2 Possible Generalizations of Cayley's Theorem? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:11:07Z http://mathoverflow.net/feeds/question/108145 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108145/2-possible-generalizations-of-cayleys-theorem 2 Possible Generalizations of Cayley's Theorem? Jon Cohen 2012-09-26T09:45:56Z 2012-09-27T14:43:54Z <p>I'm wondering about the following 2 generalizations of Cayley's Theorem (every group embeds in a symmetric group). If these are known to be true/false, references would be appreciated.</p> <p>1) (Weak Version) Given any finite collection of (not necessarily distinct) finite groups, can we embed them simultaneously in a (finite) group so that they have pairwise-disjoint intersection (i.e., intersect only at the identity)?</p> <p>2) (Strong Version) Given an arbitrary set of arbitrary groups, can we embed them simultaneously in some group so that they have pairwise-disjoint intersection? It seems like this version may run into set-theoretic difficulties; if so an explanation/reference for those would also be welcome. </p> <p>Note of course that if such a group exists for a given collection, then we can embed it in a symmetric group by Cayley's Theorem (so these are indeed generalizations). </p> <p>Edit: To make it more interesting / rule out the obvious answer pointed out in the comments, can we arrange the embeddings so that the groups' normalizers are pairwise disjoint?</p> http://mathoverflow.net/questions/108145/2-possible-generalizations-of-cayleys-theorem/108250#108250 Answer by Derek Holt for 2 Possible Generalizations of Cayley's Theorem? Derek Holt 2012-09-27T14:43:54Z 2012-09-27T14:43:54Z <p>So the question now is, given two nontrivial finite groups $G,H$, can we embed them both in a finite group $X$ such that the normalizers of $G$ and $H$ in $X$ intersect trivially?</p> <p>I think we can do that as follows. Suppose that we can find a module $V$ for $G \times H$ over a finite field, such that neither $G$ nor $H$ fix any nonzero vectors $v \in V$, and such that there exists a vector $v \in V$ such that no nontrivial element of $G \times H$ stabilizes $v$. Then we let $X$ be the semidirect product of $V$ by $G \times H$. Because of the first property, the normalizers of $G$ and $H$ in $X$ have trivial intersection with $V$, so they are both equal to $G \times H$. The second property ensures that the complements $G \times H$ and $(G \times H)^v$ intersect trivially. So the subgroups $G$ and $H^v$ have the desired property, because their normalizers in $X$ are $G \times H$ and $(G \times H)^v$.</p> <p>We can contruct such a module $V$ as follows. Choose a prime $p$ not dividing $|G|$ or $|H|$ and let $V_1$ and $V_2$ be the deleted permutation modules for the regular permutation representations of $G$ and $H$ over ${\mathbb F}_p$. (So $V_1$ and $V_2$ have dimensions $|G|-1$ and $|H|-1$.) These have vectors $v_1$ and $v_2$ that are not stabilised by any nontrivial element of $G$ and $H$, respectively. Now let $V$ be the $G \times H$ module $V_1 \otimes V_2$ and let $v =v_1 \otimes v_2$.</p>