Number of generators of a subgroup of a finite simple group

Question

For a finite group $G$ we denote $d(G)$ the minimal size of a set of generators of $G$. We define $D(G) = \max( d(H) \mid H\leq G)$.

Let $S$ be a finite simple group. Are there `good' bounds on $D(S)$ in terms of the size of $S$?

Every finite group $G$ embeds into $A_n$ for some $n>4$. This gives some (not very good) bound for $D(S)$. — , Jun 23 '12 at 13:08
$\log_2 |S|$ is an easy upper bound. I would be surprised if the large elementary abelian groups of rank about $n/2$ inside the alternating groups $A_n$ didn't provide the asymptotic maximum. — Douglas Zare, Jun 23 '12 at 13:10
@Duglas: You are probably right that the Abelian subgroups have maximal rank, but perhaps one should consider groups of Lie type instead of $A_n$. — , Jun 23 '12 at 13:56
No it seems that $A_n$ is better asymptotically ($n!$ vs $q^{n^2}$). — , Jun 23 '12 at 14:24
by mistake I rolled back, and can't find how to undo. HELP!!! — Lior Bary-Soroker, Nov 25 '12 at 10:12
Mail the administrators. Alternatively, check the edit list (click on the hours ago) and see if you can rollback to a desired version. Gerhard "Ask Me About System Design" Paseman, 2012.11.25 — Gerhard Paseman, Nov 25 '12 at 22:15

Geoff Robinson · Accepted Answer · 2015-10-14T12:12:13.987

14

By a Theorem of R. Guralnick and A. Lucchini (see MR1015993 and MR 1023965),(which does require the classification of finite simple groups) the minimum number of generators for a finite group $G$ can exceed by at most one the maximum (over all its Sylow subgroups $P$) of the minimum number of generators of $P$. It follows that the value of $D(G)$ is between $d(H)$ and $d(H)+1$ for some $p$-subgroup $H$ of $G.$ Thus for a finite simple group $S,$ the question does essentially come down to bounding the minimum number of generators of subgroups of $S$ of prime power order, as was suggested might be the case in some comments. The sectional $p$-rank of a finite group $G$ is the maximum over all section of $G$ which are $p$-groups, of the minimum number of generators of that section (a section of $G$ is a group of the form $H/K$ where $H$ is a subgroup of $G$ and $K \lhd H ).$ Hence if we define the sectional rank of $G$ to be the maximum of the minimum number of generators of an Abelian section of $G$, and denote it by $ar(G),$ then we see that for any finite group $G$, simple or not, we have $ ar(G) \leq D(G) \leq ar(G)+1.$

edited Oct 14 '15 at 12:12

answered Jun 23 '12 at 22:22

Geoff Robinson

42,683

For finite simple groups sectional rank and the ordinary rank (the biggest rank of an abelian subgroup) almost coincide as well? – Jun 24 '12 at 05:08
Thank you very much for the answer. Is there something known about $ar(S)$ for finite simple groups? – Lior Bary-Soroker Jun 24 '12 at 06:28
@Mark: I am not sure. – Geoff Robinson Jun 24 '12 at 10:05
Note that the minimum number of generators of a finite Abelian group is the maximum of the same quantity over its Sylow $p$-subgroups. The structure of Sylow $p$-subgroups of a finite simple group can be analyzed in any particular case. – Geoff Robinson Jun 24 '12 at 10:11
Write $D'(G)$ for the sup of $D(G)$ over Sylow subgroups of $G$. That is, $D'(G)=\sup_p D_p(G)$, where $D_p(G)$ is the rank of a $p$-Sylow in $G$. So the theorem you mention says that $D(G)\in{D'(G),D'(G)+1}$. A natural question is whether for $G$ of Lie type relative to a prime $p$, $D'(G)$ is always (up to maybe finitely many exceptions or "bounded error") attained by precisely the $p$-Sylow of $G$. – YCor Apr 17 '23 at 11:45
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@GeoffRobinson no, you misunderstood my comment. I your example $D(G)$ is equal to $D(H)$ where $H$ is a $p$-Sylow in $G$. (The point of your example is that $D(H)$ is not equal to the rank of $H$ itself, which is smaller.) Here's an equivalent reformulation of my comment. Let $D(G)$ be the sup of ranks of all subgroups of $G$, and $D'(G)$ the sup of ranks of all subgroups of prime-power order of $G$. Is the inequality $D'(G)\le D(G)$ always an equality? (or "close" to an equality: bounded error, few exceptions, etc) – YCor Apr 23 '23 at 10:52

score 2 · Answer 2 · answered Jun 23 '12 at 11:51

2

Every finite simple group can be generated by two elements. Except in the case of prime order, one of the elements can have order 2. See here for example.

answered Jun 23 '12 at 11:51

Brendan McKay

37,203

7

This does not answer the question. – Andreas Thom Jun 23 '12 at 12:38
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@Andreas: Serves me right for not reading it properly. Thanks. – Brendan McKay Jun 24 '12 at 02:53
but interesting nonetheless! – Will Chen Feb 10 '15 at 23:04
The link to springerlink.com is broken. I'm also unable to find any snapshot saved on the Wayback Machine. – The Amplitwist Apr 17 '23 at 10:06
https://mathoverflow.net/questions/59213/generating-finite-simple-groups-with-2-elements seems to be relevant. – Salvo Tringali Apr 17 '23 at 12:20

score 2 · Answer 3 · edited Apr 17 '23 at 10:41

Nobody seems to have mentioned the work of Burness, Liebeck and Shalev yet:

Burness, Timothy C.; Liebeck, Martin W.; Shalev, Aner, Generation and random generation: from simple groups to maximal subgroups., Adv. Math. 248, 59–95 (2013). Zbl 1292.20013. PDF

They prove that if $S$ is a non-abelian finite simple group and $H$ is a maximal subgroup of $S$ then $d(H)\leq4$. Furthermore, there are infinitely many examples that attain this bound.

Number of generators of a subgroup of a finite simple group

3 Answers3