Maximal subgroups of abelian groups and Q-algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:36:57Z http://mathoverflow.net/feeds/question/13258 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13258/maximal-subgroups-of-abelian-groups-and-q-algebras Maximal subgroups of abelian groups and Q-algebras DoubtingThomas 2010-01-28T15:43:55Z 2010-01-28T20:31:49Z <p>Let $G$ be an abelian group which does not have a maximal subgroup. Does it follow that $G$ is a $\mathbb{Q}$-algebra?</p> <p>It is easy to see that $\mathbb{Q}$-algebras do not admit any maximal subgroups. I am looking for an example which is not of this kind. If such an example does not exist, it would be nice to see a proof that $\mathbb{Q}$-algebras are the only ones with this property.</p> http://mathoverflow.net/questions/13258/maximal-subgroups-of-abelian-groups-and-q-algebras/13277#13277 Answer by Rishi Vyas for Maximal subgroups of abelian groups and Q-algebras Rishi Vyas 2010-01-28T20:31:49Z 2010-01-28T20:31:49Z <p>Assuming that by a $\mathbb{Q}$ algebra you mean a $\mathbb{Q}$ vector space, the abelian groups that admit a $\mathbb{Q}$ - vector space structure are precisely the divisible torsion-free abelian groups, i.e. torsion-free abelian groups A such that $\forall$ $x\in A, n\in \mathbb{N}$, $\exists$ $y\in A$ s.t. $ny=x$. The condition is clearly necessary, and for sufficiency, consider the canonical mapping of $A$ into $\mathbb{Q}\otimes_{\mathbb{Z}}A$. This will be an isomorphism precisely when $A$ satisfies the above mentioned condition. </p> <p>As for the question regarding maximal subgroups, one can show that an abelian group has no maximal subgroups if and only if it is divisible, in the sense mentioned above. If a group is not divisible, then there will exist a prime $p$ such that $pA$ is a proper subgroup of $A$; we can then use the fact that $A/pA$ is a vector space over $\mathbb{Z}/p\mathbb{Z}$ to pick out a maximal subgroup. On the other hand, maximal subgroups in abelian groups must always be of finite (in fact prime) index. Thus, if we have a maximal subgroup $B$ in $A$, of index $p$, then $pA\subseteq B$. However, by divisibility, $pA=A$, a contradiction.</p>