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 Sylowp subgroup of some group in such families of groups? ( For example, the nonabelian groups of order 8 are Sylow2 subgroups of SL(2,3) and S4; there are five nonabelian groups of order 16, having no element of order 8, and one of them, namely $D_8 \times C_2$, is Sylow2 subgroup of $S_6$. Also $SD_{16}$ is Sylow2 subgroup of GL(2,3).)

The question is somewhat loosely stated, leading to various answers and comments which are at crosspurposes. 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 builtin 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. 


No. If the Sylow 2subgroup of a group is cyclic then the group has a homomorphism onto its Sylow 2subgroup. (I don't have a reference at hand, look up $p$nilpotent groups. Or just prove that if the Sylow 2subgroup 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 2subgroup. As already noted, every $p$group occurs as a subgroup of a Sylow psubgroup 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. 


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 


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. 


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

