most recent 30 from http://mathoverflow.net 2013-05-19T20:00:49Z http://mathoverflow.net/feeds/question/95090 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95090/counting-non-isomorphic-groups-of-a-given-cardinality Igor Rivin 2012-04-24T23:18:58Z 2012-04-25T01:46:22Z

<p>Given an infinite cardinal $\kappa,$ is there some nice way to construct $2^\kappa$ non-isomorphic groups of that cardinality? In the answer to <a href="http://math.stackexchange.com/questions/119642/how-many-non-isomorphic-abelian-groups-of-order-kappa-are-there-for-kappa" rel="nofollow">this stackexchange question</a>, there is a fairly high-powered argument to show that that many (abelian) groups do exist, but it seems shocking that there is not a direct construction, like there is for $\aleph_0.$ (by the way, I doubt that allowing arbitrary, instead of just abelian, groups helps that much, but one never knows.) This question came from a conversation with our own @Joel David Hamkins.</p>

http://mathoverflow.net/questions/95090/counting-non-isomorphic-groups-of-a-given-cardinality/95104#95104 Simon Thomas 2012-04-25T01:46:22Z 2012-04-25T01:46:22Z

<p>If you just want a direct construction which avoids nontrivial set theory such as stationary sets etc., how about this?</p> <p>Step One: For each subset $S \subseteq \kappa$, let $M(S)$ be the structure $\langle \kappa; &lt; , S \rangle$, where $S$ is regarded as a unary relation. Obviously, if $S \neq T$, then $M(S)$ and $M(T)$ are non-isomorphic. </p> <p>Step Two: For each subset $S \subseteq \kappa$, encode $M(S)$ into a corresponding graph $\Gamma(S)$ so that if $S \neq T$, then $\Gamma(S)$ and $\Gamma(T)$ are non-isomorphic. (This is an easy exercise.)</p> <p>Step Three: For each subset $S \subseteq \kappa$, encode the graph $\Gamma(S)$ into a suitable group $G(S)$ with generators $\Gamma(S)$ and relations $R(S)$ which encode the adjacency relation. (This can be done using small cancellation theory.)</p>