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Here is a very natural question:

Q: Is it always possible to generate a finite simple group with only $2$ elements?

In all the examples that I can think of the answer is yes.

If the answer is positive, how does one prove it? Is it possible to prove it without using the classification of finite simple groups?

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Upon seeing this question, I initially parsed it as an easy question about the cyclic group $C_2$. –  James Cranch Aug 26 '12 at 8:38
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5 Answers

up vote 25 down vote accepted

The answer to your question is yes. Moreover, if you pick two random elements from a finite simple group, then they generate the whole group with probability which tends to 1 as the size of the group grows. There are even stronger results in this direction, but I am not an expert in the subject so you will have to look for it yourself. You should look for papers by Liebek and Shalev, Lubotzky, Kantor, and there are others who I am not sure about now.

All of these results require the classification. There are very few results regarding finite simple groups which do not require the classification.

Edit: here is a link to a fairly old survay paper in the notices: http://www.ams.org/notices/200104/fea-shalev.pdf. There are many new developments in the last decade.

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Thanks for the reference –  Hugo Chapdelaine Mar 22 '11 at 18:40
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A group theory guru once told me that this is almost equivalent to the classification, in that the classification could be tremendously simplified if this (every finite simple group can be generated by two elements) could be taken as a lemma. –  Kevin O'Bryant Mar 23 '11 at 2:37
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I guess the classification of finite simple groups is among these few results :-) –  Torsten Ekedahl Mar 23 '11 at 11:44
    
@Kevin: Can you maybe reveal who this Guru is? –  j.p. Mar 23 '11 at 13:50
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Indeed Group Theory is a cult and we have gurus. Moreover, we sometimes sacrifice mathematical virgins to the monster. –  Yiftach Barnea Mar 23 '11 at 14:05
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Since I happen to know the OP is number-theoretically inclined, let me add the following remark:

For "most" finite simple groups $G$ it is indeed the case that $G = \langle x, y \rangle$ where $x$ has order $2$ and $y$ has order $3$. Equivalently, $G$ is a quotient of the free product $\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/3\mathbb{Z} \cong \operatorname{PSL}_2(\mathbb{Z}) = \Gamma(1)$.

This has the following geometric consequence: there is some subgroup $\Gamma_G \subset \Gamma(1)$ such that $X_G = \Gamma_G \backslash \overline{\mathcal{H}}$ is a modular curve and $X_G \rightarrow X(1) \cong \mathbb{P}^1$ is a $G$-Galois branched covering. By taking $G$ to be something else than $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ one sees that $\Gamma(1)$ admits many non-congruence subgroups. For instance, it is well-known (added: I should have said "a well-known theorem of J.G. Thompson") that one can take $G$ to be the Fischer-Griess Monster.

I don't want to make precise what I mean by "most". Note that there are infinitely many finite simple groups with order prime to $3$ (although one has to look fairly far down the list of all finite simple groups to see them: Suzuki groups), so I definitely do not mean "all but finitely many".

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For a slightly smaller value of "most" you can even say that for most finite simple groups you can choose xy to have order 7, that is most large rank simple groups (over most fields) are Hurwitz Groups: en.wikipedia.org/wiki/… — There are definitely lots of groups where this doesn't work, but I find it surprising how often it does work. – Jack Schmidt 0 secs ago –  Jack Schmidt Mar 23 '11 at 14:19
    
@Jack: my definition of "most" does not extend to the statement "Most finite simple groups are Hurwitz groups", I think. But I'll think further about this... –  Pete L. Clark Mar 23 '11 at 14:38
    
Hi Pete, very cool result! –  Hugo Chapdelaine Mar 23 '11 at 16:30
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In addition to the two answers already given it might be worth to mention that the generating graph of a finite simple group has no isolated vertices: This means that for every nonidentity element $x\in G$, there is some other element $y$ such that $G=\langle x, y\rangle$. (The generating graph of a group has the nonidentity elements of $G$ as vertices, where to vertices are connected if they generate the group.) This is shown in

Guralnick, Robert, Kantor, William, Probalistic generation of finite simple groups, J. Algebra 234 (2000), p. 743–792. (MR1800754)

Recently, Breuer et. al. have shown that the generating graph of every "sufficiently large" finite simple group contains a Hamiltonian cycle. You might also look at the references given in their paper:

Breuer, T., Guralnick, R. M., Lucchini, A., Maróti, A., Nagy, G. P., Hamiltonian cycles in the generating graphs of finite groups, Bull. Lond. Math. Soc. 42 (2010), p. 621–633. (MR2669683)

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Yes. See this.

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There is a paper in arxiv by Robert Guralnick and Gunter Malle that answers your question in a stronger way. Their aim is to prove existence of algebraic surfaces obtained in a specific way as a quotient of finite group actions on products of curves of genus > 1. They prove the existence of two conjugacy classes in a finite simple group with the property that picking one element each from these classes always generates the group.

Here is the link:

http://arxiv.org/abs/1009.6183

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Thanks Vanchinathan for the reference –  Hugo Chapdelaine Apr 18 '12 at 12:36
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