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Thanks for any help or comments.

In my research I need a characterization for special linear groups over finite fields by some information of its subgroups, especially centralizers. I saw the following on page 56 of the book The Classification of the Finite Simple Groups by Gorenstein, Lyons and Solomon.

Let $G$ be a finite group such that for $x,y\in G$ and $K,J$ are components of $C_G(x), C_G(y)$ respectively. If we have the following properties, then $G$ is isomorphic to $SL_n(F)$.

1) $K,J\cong SL_{n-1}(F)$.

2) $I:=E(K\cap J)\cong SL_{n-2}(F)$.

3) $C_G(K), C_G(J)$ have cyclic Sylow $p$-subgroups.

4) $I$ is contained in a single component of $C_G(u)$ for every $u\in \langle x,y\rangle$

Do you think the above argument is true or needs some new properties or there exists a better characterization?

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    $\begingroup$ GLS = Gorenstein, Lyons, Solomon, " The Classification of the Finite Simple Groups". See here: ams.org/publications/authors/books/postpub/surv-40 $\endgroup$ – Max Horn Sep 23 '15 at 7:53
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    $\begingroup$ @Maryam, I browsed the bit of GLS that you mention and it wasn't completely clear to me that they were asserting that the four properties that you list above actually determine the group $G$ up to isomorphism. Instead they assert that the tuple $(x,y,I,J,K)$ determines the neighbourhood of $(x,K)$. This last concept is something I'm not familiar with... $\endgroup$ – Nick Gill Sep 24 '15 at 9:03
  • $\begingroup$ @Nick. Thanks. I myself could not understand the concept of neighberhood. Anyway, is there any characterization? $\endgroup$ – Maryam Sep 24 '15 at 12:50
  • $\begingroup$ @Maryam, it seems plausible that the tuple $(x,y,I,J,K)$ determines $G$ up to isomorphism... but I would need to think a little before I could see how hard it is to prove.... One other thing: it is worth mentioning that GLS take $q$ to be a power of a prime $r$. In particular the prime $p$ mentioned in item (3) is any prime $p$ dividing the order of the relevant centralizer. $\endgroup$ – Nick Gill Sep 25 '15 at 11:49
  • $\begingroup$ @Nick. Many thanks. I am looking forword for any comments. $\endgroup$ – Maryam Sep 25 '15 at 19:10

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