# Is a non-abelian free group fully residually a finite non-abelian simple group?

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.

My question is this:

Is a non-abelian free group fully residually a finite non-abelian simple group?

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.

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.

## 4 Answers

Yes! See my recent preprint Alternating quotients of free groups.

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:

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 $\{g_1,\ldots,g_n\}$ 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 $i$.

• Ooh, very nice! It had not occurred to me to think about the stronger result you proved. Thanks! – James May 7 '10 at 22:53

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 http://www.ams.org/mathscinet/search/publdoc.html?amp=&loc=refcit&refcit=1237075%201157915%201194786&vfpref=html&r=5&mx-pid=1971144. If you are interested in pro-p groups you might like to look at my paper http://www.ams.org/mathscinet/search/publdoc.html?amp=&loc=refcit&refcit=1971144&vfpref=html&r=11&mx-pid=1844555. 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.

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.

Okay I have found an online version of the paper by Dixon, Pyber, Seress, and Shalev at http://mathstat.carleton.ca/~jdixon/Residual.pdf so I was able to check things. We need the following:

Theorem 3: 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 $\infty$.

Now, clearly replacing $w$ by any finite set of words, the theorem is still true. Therefore, the fully residually case is true.

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.

Long and Reid's paper

Simple quotients of hyperbolic 3-manifold groups, PAMS, Volume 126, Number 3, 877–880

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.

At any rate, it's a nice paper.

• Their proof does indeed show that free groups are fully residually PSL_2. I should have remembered that above. – HJRW May 7 '10 at 21:38