Nitpicky Sylow Subgroup Question Would we call the trivial subgroup of a finite group $G$ a Sylow-$p$ subgroup if $p \nmid |G|$? Or do we just only look at Sylow-$p$ subgroups as being at least the size $p$ (knowing that a Sylow-$p$ subgroup is a subgroup of $G$ with order $p^k$ where $k$ is the largest power of $p$ that has $p^k \mid |G|$)?
 A: For what it is worth, I consider all primes $p$, not just those that divide the group order.
This makes many statements smoother. For instance, the defect group of the principal block is the Sylow $p$-subgroup, and a block is semisimple if and only if the defect group is trivial. Thus the principal block is semisimple iff $p$ does not divide the order of the group. It would be awkward to state the theorem only for non-principal blocks to avoid mentioning size $p^0$ Sylow $p$-subgroups.
Another reason is induction. For instance, a group is called $p$-closed if it has a normal Sylow $p$-subgroup. Subgroups and quotient groups of $p$-closed groups are $p$-closed. Except that if we only allow Sylows for $p$ that divides $G$ we have to redefine $p$-closed to be “normal Sylow $p$-subgroup or $p$ does not divide the order of the group” and now every time we consider a subgroup or quotient group we have to consider two cases: normal Sylow $p$-subgroup or $p$ does not divide the order of the group.
For this reason, most finite group theorists allow the trivial primes as well. For instance: Alperin, Aschbacher, Gorenstein, Huppert, Kurzweil and Stellmacher, Suzuki, etc. all explicitly allow primes that do not divide the order of the group.
A: I believe it is customary among most authors to take only primes that divide the group's order. Now, if you're doing some calculations over a large or infinite number of primes then you could always define a Sylow subgroup of a prime not dividing the group's order as the trivial subgroup.
