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Thanks for any comments

Let $G=GL_n(F)$ be general linear group over finite field $F$. Consider two isomorphic subgeoup $H_1,H_2$ of $G$ such that $H_i\cong GL_k(\bar{F})$, where $\bar{F}$ is an extention of $F$ of prime degree. Is it possible $H_1\cap H_2\neq Z(G)$

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  • $\begingroup$ Where does this question come from? What if $H_1=H_2$? It seems that you assert that any single subgroup with your property is the center. $\endgroup$ Commented May 13, 2015 at 21:44
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    $\begingroup$ The answer to the question is clearly yes, even if you assume $H_1 \ne H_2$. For example ${\rm GL}(4,2) \cong A_8$ and ${\rm GL}(2,4) \cong A_5 \times C_3$, and there are many possible intersections of pairs of subgroups isomorphic to ${\rm GL}(2,4)$. $\endgroup$
    – Derek Holt
    Commented May 13, 2015 at 21:58
  • $\begingroup$ I'm voting to close this question because it has been answered in a comment. $\endgroup$
    – Stefan Kohl
    Commented May 13, 2015 at 22:29
  • $\begingroup$ @Alex. It is clear that $H_1$ and $h_2$ must be distinct. $\endgroup$
    – maryam
    Commented May 14, 2015 at 5:55
  • $\begingroup$ @Professor Holt. Thanks. For other finite fields and other integer $n$, $GL(n,q)$ there exist such a counterexample? $\endgroup$
    – maryam
    Commented May 14, 2015 at 6:04

1 Answer 1

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In addition to @DerekHolt's observation, it is easily possible for the centralizer of part of a rational representation of a $GL_r(F')$ in $GL_n(F)$ (with $F'$ a finite extension of $F$) to be non-trivial, while the centralizer of the whole image is smaller. Then the conjugate of the image by anything outside the centralizer of the part will create a copy of $GL_r(F')$ that intersects the original in a proper subgroup, and that subgroup is at least as large as the centralizer, so properly larger than the center...

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  • $\begingroup$ I do not understand your statement. In general $GL_n(F)$ have subgroups isomorphic with $GL_r(F')$ and are distinct. For all $r$ and $n$ and for every pair of such subgroups, intersection is trivial or not? $\endgroup$
    – maryam
    Commented May 14, 2015 at 6:11

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