most recent 30 from http://mathoverflow.net 2013-05-24T18:16:37Z http://mathoverflow.net/feeds/question/46046 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46046/a-question-about-finite-groups Andrei Moroianu 2010-11-14T14:46:45Z 2010-11-14T17:40:48Z

<p>Let $k$ be a positive integer. Is it true that any finite group $H$ of cardinal $4k+2$ whose center contains an element $h$ of order $2$ is isomorphic to the direct product $H=(\mathbb{Z}/2\mathbb{Z})\times G$, where $G=H/\{1,h\}$? </p> <p>An equivalent statement would be: Let $G$ be a finite group of odd cardinal. Is it true that the second cohomology group $H^2(G,\mathbb{Z}/2\mathbb{Z})$ with respect to the trivial action of $G$ on $\mathbb{Z}/2\mathbb{Z}$, vanishes? </p>

http://mathoverflow.net/questions/46046/a-question-about-finite-groups/46047#46047 Tim Dokchitser 2010-11-14T15:08:44Z 2010-11-14T15:08:44Z

<p>Yes, because $G$ and $Z/2Z$ have coprime order, every extension of one by the other splits by the Schur-Zassenhaus theorem (http://en.wikipedia.org/wiki/Schur-Zassenhaus_theorem). If you prefer, $H^2(G,M)$ is killed both by $|G|$ and by $|M|$, so it must be trivial in your case.</p>

http://mathoverflow.net/questions/46046/a-question-about-finite-groups/46056#46056 Derek Holt 2010-11-14T17:40:48Z 2010-11-14T17:40:48Z

<p>There is a reasonably simple argument that any group $H$ of twice odd order has a normal subgroup $N$ of index 2. Given that, if you know also that $H$ has a central subgroup $Z$ of order 2, then it is straightforward to show that $H \cong N \times Z$.</p> <p>The argument goes like this. By Cayley's Theorem, there is an isomorphism $\phi$ that maps $G$ to a permutation group of degree $|G|$ in which all non-identity elements of $G$ act without fixed points. So the elements of order 2 in $G$ are mapped by $\phi$ onto odd permutations, and hence the inverse image $N$ in $G$ of the intersection of the image of $\phi$ with the alternating group has index 2 in $G$.</p>