Groups of order $2520$ Suppose that $G$ is a group of order $2520 = 2^3 \cdot 3^2 \cdot 5 \cdot 7$. The property that I want to check is this: 

Must $G$ contain an abelian subgroup of order at least $12$?

If $G$ is soluble then, yes. A Hall $\{5,7\}$-subgroup of $G$ (which exists by Hall's theorem) is necessarily abelian. On the other hand, if $G$ is simple, then $G \cong A_7$, since $A_7$ is the only simple group of that order (in fact, it is the only perfect group of that order). Also, $A_7$ has an abelian subgroup of order $12$.
Some further observations: 


*

*If $G$ is a candidate counterexample to the (implicit) assertion, then for $p \in \{3,7\}$ a Sylow $p$-subgroup of $G$ must be self-centralising. Using the $N/C$ theorem and the standard $n_p \equiv 1\,(\operatorname{mod} p)$, we arrive at the only possibilities $n_7 = 2^3 \cdot 3 \cdot 5$ and $n_3 = 2 \cdot 5 \cdot 7$.

*For $p=5$ we cannot argue that a Sylow $5$-subgroup of $G$ must be self-centralising, because the possibility that its centraliser has order $2 \cdot 5$ cannot be excluded. At least not immediately. In any case though, $|C_G(P):P| \in \{1,2\}$ and $n_5 = 2 \cdot 3^2 \cdot 7$. Here $P$ is a Sylow $5$-subgroup of $G$.

*Since $G$ cannot be soluble, it must have a composition factor isomorphic to one of $\{A_5, \operatorname{PSL}_3(2), \operatorname{PSL}_2(8), A_6\}$. (We have already argued the case $G \cong A_7$.) That composition factor, however, cannot be direct.
Thoughts?

MatheinBoulomenos notices that I had missed one possibility for a non-abelian composition factor of $G$, namely $\operatorname{PSL}_2(8)$. I have now included this in the list.
 A: If you take a closer look at the $3$-subgroups, you do not have to rely
on the classification of finite simple groups of order dividing $2520$:
Let's assume that $G$ is a group of order $2520$ without any abelian
subgroups of order $\ge 12$.
We'll frequently use the fact that
(*) elements of order $5$ or $7$ cannot
normalize any non-trivial $3$-subgroup,
as they would centralize it yielding an abelian subgroup of order $\ge 12$.
In particular, $G$ has at least two different Sylow $3$-subgroups.
Claim: The intersection $S\cap T$ of any two (different) Sylow
$3$-subgroups $S$ and $T$ of $G$ is trivial.
Otherwise $S\cap T$ has order $3$ and its centralizer $C = C_G(S\cap T)$
has at least two different Sylow $3$-subgroups ($S$ and $T$, as they are
abelian). As $C$ cannot have elements of order $5$ or $7$, its order $|C|$
divides $2^3\cdot 3^2$ and $C$ has four Sylow $3$-subgroups. So it has an
abelian subgroup of order $4$ centralizing $S\cap T$ leading to a
contradiction.
Claim: The number $n_3$ of Sylow $3$-subgroups of $G$ is $280$.
By (*) the number $n_3$ is a multiple of $35$, so it is either $70$ or
$280$ by Sylow.
In case $n_3 = 70$ take a look at the action by conjugation of a fixed
Sylow $3$-subgroup $S$ on the set of all Sylow $3$-subgroups.
If an element $1\ne s\in S$ fixes a Sylow $3$-subgroup $T$, i.e.,
$T^s = T$ then it is contained in $T$, $s\in T$, as otherwise $s$ and $T$
generate a $3$ subgroup $\langle s, T\rangle$ properly containing the
Sylow $3$-subgroup, which is absurd.
As therefore every non-trivial element of $S$ fixes only $S$, all other
orbits have length $9$ contradicting $63 \ne 1 \bmod 9$.
Final contradiction: $n_3$ cannot be $280$ either.
As you already noted, one can obtain a contradiction using the
$N/C$-theorem.

An alternative argument (with three different endings) for the final
contradiction can be given using Frobenius groups:
As $n_3 = 280$ implies that the Sylow $3$-subgroups are self-normalizing
(i.e., equal to their normalizers) and intersect trivially by the first
claim, $G$ is a Frobenius group with Frobenius complement any Sylow
$3$-subgroup. By Frobenius $G$ has a normal subgroup $K$ (the Frobenius
kernel) of order $2^3\cdot 5\cdot 7$.
Ending 1: By John Thompson's famous thesis $K$ is nilpotent, which implies
that $K$ has an abelian subgroup of order $140$.
Ending 2: One can repeat the argument just given by looking at the Sylow
$5$-subgroups of $K$ to show that $K$ is a Frobenius group with Frobenius
complement any Sylow $5$-subgroup and Frobenius kernel $N$ of order
$2^3\cdot 7$. $N$ is normal in $G$ and has either a normal Sylow
$7$-subgroup or is a Frobenius group with Frobenius complement any Sylow
$7$-subgroup and Frobenius kernel the normal Sylow $2$-subgroup.
Both cases easily lead to abelian subgroups of order $\ge 12$.
Ending 3: An even easier way to finish the proof is to look at the action
by conjugation of a Sylow $3$-subgroup $S$ of $G$ on the Sylow
$5$-subgroups of $K$ ($G$'s Frobenius complement). As the number of Sylow
$5$-subgroups of $K$ is not a multiple of $3$, there is at least one
fixed point $U$, which is normalized and hence centralized by $S$ giving
an abelian subgroup $US$ of order $45$.
A: Here is a simpler approach. In all cases, we have a unique nonabelian composition factor $T$. Let $R$ be the soluble radical of $G$. Then $G/R$ has trivial soluble radical, and a unique nonabelian composition factor $T$, so it must be almost simple with socle $T$.
If $T=PSL(2,7)$, then $G/R=PSL(2,7)$. (It cannot be $PGL(2,7)$ by order considerations.) That means that $|R|=15$ and so $R$ is cyclic.
If  $T=A_5$, then $R$ has order $21$ or $42$ (depending whether $G/R$ is $A_5$ or $S_5$). In any case, $R$ has a normal Sylow $7$-subgroup $P$. $P$ is characteristic in $R$, so normal in $G$. Now, a Sylow $5$-subgroup of $G$ together with $P$ generate a (cyclic) group of order $35$.
If  $T=A_6$, then $R=C_7$ and, as in the last case, we get a (cyclic) group of order $35$.
Finally, if  $T=PSL(2,8)$, then $R=C_5$  and again we get a (cyclic) group of order $35$.
(Note: In the above, I'm assuming knowledge of the size of outer automorphism groups of the relevant simple groups, but I don't need to know about Schur multipliers, or Schur-Zassenhaus.)
