There are not simple groups of order 2016. [Scott, Ex.10.7.13.]

1. Hard times to come. There are not
simple groups of order 2016. [Scott,
Ex.10.7.13.]
Proof. Let G be a simple groupof order 2016.
Along the lines of the analogue statement for
1008, 10.7.5. in Scott. Let n7 be the number of 7-
Sylow subgroups P. One has that n7=8,36 or 288
by Sylow (must divide 288 and congruent to 1
mod 7). If n7=8 then 3 divides |C(P)| so there is
an element of order 21 which is impossible in the
symmetric group S8.

2. Let n7=36, |N(P)|=56, N/C embeds into Aut(P) of
order 6 hence |C(P)|=28. There is an element of
order 2 in C(P) hence also an element x in C(P) of
order 14. By 10.2.10 in the faithful representation on
N(P) of degree 36, x^2 is of cycle type 1-7-7-7-7-7
so x has type 1-14-14-7 and Ch(x^7)=8. If y is in
N(P) - C(P) then by 10.2.11 Ch(y)<=6 and x and y
are not conjugates. We conclude that all conjugates
by elements in N(P) of involutions in C(P) remain in
C(P).

3. Let a be an involution in C(P) and denote by f(a) the order of
Cl_N(P)(a). Appplying 10.7.4, |Cl_G(a)|=36*f(a)/8 hence f(a)
must be even. A group of order 28 is either abelian of type
C28 or C2xC2xC7, or nonabelian either dihedral or the
extension of C14 by C4 with amalgamation. C(P) must be
abelian as P is central, which is the case neither in D14 nor
in the extension of C14 by C4 with amalgamation. If C(P) is
Abelian of type C28 then there is only one involution which is
impossible. If C(P) is Abelian of type C2xC2xC7 then there
are 3 involutions of which necessarily at least one is central
in N(P) as N/C is of order 2, which is again impossible.

4. There remained the case n7=288. In this case G is
Frobenius. Indeed, P is such that P ∩ P^g is the
identity subgroup for every g in G − P. Then also
the Frobenius kernel of order 288 is nilpotent,
and both Sylow for 2,3 are also normal. The proof
is complete.
In spite of this uneasy fact, prosperous new year to
all at LinkedIn and the StackExchange.