Let $G=GL(n,q)$ be a generalized linear group whose center is a cyclic group of prime order $p$. Does there always exists an element $x\in G$ such that $C_G(x)$ is a group of exponent $p$?
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$\begingroup$ If the center has order a prime, then $q=3$ or $q$ is even. One can examine $GL(2,3)$ directly, so you really only need to worry about $q$ even.... You'll be done if you can show that $G$ contains a regular semisimple element with eigenvalues in $\mathbb{F}_q$, and this is always the case when $n>q$. If $n\leq q$, then I think the answer will be no. $\endgroup$– Nick GillCommented Nov 27, 2013 at 10:21
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$\begingroup$ @Nick: have you got this the right way round? It seems to be true in ${\rm GL}(3,4)$ but false in ${\rm GL}(4,4)$ and ${\rm GL}(5,4)$, for example. $\endgroup$– Derek HoltCommented Nov 27, 2013 at 11:18
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$\begingroup$ @DerekHolt Sorry, you're quite right! I believe the answer is yes with $q>n$ but not with $q\leq n$. $\endgroup$– Nick GillCommented Nov 27, 2013 at 12:00
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1 Answer
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As I mentioned in comments, either $q=3$ or $q$ is even.
Suppose that $q>n$. Then $G$ contains a regular semisimple element $x$ with eigenvalues in $\mathbb{F}_q$, then its centralizer is a maximal split torus of $G$, which is elementary abelian of order $(q-1)^n$. Thus the answer is YES in this case.
Suppose that $q\leq n$. Let's try to find an element $x$ such that $C_G(x)$ has exponent $p$. Clearly $x$ must be semisimple. If $x$ has eigenvalues outside $\mathbb{F}_q$, then it lies in a torus which contains a cyclic subgroup of order $q^i-1$ where $i>1$. This subgroup lies in $C_G(x)$ which is a contradiction. Thus all eigenvalues of $x$ lie in $\mathbb{F}_q$. Since $q\leq n$ we know that at least one of these eigenvalues is repeated. Now one can check that $|C_G(x)|$ is divisible by $r$, the field characteristic, so again we have a contradiction. Thus the answer is NO in this case.
