Background: Suppose we have a differential equation $$ y^{(n)}=\omega(x,y,y',\ldots,y^{(n-1)}) $$ that satisfies two symmetries $\mathbf{X}_1$ and $\mathbf{X}_2$. That is, if $$ \mathbf{A}:=\frac{\partial}{\partial x}+y'\frac{\partial}{\partial y}+\cdots+\omega\frac{\partial}{\partial y^{(n-1)}}, $$ then we have $$ [\mathbf{X}_1,\mathbf{A}]=\lambda_1\mathbf{A},\quad[\mathbf{X}_2,\mathbf{A}]=\lambda_2\mathbf{A} $$ for some functions $\lambda_1$, $\lambda_2$ of $x,y,\ldots,y^{(n-1)}$. $[\cdot,\cdot]$ is the commutator and is defined by $[A,B]=AB-BA$.
Question: How can we show that $[\mathbf{X}_1,\mathbf{X}_2]$ is again a symmetry?
Atempt: We must show that $[[\mathbf{X}_1,\mathbf{X}_2],\mathbf{A}]=\rho\mathbf{A}$ for some function $\rho$. By the Jacobi identity, we have $$ [[\mathbf{X}_1,\mathbf{X}_2],\mathbf{A}]=[\mathbf{X}_1,\lambda_2\mathbf{A}]-[\mathbf{X}_2,\lambda_1\mathbf{A}] $$ In my book, they continue from here and just write $$ \begin{eqnarray} [[\mathbf{X}_1,\mathbf{X}_2],\mathbf{A}] &=& [\mathbf{X}_1,\lambda_2\mathbf{A}]-[\mathbf{X}_2,\lambda_1\mathbf{A}] \\ &=&\lambda_2\lambda_1\mathbf{A}+(\mathbf{X}_1\lambda_2)\mathbf{A}-\lambda_2\lambda_1\mathbf{A}-(\mathbf{X}_2\lambda_1)\mathbf{A} \\ &=&\rho\mathbf{A} \end{eqnarray} $$ My Problem: I don't see how can we go from the first line to the second.
If we use the definition of commutators, we have
$$ \begin{eqnarray} [[\mathbf{X}_1,\mathbf{X}_2],\mathbf{A}] &=& [\mathbf{X}_1,\lambda_2\mathbf{A}]-[\mathbf{X}_2,\lambda_1\mathbf{A}] \\ &=&\mathbf{X}_1\lambda_2\mathbf{A}-\lambda_2\mathbf{A}\mathbf{X}_1-\mathbf{X}_2\lambda_1\mathbf{A}+\lambda_1\mathbf{A}\mathbf{X}_2 \end{eqnarray} $$ and they say that this is equal to $$\lambda_2\lambda_1\mathbf{A} + (\mathbf{X}_1\lambda_2)\mathbf{A}-\lambda_2\lambda_1\mathbf{A}-(\mathbf{X}_2\lambda_1)\mathbf{A} = (\mathbf{X}_1\lambda_2)\mathbf{A}-(\mathbf{X}_2\lambda_1)\mathbf{A},$$ so in other words, they claim that $$\lambda_1\mathbf{A}\mathbf{X}_2=\lambda_2\mathbf{A}\mathbf{X}_1$$ or equivalently $$ \lambda_1 \mathbf{X}_2 \mathbf{A}=\lambda_2\mathbf{X}_1\mathbf{A} $$ Why?
If you remember the Jacobi identity then you just have to remember how the commutator behaves when you have a product by a function: $$ [X,fY]=(Xf)Y+f[X,Y] $$ and then the result follows