Is a non-abelian free group fully residually a finite non-abelian simple group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:01:15Z http://mathoverflow.net/feeds/question/23898 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23898/is-a-non-abelian-free-group-fully-residually-a-finite-non-abelian-simple-group Is a non-abelian free group fully residually a finite non-abelian simple group? James 2010-05-07T21:12:19Z 2010-05-08T08:39:11Z <p>It is well known that a non-abelian free group is residually a finite simple group. Katz and Magnus proved, in fact, that non-abelian free groups are residually alternating and residually $PSL_{2}$. S. J. Pride has some nice results along these lines as well. The best result that I know of is the theorem of Weigel that can be formulated as follows. If $\mathfrak{X}$ is a group-theoretic class containing an infinite set of pairwise non-isomorphic finite non-abelian simple groups, then every non-abelian free group is residually an $\mathfrak{X}$-group.</p> <hr> <p>My question is this:</p> <blockquote> <p>Is a non-abelian free group fully residually a finite non-abelian simple group?</p> </blockquote> <p>It seems likely that the answer to such an obvious question is known, but I have not been able to find it in the literature.</p> <p>I should probably add that I suspect we can probably replace "finite non-abelian simple" with "alternating", but I haven't yet given any thought to the other infinite series. I'd like to learn whether anything is known before spending more time on this.</p> http://mathoverflow.net/questions/23898/is-a-non-abelian-free-group-fully-residually-a-finite-non-abelian-simple-group/23899#23899 Answer by HW for Is a non-abelian free group fully residually a finite non-abelian simple group? HW 2010-05-07T21:28:59Z 2010-05-07T21:36:17Z <p>Yes! See my recent preprint <a href="http://arxiv.org/abs/1005.0015" rel="nofollow">Alternating quotients of free groups</a>.</p> <p>I expect that what you want is well known, but I too couldn't find it in the literature. In fact, I prove the much stronger result that free groups are something like 'locally extended fully residually alternating'. Specifically:</p> <blockquote> <p>Let $F$ be any free group of rank at least two, let $H$ be a finitely generated subgroup of infinite index in $F$ and let <code>$\{g_1,\ldots,g_n\}$</code> be a finite subset of $F\smallsetminus H$. Then there is a surjection $f$ from $F$ to a finite alternating group such that $f(g_i)$ is not in $f(H)$ for any $i$. </p> </blockquote> http://mathoverflow.net/questions/23898/is-a-non-abelian-free-group-fully-residually-a-finite-non-abelian-simple-group/23900#23900 Answer by Yiftach Barnea for Is a non-abelian free group fully residually a finite non-abelian simple group? Yiftach Barnea 2010-05-07T21:33:39Z 2010-05-07T22:58:51Z <p>Indeed Thomas Weigel was the first to prove the full result. However, there is a probablistic proof due to Dixon, Pyber, Seress, and Shalev, see <a href="http://www.ams.org/mathscinet/search/publdoc.html?amp=&amp;loc=refcit&amp;refcit=1237075%201157915%201194786&amp;vfpref=html&amp;r=5&amp;mx-pid=1971144" rel="nofollow">http://www.ams.org/mathscinet/search/publdoc.html?amp=&amp;loc=refcit&amp;refcit=1237075%201157915%201194786&amp;vfpref=html&amp;r=5&amp;mx-pid=1971144</a>. If you are interested in pro-p groups you might like to look at my paper <a href="http://www.ams.org/mathscinet/search/publdoc.html?amp=&amp;loc=refcit&amp;refcit=1971144&amp;vfpref=html&amp;r=11&amp;mx-pid=1844555" rel="nofollow">http://www.ams.org/mathscinet/search/publdoc.html?amp=&amp;loc=refcit&amp;refcit=1971144&amp;vfpref=html&amp;r=11&amp;mx-pid=1844555</a>. Now, it has been a while since I thought about it, but I would guess that the probablistic argument will actually show also the fully residually case.</p> <p>Thinking about it a bit more, I cannot see immediately how to prove the fully residually case for pro-p groups. However, if I remembr correctly the probablistc proof in the discrete case, then the fully residually case should still be fine. The issue is: given $\mathfrak{X}$ an infinite family of non isomorphic non-abelian finite simple groups, fix an identity (i.e. an element of the free group), what is the probabilty that two random elments of a group in $\mathfrak{X}$ do not satisfy the identity? If it tends to 1 with the size of the group, then the fully residually case should be true.</p> http://mathoverflow.net/questions/23898/is-a-non-abelian-free-group-fully-residually-a-finite-non-abelian-simple-group/23901#23901 Answer by Richard Kent for Is a non-abelian free group fully residually a finite non-abelian simple group? Richard Kent 2010-05-07T21:34:06Z 2010-05-07T21:34:06Z <p>Long and Reid's paper </p> <p>Simple quotients of hyperbolic 3-manifold groups, PAMS, Volume 126, Number 3, 877–880</p> <p>contains a proof of the theorem that free groups are residually PSL_2, and I think their proof shows that they are fully residually PSL_2.</p> <p>At any rate, it's a nice paper.</p> http://mathoverflow.net/questions/23898/is-a-non-abelian-free-group-fully-residually-a-finite-non-abelian-simple-group/23921#23921 Answer by Yiftach Barnea for Is a non-abelian free group fully residually a finite non-abelian simple group? Yiftach Barnea 2010-05-08T07:58:49Z 2010-05-08T08:39:11Z <p>Okay I have found an online version of the paper by Dixon, Pyber, Seress, and Shalev at <a href="http://mathstat.carleton.ca/~jdixon/Residual.pdf" rel="nofollow">http://mathstat.carleton.ca/~jdixon/Residual.pdf</a> so I was able to check things. We need the following:</p> <p><strong>Theorem 3</strong>: Let $S$ be a finite simple group and let $w$ be a non-trivial element of the free group $F_2$ on $X,Y$. Then the probability that two randomly chosen elements $x$ and $y$ of $S$ satisfy both that $x$ and $y$ generate $S$ and $w(x,y) \neq 1$ tends to $1$ as $|S|$ tends $\infty$.</p> <p>Now, clearly replacing $w$ by any finite set of words, the theorem is still true. Therefore, the fully residually case is true.</p> <p>Now, the pro-$p$ case is more difficult because while the random generation has probability $1$ the not satisfying an identity has only positive probability. But it may be that the proof itself still works for finite number of words.</p>