Equivalent condition for involutivity for $E:=M^k\times\mathbb{R}^k$ on the distrbution defined by the forms $\alpha_j$. Apologies for the title. I had trouble coming up with one as the set up for this problem is relatively involved.
I'm working on some problems that were on an old differential geometry qualifying exam at my university. One I'm having trouble with is as follows:

Let $\{\omega_{ij}\}_{1\leq i,j\leq k}$ be the set of $k^2$ differential 1-forms on a manifold $M$. Let $E=M\times\mathbb{R}^k$ and $(y_1,\ldots,y_k)$ be the standard coordinates in $\mathbb{R}^k$. Also let $\pi :E\to M$ be the projection onto the first factor. For every $1\leq j\leq k$ define the following $1$-form $\alpha_j$ on $E$:
  $$
\alpha_j=dy_j+\sum_{l=1}^{k}y_{l}\pi^*\omega_{jl}.
$$
  Consider the distribution $H$ on $E$ defined by the forms $\alpha_1,\ldots \alpha_k$, i.e such that
  $$
H(p)=\{X\in T_pE:\alpha_1|_p(X)=\ldots =\alpha_k|_p(X)=0\}
$$
  for every $p\in E$. Prove that the distribution $H$ is involutive if and only if
  $$
d\omega_{ij}+\sum_{l=1}^{k}\omega_{il}\wedge\omega_{lj}=0
$$
  for all $1\leq i,j\leq k$. (in your solution you can use the description of involutivity of the distributions $H$ in terms of the ideal of forms, annihilating $H$.)

As the parenthetical remark states, I've tried to use the fact that $H$ is involutive if and only if there exist 1-forms $\beta_{ij}$ such that
\begin{align*}
d\alpha_j&=\sum_{l=1}^{k}\big[dy_l\wedge\pi^*\omega_{jl}-y_j\wedge\pi^*d\omega_{jl}\big]=\sum_{s=1}^{k}\alpha_s\wedge\beta_{sj}\\
&=\sum_{s=1}^{k}\left(dy_s+\sum_{l=1}^{k}y_l\pi^*\omega_{sl}\right)\wedge\beta_{sj}\\
&=\sum_{s=1}^{k}\left(dy_s\wedge\beta_{sj}+\sum_{l=1}^{k}y_l(\pi^*\omega_{sl}\wedge\beta_{sj})\right),
\end{align*}
but it just seems to get hopelessly messy. (I suppose could written significantly neater via Einstein's notation, but I'm trying to be consistent with the statement of the problem which doesn't use it.)
I have a feeling one might be able to rearrange and apply Cartan's lemma, but I don't seem to see exactly how.
Any hints or help is greatly appreciated.
 A: HINT: Look at the equation (from what you have, but I've changed indices—also, note you have a sign error in your first differentiation)
$$\sum_{\ell=1}^k dy_\ell\wedge\pi^*\omega_{j\ell}+y_\ell\pi^*d\omega_{j\ell} = \sum_{\ell=1}^k \big(dy_\ell + \sum_{i=1}^k y_i\pi^*\omega_{\ell i}\big)\wedge\beta_{\ell j}.$$
Because the $dy_\ell$ are linearly independent (and don't depend on the coordinates on $M$), you can see that we must have $\beta_{\ell j} = \pi^*\omega_{j\ell}$. [This is where the Cartan lemma comes in, to get rid of any $dy_k$ terms. Offhand you could have a contribution of $\sum a_{j\ell i}dy_i$, with symmetry in $i$ and $\ell$, but then the equation forces skew-symmetry as well in $i$ and $\ell$.]
EDIT: This was a bit glib, and I cannot reconstruct what I was thinking about  a few weeks ago with that last sentence. How about this?
You were exactly right (in your comment) to rewrite the equation thusly:
$$\sum_{\ell=1}^k dy_\ell\wedge(\pi^*\omega_{j\ell}-\beta_{\ell j}) = \sum_{\ell=1}^k y_\ell \big({-}\pi^*d\omega_{j\ell} + \sum_{i=1}^k \pi^*\omega_{i\ell}\wedge\beta_{ij}\big).$$
Let's set $\psi_{\ell j} = \pi^*\omega_{j\ell}-\beta_{\ell j}$. Then we have
$$\sum_{\ell=1}^k \big(dy_\ell + \sum_{i=1}^k y_i\pi^*\omega_{\ell i}\big)\wedge \psi_{\ell j} = \sum_{\ell=1}^k y_\ell\pi^*\big(\sum_{i=1}^k \omega_{\ell i}\wedge\omega_{ij} - d\omega_{\ell j}\big).$$
Since the $dy_\ell$ are linearly independent from the $1$-forms on $M$, and since the RHS is a $2$-form coming from $M$, we deduce that the $M$-part of $\psi_{\ell j}$ must vanish, and thus the RHS must be $0$. (We can use a Cartan lemma calculation to express the $dy_\ell$ part of $\psi_{\ell j}$ in the usual way, but it's really not relevant.) Now, setting $y_\ell = 1$ and $y_i=0$ otherwise, we recapture the desired equations $d\omega_{\ell j} = \sum\limits_{i=1}^k \omega_{\ell i}\wedge\omega_{ij}$. 
Most likely there are typos in here, as I've used different notations on different pieces of paper. Good luck! Maybe we're missing something obvious, but — if not — I don't think this makes a reasonable qualifying exam question.
