User user1257 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:23:35Z http://mathoverflow.net/feeds/user/26052 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121912/reference-for-the-set-of-orders-of-its-elements Reference for the set of orders of its elements User1257 2013-02-15T16:11:08Z 2013-02-16T11:22:15Z <p>I am looking for a reference for the maximal order of an element in PSL(2, $q$), where $q$ is prime power.</p> http://mathoverflow.net/questions/105834/reference-request Reference request User1257 2012-08-29T13:37:07Z 2012-10-13T15:37:00Z <p>I am looking for a reference or proof for the following problem:</p> <p>Problem: Let $r$ be prime, then $2r$ is a Sylow $p$-number if and only if $2r=1+p^{2^n}$. Thanks in advance.</p> http://mathoverflow.net/questions/106909/finite-solvable-group/106914#106914 Answer by User1257 for Finite solvable group User1257 2012-09-11T13:58:29Z 2012-09-11T13:58:29Z <p>We note that the Suzuki groups are only non-Abelian simple groups of order prime to $3$ and $5$ is a prime divisor of the Suzuki groups. Let $G$ be unsolvable group, then $G$ has the following normal series: $1\unlhd K\lhd M\unlhd G$ such that $M/K$ is a non-Abelian simple group (and or $M/K\cong S\times$ $S\cdot \cdot \cdot \times S$ where $S$ is non-Abelian simple group). As $3\nmid |G|$, then $M/K$ is a Suzuki group ( and or $M/K\cong S\times$ $S\cdot \cdot \cdot \times S$ where $S$ is a Suzuki group). On the other hand $5\nmid |G|$, then $M/K$ is not isomorphic to a Suzuki group (and or $M/K\ncong S\times$ $S\cdot \cdot \cdot \times S$ where $S$ is a Suzuki group), a contradiction.</p>