \Omega}d\lambda+s\int_0^1e^{\lambda s \Omega}d\lambda &=e^{s\Omega}\int_0^te^{\lambda\Omega}d\lambda+\int_0^se^{\lambda \Omega}d\lambda\\ &=\int_0^te^{(\lambda+s)\Omega}d\lambda+\int_0^se^{\lambda \Omega}d\lambda\\ &=\int_s^{s+t}e^{\lambda\Omega}d\lambda+\int_0^se^{\lambda \Omega}d\lambda\\ &=\int_0^{s+t}e^{\lambda\Omega}d\lambda\\ &=(s+t)\int_0^{1}e^{\lambda(s+t)\Omega}d\lambda\\ &=(s+t)B_{s+t}\\ &=\text{RHS} \end{align}