Let $G$ be a subgroup of $SL(n,q)$ such that $G$ is not a subgroup of $GL(n-1,q)$, where $q=p^\alpha$. If $G$ is a $C_{pp}$ group, i.e. the centralizer of each $p$-element is a $p$-group, can we say that there exists a primitive prime divisor $r$ of $q^n-1$ or $q^{n-1}-1$ such that $r$ divides $|G|$?
We call $r$ a primitive prime divisor of $q^n-1$, where $r\mid (q^n-1)$ but $r\nmid (q^i-1)$ for each $1\leq i\leq n-1$.
I believe the answer is NO in general. For $n>3$ one obtains a counter-example by taking any maximal torus $T$ corresponding to a partition $\{k, n-k\}$ of $n$ such that $2\leq k \leq n-2$. Clearly $T$ is a $C_{pp}$ group because it doesn't contain any $p$-elements. On the other hand it clearly doesn't preserve an $(n-1)$-dimensional space. Finally it is clear that its order is not divisible by a primitive prime divisor of $q^n-1$ or $q^{n-1}-1$.
Similarly if $n=3$ and $6$ divides $q-1$, then the normalizer of a split torus provides a counter-example. If $6$ does not divide $q-1$, then I'd have to think some more.
If $n=2$, then a counter-example is given by a Sylow $p$-subgroup where $p$ is the characteristic of the field.
Edit: In the comments, the OP has asked for a counter-example where $G$ is simple. For this take $G$ to be $PSL_2(q')$ where $q'$ is a power of $p$. Then $G$ is a simple group and a $C_{pp}$-group. Let $\phi:PSL_2(q')\to GL_n(k)$ be an irreducible representation over $k$, the algebraic closure of the field of order $q'$. This will yield an embedding of $PSL_2(q')$ in $GL_n(q)$ for any $q$ bigger than some constant. Because of irreduciblity $G$ does not lie in $GL_{n-1}(q)$. Now provided we choose $q$ large enough so that $q'^2 < q^{n-1}$, the condition on primitive prime divisors is violated, as required.
