## p-groups as Sylow subgroups

There are some nice families of groups as $S_n, A_n$, $GL(n,q)$, $SL(n,q)$, and they are useful; we know their elements, and we can get small groups as subgroups of these groups. Is it possible to get every $p$ group as a Sylow-p subgroup of some group in such families of groups? ( For example, the non-abelian groups of order 8 are Sylow-2 subgroups of SL(2,3) and S4; there are five non-abelian groups of order 16, having no element of order 8, and one of them, namely $D_8 \times C_2$, is Sylow-2 subgroup of $S_6$. Also $SD_{16}$ is Sylow-2 subgroup of GL(2,3).)

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I'm pretty confident that the answer is **no**: this would make the set of finite $p$-groups much simpler than it actually is. One reasonable strategy for proving this would be a counting approach: write down an upper bound for the number of $p$-groups appearing as Sylow $p$-subgroups of one of these four families of groups, and then compare this to a lower bound for the number of isomorphism classes of $p$-groups of given order. I would expect the second quantity to be much larger than the first. – Pete L. Clark Feb 7 2011 at 14:02
Obviously enough, after Cayley's theorem, any p-group is isomorphic with a subgroup of the Sylow p-subgroup of a symmetric group? – Charles Matthews Feb 7 2011 at 14:09
@Charles thats right... – RDK Feb 7 2011 at 14:11
I share Pete's view, though it would take some searching of the vast literature on finite simple groups to document the answer clearly. For instance, the third part of the ongoing AMS series of books by Gorenstein-Lyons-Solomon collects a large amount of information about the Sylow structure of finite groups of Lie type. – Jim Humphreys Feb 7 2011 at 14:34
@Torsten Ekedahl: Isn't $\mathrm{Z}_{49}$ a $7$-Sylow of $\mathrm{SL}_2(\mathbb{F}_{197})$? – Someone Feb 7 2011 at 15:16

The question is somewhat loosely stated, leading to various answers and comments which are at cross-purposes. Some specific families of finite groups are mentioned, but the list seems to be left open (?) Among these families, the symmetric and alternating groups have no built-in prime $p$ to favor. Moreover, Burnside's theorem (as people have noted) always allows one to embed a given $p$-group in some symmetric group; if the prime is odd, this is the same as embedding into a (usually simple) alternating group. But in such cases there would probably be only a minuscule chance that the given $p$-group can be embedded as a Sylow $p$-subgroup. I can't prove this offhand but suspect it's known to finite group theorists. As Pete suggests, some numerical estimates might be a good tool. Work by P. Hall and others has shown how rapidly the number of different groups of order $p^n$ grows for a given prime as $n$ grows; on the other hand, there is a lot of classical literature on the Sylow structure of symmetric groups.

The other families mentioned are among those of Lie type, defined relative to a specific prime $p$. Here the structure of a Sylow $p$-subgroup is severely constrained by the root system involved, whereas the order of such groups is easy enough to specify. So again it seems most unlikely outside limited cases that a given $p$-group will be on the list of Sylow subgroups even if you take all groups of Lie type into account. (This too might be in the literature.) For these families it doesn't even seem plausible to me that one can embed an arbitrary $p$-group into such a Sylow subgroup over a large enough field of characteristic $p$. But that wasn't the question asked.

Note that the study of a Lie family changes radically if one wants to say something about the Sylow $r$-subgroups for $r \neq p$. This has been an active topic in the study of modular representations, but gets extremely complicated to organize in a meaningful way.

I think the bottom line is that not much can be learned about the huge world of $p$-groups by trying to embed them in any of these familiar families of groups, even though that is of course possible in some very special cases.

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 You should be able to embed any finite $p$-group $P$ into $\mathrm{GL}_n(\mathbb{F}_p)$ for $n := |P|$ by letting $P$ act regularly on the base (or take any other faithful action). – Someone Feb 8 2011 at 9:16 @Someone: I guess this amounts to embedding $P$ into a symmetric group and then embedding this symmetric group into a finite general linear group. What can all of this tell you about $P$? Anyway the original question asks for $P$ to be embedded into one of the listed groups as a Sylow $p$-subgroup. – Jim Humphreys Feb 8 2011 at 14:21 @Jim: Yes, I agree. My comment referred to your "For these families it doesn't even seem plausible to me that one can embed an arbitrary $p$-group into such a Sylow subgroup over a large enough field of characteristic $p$." – Someone Feb 8 2011 at 15:52 @Someone: I agree that my offhand comment went too far. Of course, embedding a $p$-group in some huge Sylow $p$-subgroup won't tell you much about the smaller group. – Jim Humphreys Feb 9 2011 at 22:57

The answer to the original question is a definite no. There are many $p$-groups which do not occur as Sylow $p$-subgroups of classical groups, symmetric groups, or close relatives. As noted in various comments, embedding a finite $p$-group in a symmetric or classical group is an easy matter. But it is very difficult for a finite $p$-group $P$ to be a Sylow $p$-subgroup of a group $G$ with no factor group of order $p$. The most general results in this direction are probably by George Glauberman. For example, when $p \geq 5$ and $P$ is a finite $p$-group whose outer automorphism group is a $p$-group, then any finite group $G$ with $P$ as a Sylow $p$-subgroup has a factor group of order $p$. There are meaningful senses in which most" finite $p$-groups have outer automorphism group a $p$-group.

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Just a remark on symmetric groups: The structure of the Sylow $p$-subgroups of $S_n$ is indeed known: The Sylow $p$-subgroup of the symmetric group on $n=p^k$ letters is
$$\underbrace{C_p \wr C_p \wr \cdots \wr C_p}_{k},$$ that is, an $k$-fold wreath product of cyclic groups of order $p$. (See Huppert, Endliche Gruppen I, Satz III.15.3.) If the number of letters, $n$, is not a prime power, the Sylow subgroup is a direct product of groups of this type.
A simpler remark: There is no $n$ such that the Sylow $p$-subgroup of $S_n$ has order $p^p$, as is easy to see.
So symmetric groups are of no use here.

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To be clear, the first answer also contradicts Pete's second assertion, as each p-subgroup of a group embeds in a Sylow p-subgroup of that group.

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 (To be clear, I have now deleted the obvious erroneous second comment, although I left a comment thanking Charles for pointing out why it cannot be correct.) – Pete L. Clark Feb 7 2011 at 21:00

No. If the Sylow 2-subgroup of a group is cyclic then the group has a homomorphism onto its Sylow 2-subgroup. (I don't have a reference at hand, look up $p$-nilpotent groups. Or just prove that if the Sylow 2-subgroup is cyclic, there is a homomorphism on $\mathbb{Z}_2$ and use induction.) So, a few small cases aside, neither a symmetric group nor a general linear group can have a cyclic Sylow 2-subgroup.

As already noted, every $p$-group occurs as a subgroup of a Sylow p-subgroup of the symmetric group. Any group of order $n$ can be represented a permutation group of order $n$ (Cayley's theorem), hence any group of order $n$ occurs as a subgroup of $S_n$ and so so a group of order $p^k$ occurs as a subgroup of $S_{p^k}$. If you now replace permutations by permutation matrices, you can embed any group in the general linear group and so any $p$-group is a subgroup of a linear group.

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The first assertion follows from Burnside's normal $p$-complement theorem: if a Sylow $p$-subgroup of $G$ is contained in the center of its normalizer, then $G$ has a normal $p$-complement. This theorem and its proof are contained in nearly any group theory book. – F. Ladisch Feb 7 2011 at 15:03
A simpler proof is to look at the action of an element of order 2 via conjugation, it gives a homomorphism to the symmetric group S_|G| which is not contained in A_|G|, then proceed by induction. – Steve D Feb 7 2011 at 17:22
A somewhat silly comment: The cyclic group of order $2^n$ is a Sylow $2$-subgroup of $GL(1,p)=\mathbb{F}_p^{*}$, if $p$ is a prime of the form $p=1+2^n+r2^{n+1}$. (By Dirichlet, there are such primes.) – F. Ladisch Feb 7 2011 at 18:47
The abelian group 4×2 is not a Sylow 2-subgroup of GL(n,q), SL(n,q), Sym(n), or Alt(n) for the reasons stated in this answer: it is 2-nilpotent forcing. However, one takes advantage of [GL,GL]=SL and [Sym,Sym]=Alt (in virtually all cases) to see an abelian Sylow 2-subgroup of a 2-nilpotent group from these families has to be cyclic. – Jack Schmidt Feb 8 2011 at 4:54