Smith equivalence and finite Oliver groups with Laitinen number 0 or 1

In 1960, Paul A. Smith asked the following question. If a finite group $G$ acts smoothly on a sphere with exactly two fixed points, is it true that the tangent $G$ –modules at the two points are always isomorphic? We focus on the case $G$ is an Oliver group and we present a classification of finite Oliver groups $G$ with Laitinen number $a_{G} = 0$ or $1$ . Then we show that the Smith Isomorphism Question has a negative answer and $a_{G} \geq 2$ for any finite Oliver group $G$ of odd order, and for any finite Oliver group $G$ with a cyclic quotient of order $p q$ for two distinct odd primes $p$ and $q$ . We also show that with just one unknown case, this question has a negative answer for any finite nonsolvable gap group $G$ with $a_{G} \geq 2$ . Moreover, we deduce that for a finite nonabelian simple group $G$ , the answer to the Smith Isomorphism Question is affirmative if and only if $a_{G} = 0$ or $1$ .

Keywords

finite group, Oliver group, Laitinen number, smooth action, sphere, tangent module, Smith equivalence, Laitinen–Smith equivalence

Mathematical Subject Classification 2000

Primary: 57S17, 57S25, 20D05

Secondary: 55M35, 57R65.

References

Forward citations

Publication

Received: 15 September 2001

Accepted: 17 June 2002

Published: 15 October 2002

Authors

Krzysztof Pawałowski
Faculty of Mathematics and Computer Scienc Adam Mickiewicz University ul. Umultowska 87 61-614 Poznań Poland

Ronald Solomon
Department of Mathematics The Ohio State University 231 West 18th Avenue Columbus, OH 43210–1174 USA

Volume 2, issue 2 (2002)

Krzysztof Pawałowski and Ronald Solomon