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May 14, 2015 at 7:54 comment added Derek Holt Yes, as explained in the answer by paul garrett, it is very easy to construct counterexamples.
May 14, 2015 at 6:04 comment added maryam @Professor Holt. Thanks. For other finite fields and other integer $n$, $GL(n,q)$ there exist such a counterexample?
May 14, 2015 at 5:55 comment added maryam @Alex. It is clear that $H_1$ and $h_2$ must be distinct.
May 13, 2015 at 23:58 history closed Alex Degtyarev
Derek Holt
Stefan Kohl
YCor
Igor Rivin
Needs details or clarity
May 13, 2015 at 22:35 answer added paul garrett timeline score: 0
May 13, 2015 at 22:29 comment added Stefan Kohl I'm voting to close this question because it has been answered in a comment.
May 13, 2015 at 22:03 review Close votes
May 14, 2015 at 0:02
May 13, 2015 at 21:58 comment added Derek Holt 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 ${\rm GL}(2,4)$.
May 13, 2015 at 21:44 comment added Alex Degtyarev 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.
May 13, 2015 at 21:41 review First posts
May 13, 2015 at 21:44
May 13, 2015 at 21:40 history asked maryam CC BY-SA 3.0