Suppose $G$ is a group with a chief series $G = G_0 ≥ G_1 ≥ · · · ≥ G_r = {1}$. If $|G| = 336$, how could I find what all the possible values of $r$ are?
(Note that $H$ is the only non-abelian simple group whose order divides 336)
I now know $r$ cannot equal $1$, due to $H$. But it can equal $2$ and $6$ however, I don't know how to determine if it could be $3$, $4$ or $5$, I have tried by eye to see if I can create a series of these lengths and haven't managed to find one, I wondered if there was a better way?
See here for previous information that may help with this question.
For $r=2$ there is the example $\mathbb{Z}_2\times \text{GL}(3,2)$.
For $r=4$ there is the example $\mathbb{Z}_2\times \mathbb{Z}_3\times \text{Frob}_{56}$ where $$ \text{Frob}_{56}=\left\{\left[\begin{matrix}\zeta^j & 0\\a & \zeta^{-j}\end{matrix}\right]: a\in\mathbb{F}_8, \zeta\ \text{a generator of}\ \mathbb{F}^{*}_8 \right\}.$$
For $r=5$ there is the example $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_7\times\text{A}_4$.
For $r=6$ there is the example $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3 \times\mathbb{Z}_7$.
But, $r=1$ is not possible since there is no simple group of order $336$.
And $r=3$ is not possible since the chief factors would have to be $(\mathbb{Z}_2)^4$, $\mathbb{Z}_3$ and $\mathbb{Z_7}$; which would force the $(\mathbb{Z}_2)^4$ to be normal, and then there is no irreducible action of either of the groups of order $21$ on a $4$-dimensional space over $\mathbb{F}_2$.