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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 $1\leq i\leq n-1$.

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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.

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Dear Nick Thank you very much for your complete and useful answer. Is it still true if $p$ divides the order of $G$? –  BHZ Jun 20 '13 at 6:28
    
@BHZ Before I could give an answer I'd need you to address the issues in my "remark" above. Because if the centre of $SL_n(q)$ is non-trivial, there will be NO $C_{pp}$-groups according to your definition, and so the answer will be trivially YES. If we use the definition that I proposed in the remark, then things become more tricky. –  Nick Gill Jun 20 '13 at 8:29
    
Dear Nick Thanks for your answers. Let me state our question in a different manner. Let $G$ be a subgroup of $PSL(n,q)$ such that $G$ is not a subgroup of $PSL(n-1,q)$, where $q=p^\alpha$. If $G$ is a simple group and also is a $C_{pp}$ group with this condition that $p$ divides the order of $G$, can we say that there exists a primitive prime divisor $r$ of $q^n-1$ or $q^{n-1}-1$ such that $r$ divides $|G|$? Now we use the same definition of $C_{pp}$ as usual. Thanks –  BHZ Jun 20 '13 at 10:28
    
I've just realised that my "remark" was somewhat misguided - you are talking about the centralizer in $G$, not in $SL_n(q)$, thus my concerns about $Z(SL_n(q))$ don't really make any difference, sorry. And this allows me to assert a different counter-example to your original question, namely by taking $G$ to be a Sylow $p$-subgroup of $SL_n(q)$. –  Nick Gill Jun 20 '13 at 12:15
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The existence of $\varphi$ is just standard rep theory. Perhaps it would be more transparent if you took $k$ to be a field of characteristic $r$ where $r$ does not divide the order of $G$, your simple group. Then $G$ is trivially a $C_{pp}$-group and the representation theory of $G$ over $k$ is ``the same" as representation over $\mathbb{C}$, so you can just consult character tables for $G$ to get irreducible representations of the type I've described. –  Nick Gill Jun 20 '13 at 13:54

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