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 $S$?
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asked Jun 23, 2012 at 11:45
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1$\begingroup$ Every finite group $G$ embeds into $A_n$ for some $n>4$. This gives some (not very good) bound for $D(S)$. $\endgroup$– user6976Commented Jun 23, 2012 at 13:08
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3$\begingroup$ $\log_2 |S|$ 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. $\endgroup$– Douglas ZareCommented Jun 23, 2012 at 13:10
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$\begingroup$ @Duglas: You are probably right that the Abelian subgroups have maximal rank, but perhaps one should consider groups of Lie type instead of $A_n$. $\endgroup$– user6976Commented Jun 23, 2012 at 13:56
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1$\begingroup$ No it seems that $A_n$ is better asymptotically ($n!$ vs $q^{n^2}$). $\endgroup$– user6976Commented Jun 23, 2012 at 14:24
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$\begingroup$ by mistake I rolled back, and can't find how to undo. HELP!!! $\endgroup$– Lior Bary-SorokerCommented Nov 25, 2012 at 10:12
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3 Answers
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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.$
answered Jun 23, 2012 at 22:22
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$\begingroup$ For finite simple groups sectional rank and the ordinary rank (the biggest rank of an abelian subgroup) almost coincide as well? $\endgroup$– user6976Commented Jun 24, 2012 at 5:08
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$\begingroup$ Thank you very much for the answer. Is there something known about $ar(S)$ for finite simple groups? $\endgroup$ Commented Jun 24, 2012 at 6:28
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$\begingroup$ 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. $\endgroup$ Commented Jun 24, 2012 at 10:11
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1$\begingroup$ @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) $\endgroup$– YCorCommented Apr 23, 2023 at 10:52
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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, 2012 at 11:51
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7$\begingroup$ This does not answer the question. $\endgroup$ Commented Jun 23, 2012 at 12:38
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1$\begingroup$ @Andreas: Serves me right for not reading it properly. Thanks. $\endgroup$ Commented Jun 24, 2012 at 2:53
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$\begingroup$ The link to
springerlink.comis broken. I'm also unable to find any snapshot saved on the Wayback Machine. $\endgroup$ Commented Apr 17, 2023 at 10:06 -
$\begingroup$ mathoverflow.net/questions/59213/… seems to be relevant. $\endgroup$ Commented Apr 17, 2023 at 12:20
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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.
