Can nonabelian groups be detected "locally"? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:14:41Z http://mathoverflow.net/feeds/question/77883 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77883/can-nonabelian-groups-be-detected-locally Can nonabelian groups be detected "locally"? Gjergji Zaimi 2011-10-12T01:42:48Z 2011-10-13T21:43:55Z <p>Suppose $m,n\geq 2$ are two integers. Is it true that for every sufficiently large nonabelian group $G$, one can find a set $A\subset G$, with $|A|=n$, so that $|A^m| >\binom{n+m-1}{m}$?</p> <p>(Edit) Let's also add the condition $m\le n$ since the answer below provides a counter-example for large enough $m$. In general it would be interesting to know the range of $(m,n)$ for which the statement holds.(/Edit)</p> <p>Here $A^k=\lbrace a_1a_2\cdots a_k| a_1,a_2,\dots,a_k\in A\rbrace$ is a product set. It is obvious that in every abelian group one has $|A^m| \le\binom{n+m-1}{m}$, for every $A$. </p> <p>I don't have an application in mind, I was trying the case $m=2$ and I think I have a proof (still haven't checked all the steps, but it's not particularly enlightening since it splits into many cases). I'm wondering if this is true in general and if there is a slick proof, or if there is a counter-example.</p> http://mathoverflow.net/questions/77883/can-nonabelian-groups-be-detected-locally/77903#77903 Answer by Guntram for Can nonabelian groups be detected "locally"? Guntram 2011-10-12T07:46:57Z 2011-10-12T08:23:27Z <p>The answer is no. Consider the family of groups $G_k:=(\mathbf Z_2{}^k)\rtimes \mathbf Z_2$, where the group on the right acts by interchanging the first and second coordinate. Then the commutator subgroup $G_k'$ is generated by $g_k:=((1,1,0,\ldots, 0),0)$, i.e. is of order two. So $G_k$ is non-abelian (of arbitrarily large order), but as little non-abelian as possible. Note that each element in $G_k$ has order dividing 4. </p> <p>Now let $A=\lbrace a_1,\ldots, a_n \rbrace \subseteq G_k$. Then any element in $A^m$ can be written as $a_1^{e_1} \cdots a_n^{e_n} g_k^{e_k}$, where $0\leq e_i \leq 3$ for $i=1,\ldots n$ and $e_k=0,1$. This follows by applying the identity $ab=[a,b]ba$ repeatedly. In particular, $|A^m|$ is bounded by a constant depending on $n$ only, not on $m$.</p> <p>Edit: To be explicit, any two elements in $G_k$ will generate a subgroup of order at most 32, so $(n,m)=(2,31)$ is one counter-example.</p> http://mathoverflow.net/questions/77883/can-nonabelian-groups-be-detected-locally/78005#78005 Answer by S. Carnahan for Can nonabelian groups be detected "locally"? S. Carnahan 2011-10-13T07:54:17Z 2011-10-13T07:54:17Z <p>If I'm not mistaken, the family of groups <code>$\left\{ Q_8 \times (\mathbb{Z}/2\mathbb{Z})^{\times k} \right\}_{k \geq 0}$</code> is a counterexample to the case $m=n=2$, where $Q_8$ is the quaternion group with 8 elements.</p> <p><strong>Claim:</strong> Any subset $A$ with 2 elements yields $A^2$ with at most 3 elements.</p> <p>Write two elements as $(g,x)$ and $(h,y)$, with $g,h \in Q_8$ and $x,y \in (\mathbb{Z}/2\mathbb{Z})^{\times k}$. If $g$ and $h$ commute, then $(g,x)(h,y) = (h,y)(g,x)$ and we get at most 3 elements. If $g$ and $h$ don't commute, then $g^2 = h^2$ by the special property of $Q_8$, so $(g,x)^2 = (h,y)^2$ and we get exactly 3 elements.</p> http://mathoverflow.net/questions/77883/can-nonabelian-groups-be-detected-locally/78069#78069 Answer by Gjergji Zaimi for Can nonabelian groups be detected "locally"? Gjergji Zaimi 2011-10-13T21:43:55Z 2011-10-13T21:43:55Z <p>Here is a collection of what I have so far thanks to the answers by Guntram and S. Carnahan. Let's denote by $P(n,m)$ the property that $|A^m| \le\binom{n+m-1}{m}$ for all subsets $|A|=n$. </p> <p>We have that the only nonabelian $P(2,2)$ groups are of the form $Q_8\times G$ where $G$ is an elementary abelian 2-group, and that $P(3,2)$ groups have to be abelian by Freiman's paper <a href="http://www.springerlink.com/content/3040n1420586v676/" rel="nofollow">"On two- and three-element subsets of groups"</a>. </p> <p>In <a href="http://www.jstor.org/pss/2159119" rel="nofollow">"A characterization of abelian groups"</a>, Brailovsky proves that large enough $P(n,2)$ are abelian by showing that $P(n,2)\implies P(n',2)$ for all $n\geq n'\geq 2$, so that the result follows from the previous paragraph.</p> <p>In "Small squaring and cubing properties for finite groups", Berkovich, Freiman and Praeger prove that the only nonabelian group with $P(2,3)$ is $S_3$.</p> <p>On the other hand there are nonabelian groups with $P(n,m)$ whenever $\binom{n+m-1}{m}\geq 2^{2n+1}$ as in Guntram's answer.</p>