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Definition: Let $G$ be a group, and let $H \leq G$ be a subgroup. We say that $H$ is big in $G$ if for every intermediate subgroup $H \leq L \leq G$ there exists some $x \in L$ such that $\langle H \cup \{x\} \rangle = L$.

Question: Is there a big solvable subgroup in every finite group?

Motivation: By Theorem A in M.Aschbacher and R.Guralnik, "Solvable generation of groups and Sylow subgroups of the lower central series", in every finite group $G$ there exists a pair of conjugate solvable subgroup $H_1,H_2 \leq G$ such that $\langle H_1 \cup H_2 \rangle = G$. So if $x \in G$ is such that $xH_1x^{-1} = H_2$ then $\langle H_1 \cup \{x\} \rangle = G$. In my question I am asking for a generalization of this conclusion.

Pablo
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    related: http://mathoverflow.net/questions/208776/subgroups-of-powers-of-the-alternating-group-on-5-elements ($A_4^n$ is "big" in $A_5^n$ for every $n$). You could provide a little more about what you currently know about the question; for instance whether you know the answer for all finite simple groups. – YCor Jun 09 '15 at 10:09
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    This definition might not be universally accepted. I have once learned that a proper subgroup $H$ of $G$ is big if it intersects every conjugacy class. For instance the triangular subgroup is big in ${\bf GL}_n(\mathbb C)$. In a finite group, there does not exist such a big subgroup. – Denis Serre Jun 09 '15 at 11:24
  • @YCor I do not know the answer for finite simple groups, and actually I wondered whether I should ask the question just for finite simple groups rather than for every finite group. – Pablo Jun 09 '15 at 11:49
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    No I think it's fine to ask for all finite groups, and mention that you don't know for finite simple groups. For instance for $\mathrm{Alt}_n$, you know for which values of $n$? – YCor Jun 09 '15 at 12:19
  • @YCor this is exactly what I am thinking about now. After reading the manuscript mentioned in the question, it seems natural to guess that a $2$-Sylow subgroup of $A_n$ is big. I have verified this for $n = 5$. – Pablo Jun 09 '15 at 12:38
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    I don't think $A_6$, $A_7$ and $A_8$ are out of reach... – YCor Jun 09 '15 at 17:06
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    $A_6$, $A_7$ and $A_8$ all have maximal soluble subgroups. For example, $(S_4\times S_4)\cap A_8$ is a maximal intransitive subgroup of $A_8$. You can play a similar trick in $A_9$ by taking a maximal imprimitive group. $A_{10}$ is the smallest alternating group that doesn't have a maximal soluble subgroup. – verret Jun 10 '15 at 06:47
  • @verret thanks! But is the $2$-Sylow subgroup big in $A_{10}$? – Pablo Jun 10 '15 at 07:45
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    Pedantry: I think the A_8 example should be $(S_4\wr S_2)\cap A_8$ rather than $(S_4\cap S_4)\cap A_8$. (The former is not maximal in $A_8$, although it may be big.) – Nick Gill Jun 10 '15 at 15:01
  • By taking quotients, one can assume that $F(G)$ is trivial. Thus $F^*(G)$ is a central product of a bunch of quasi-simples. If that helps. – Nick Gill Jun 10 '15 at 15:10
  • A10,11,12,13 all have (by a computer search) a big solvable subgroup, so do all simple groups of order <=2*10^6. – ahulpke Jun 10 '15 at 16:20
  • @ahulpke could you please check whether for all these finite simple groups the $2$-Sylow subgroup is big? – Pablo Jun 10 '15 at 18:25
  • @Pablo It is for A10. For A11 the computer is still running, and the larger ones are likely not going to work with my current algorithm (finding the intermediate subgroups grows with the subgroup index). – ahulpke Jun 10 '15 at 20:29
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    @Pablo Also, for A7 the 2-Sylow subgroup is not big, a sub direct product of $S_4$ with $S_3$ of order 72 cannot be generated. – ahulpke Jun 10 '15 at 20:42
  • @ahulpke Note that $H$ is big in $G$ if and only if for any intermediate subgroup $H \leq L \leq G$ the union of all $H \leq K \lneq L$ is a proper subset of $L$. Maybe this can help in accelerating your search? – Pablo Jun 10 '15 at 21:47
  • @Pablo No-- the issue is to find the possible $L$. (One could rewrite the code to use the known subgroup lattice of A11, but I have not done so.) – ahulpke Jun 10 '15 at 22:16
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    By the way, there is a sequence of primes ${p}$ such that $A_p$ contains a maximal soluble (primitive) subgroup. It includes: {13,19,29,31,37,41,43,47,53,59,61,67,71,73,79...}. I am not sure if this sequence is infinite but I guess so. – verret Jun 11 '15 at 00:27
  • @verret how do you know that these solvable maximal subgroups exist? – Pablo Jun 11 '15 at 05:39
  • $S_p$ always contains $H:=AGL(1,p)$ so I consider $H\cap A_p$. This is primitive soluble (and as large as possible with respect to this property) but it is not always maximal in $A_p$. In fact, I am not sure how to tell whether it is maximal in general (but some people probably do), but I used magma to determine which small alternating groups had maximal soluble groups... (For p>16, it is easy to see that such a group must be as above.) – verret Jun 11 '15 at 05:46
  • In fact, it seems that $H\cap A_p$ is maximal in $A_p$ for $p$ greater than 23. (The small exceptions are due to things like the Matthieu groups.) Again, this should be provable (it's probably already known in fact), and it shows that every large alternating group of prime degree has a big (maximal even) solvable subgroup. – verret Jun 11 '15 at 06:28
  • @verret It seems that this subgroup is maximal and solvable for any prime number $p$ (but maybe not always primitive). This is explained here http://mathoverflow.net/questions/112657/maximal-subgroups-of-finite-simple-groups. The problem however seems more difficult for $A_n$ where $n$ is not a prime. – Pablo Jun 11 '15 at 07:51
  • No, for $p\geq 3$, this subgroup always contains a $p$-cycle and is thus always transitive and primitive. It is not necessarily maximal. (Note that the answer you linked says that all maximals are of these types, not that all groups of these types are maximal. Sometimes, one of the groups listed there is contained in another, as happens for $AGL(1,23)\cap A_{23}$ which seems to be contained in $M_{23}$.) – verret Jun 11 '15 at 09:09
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    @verret AGL(1,p) is maximal in $A_p$ for $p$ prime with the exceptions of degree 3,7,11,17,23. This follows from a list in Liebeck,Praeger, Saxl, {\em A classification of the maximal subgroups of the finite alternating and symmetric groups}, J.Algebra 111 (1987), no. 2, 365–383. – ahulpke Jun 11 '15 at 14:58
  • @DenisSerre, is it obvious or a theorem that there are no big subgroups of finite groups in your sense? – LSpice Feb 19 '21 at 16:22
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    @LSpice. This is a theorem. $H$ is big in this sense ifs $G$ is the union of conjugates of $H$. There are $[N(H):H]$ such conjugates, which intersect at least at the neutral element. Thus their union has at most $1+N(H):H<|N(H)|\le|G|$ element. It cannot be the whole group. – Denis Serre Feb 19 '21 at 17:07

2 Answers2

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As a partial answer generalizing Pablo's example, the claim is true for a group with a $(B,N)$-pair for which $B$ is solvable. In this case, take $H:=B$. This recovers Pablo's example as $A_5$ acts doubly transitively on a set with 5 points, hence has a $(B,N)$-pair, with a solvable point stabilizer $B=A_4$. Other examples include e.g. general linear groups $GL_n(K)$.

See Wikipedia for the definition of a $(B,N)$-pair and the notation used in the sequel.

Indeed, any subgroup $L$ satisfying $B\leq L \leq G$ is a standard parabolic subgroup, $L=P_X$ for some subset $X$ of the index set of the Dynkin diagram of $G$. Write $X$ as the union of its connected components, $X=\cup C_i$, and for each $i$ choose some $g_i$ in the big cell ${B_X}^+{B_X}^-$. Then $g:=g_1\cdots g_n$ is as required, i.e. $B_g:=\langle B,g\rangle = L$. To see this, note that $B_g$ again is a standard parabolic subgroup and by the choice of $g_i$, which isn't contained in a proper parabolic subgroup of $P_{C_i}$, $B_g \cap P_{C_i} = P_{C_i}$.

Guntram
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  • Does this mean that you can somehow treat the case of $G = \mathrm{SL}_n(\mathbb{F}_q)$ where $q$ is a power of a prime > 5? – Pablo Jun 10 '15 at 22:27
  • Why is $g$ not in a proper parabolic? Does it come from, say, looking at to what Bruhat cell it belongs? – LSpice Feb 19 '21 at 16:28
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Michio Suzuki proved that every finite group is generated by a pair of conjugate solvable subgroups. See http://projecteuclid.org/download/pdf_1/euclid.hokmj/1381517825 (Open access). The subgroup $S$ which Suzuki exhibits is not a priori "big" in your sense (as far as I can see), but may suggest an approach to the question.

Later edit: Ah, I see that the existence part of Suzuki's result is no stronger than that of Aschbacher and Guralnick, it's just that his subgroup $S$ has more properties which may be useful from an inductive point of view.

Geoff Robinson
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