Even with little/no knowledge of the symmetric group of degree $n$, I think that Cayley theorem is sufficient to infer that such group must have a proper subgroup of order $n$, for any $n>2$.
I was wondering what about groups of order $n!$: do they always have a proper subgroup of order $n$? This is the case for $S_n$. This is also the case for $C_6$, which complete the survey for $n=3$, but I can't even approach $n=4$ to look for counterexamples.
A (very) partial answer, for some special values, concerning the solvable case. (Note that solvable groups tend to have more subgroups than others.)
Every solvable group of order $n!$ for $n\in\{6,10,15,21\}$ has an element of order $n$.
Indeed, write (uniquely) $n!=u_nv_n$ where $v_n$ is coprime to $n$, and every prime divisor of $u_n$ divides $n$. So $u_6=2^4.3^2$, $u_{10}=2^8.5^2$, $u_{12}=2^{10}.3^5$, $u_{14}=2^{11}.7^2$, $u_{15}=3^6.15^3$. Since every finite solvable group has a Hall subgroup for every set of prime numbers, for every $n$,
the statement "every solvable group of order $n!$ has an element of order $n$" is equivalent to: "every solvable group of order $u_n$ has an element of order $n$".
For each prime $p$, we denote by $N_p$ a maximal $p$-elementary normal subgroup of $G$. Any (solvable) group of order $n\ge 2$ has a minimal nontrivial subgroup $N$, which is elementary abelian. In other words, for some prime divisor $p$ of $n$ we have $N_{p}\neq 1$.
Next we check the above statement for $n\in\{6,10,14,15\}$. In each case, $n=pq$ with $p\neq q$ primes, and $u_n=p^aq^b$. The idea is to consider the action of a $q$-Sylow on $N_p$ and of a $p$-Sylow on $N_q$.
$\bullet$ For $n=6$, $p^aq^b=2^4.3^2$. The automorphism group of $N_3$ has order dividing the order of $\mathrm{GL}_2(\mathbf{F}_3)$, which is $3(3^2-1)(3-1)=2^3.3$. This is not divisible by $2^4$. Hence for a given $2$-Sylow, some element of order $2$ centralizes $N_3$; hence if $N_3\neq 1$ there is an element of order $6$.
We have $N_2\simeq(\mathbf{Z}/2\mathbf{Z})^s$ for some $s\le 4$. If $s\le 3$, the automorphism group of $N_2$ has order dividing the order of $\mathrm{GL}_3(\mathbf{F}_2)$, which is $2^3(2^3-1)(2^2-1)(2-1)=3.(2^3.7)$, which is not divisible by $9$, and we argue as in the previous case to get an element of order $6$. If $a=4$, the automorphism group of $N_2$ is isomorphic to $\mathrm{GL}_4(\mathbf{F}_2)$, which has order $2^6(2^4-1)(2^3-1)(2^2-1)(2-1)=3^2\ell$ with $\ell$ coprime to $3$. Since $\mathrm{GL}_2(\mathbf{F}_2)$ has an element of order $3$, we obtain a $3$-Sylow of $\mathrm{GL}_4(\mathbf{F}_2)$ by $2\times 2$ block diagonal matrices. In particular, an element in a single block (trivial on the second block) centralizes some element of order $2$, and hence there is an element of order $6$.
$\bullet$ For $n=10$, $p^aq^b=2^85^2$. We immediately see that $N_5$ has automorphism group of order not divisible by $2^8$, so there is an element of order $10$ as soon as $N_5\neq 1$.
If $|N_2|\le 2^7$, its automorphism group has order not divisible by $5^2$, so there is an element of order $10$ if $1<|N_2|<2^8$.
If $|N_2|=8$, a 5-Sylow in $\mathrm{GL}_2(\mathbf{F}_2)$, as in the previous case $n=6$, has order $25$ and is given by a product of 5-Sylow in block copies of $\mathrm{GL}_4(\mathbf{F}_2)$. As in that case, we deduce the existence of an element of order $10$.
$\bullet$ For $n=15$, $p^aq^b=3^6.5^3$. Then $3^6$ does not divide $|\mathrm{GL}_3(\mathbf{F}_5)|$, so if $N_5\neq 1$ then there is an element of order $15$. Also $5^3$ does not divide $|\mathrm{GL}_6(\mathbf{F}_3)|$ so if $N_3\neq 1$ then there is an element of order $15$.
$\bullet$ $n=21$ has $p^aq^b=3^9.7^3$ works exactly as in the previous case.
$n=12,14$ can be approached in the same way but I didn't push to a conclusion. For instance, for $n=14$ (so $p^aq^b=2^{11}.7^2$), $N_7\neq 1$ easily implies the existence of an element of order $14$; when $N_2\neq 1$, I don't immediately reach a conclusion in the case when precisely $|N_2|=2^9$.
Edit: I can also conclude the solvable case for $n=14$.
Every solvable group of order $n!$ for $n=14$ has an element of order $n$.
Here $p^aq^b=2^{11}.7^2$. If $N_7\neq 1$, as already said, $2^{11}$ does not divide $|\mathrm{GL}_2(\mathbf{F}_7)|$ so there is an element of order $14$.
If $N_2\neq 1$, write $|N_2|=2^s$, with $1\le s\le 11$. Since $2^k$ equals $1$ modulo $7$ iff $3$ divides $k$, one sees that if $s$ is not divisible by $3$, then every $7$-Sylow in $\mathrm{GL}_s(\mathbf{F}_2)$ fixes a nonzero vector, and hence we deduce in this case the existence of an element of order $14$.
Also $49$ does not divide $|\mathrm{GL}_3(\mathbf{F}_2)|$ so if $s=3$ then we also produce an element of order $14$.
If $s=6$, a $7$-Sylow in $\mathrm{GL}_6(\mathbf{F}_2)$, as in the cases $n=6,10$, has order $49$ and appears in diagonals blocks $3\times 3$, and in particular it contains nontrivial elements with nontrivial fixed points, and again we produce elements of order $14$.
Finally let us consider $s=9$: then the $7$-Sylow in $\mathrm{GL}_9(\mathbf{F}_2)$ have order $7^3$, and decompose according to diagonals blocks $3\times 3$. To be the identity on the first block defines a subgroup of index $7$. Also the action of the $7$-Sylow of $G$ (of order 49) goes through a subgroup of index set. Hence these two subgroups have nontrivial intersection, and hence we again produce an element of order $14$.
Edit 2 (August 11):
Every group of order $n!$ for $n=10$ has an element of order $n$.
The solvable case was done before. So we have to look at the possible non-abelian simple subquotients. Namely, we need the list of simple groups $S$ such that $|S|$ divides $10!$, and among them single out those for which $|S|^2$ divides $10!$.
There are $12$ such nonabelian simple groups of order dividing $10!=2^8.3^4.5^2.7$, namely,
$\mathrm{Alt}_5$ of order $60=2^2.3.5$, $|\mathrm{Out}|=2$,
$\mathrm{PSL}_2(7)$ of order $168=2^3.3.7$, $|\mathrm{Out}|=2$,
The last two have an element of order 10. For the first 10, $25$ does not divide the order of the automorphism group. We then argue as follows: first, mod out by a maximal normal subgroup of order coprime to $10$. Then the resulting quotient has order divisible by $2^8.5^2$, and has a minimal normal subgroup of order not coprime to $10$. The case of an elementary abelian $2$-group or $5$-group was already tackled. Otherwise, we have $S$ or $S^2$ as above. In case this is $S$ (not $S^2$), and assuming $S$ not $J_2/\mathrm{Alt}_{10}$ (already tackled), some element of order $5$ in the $5$-Sylow normalizes this normal subgroup (of even order), so there is an element of order $10$.
Finally, we have the case when $S^2$ is a subquotient (with kernel of order coprime to $10$). The examples for which $|S|^2$ divides $10!$ are the first five. For those of order divisible by $10$, $S^2$ has an element of order $10$ and we are done. For those of order coprime to $5$, the automorphism group of $S^2$ has order coprime to $5$, and hence we argue as in the case of $S$.
It was already checked by another user that every group of order $6!$ has a subgroup of order $6$. Still I get an element of order $6$, except possibly in a subgroup of index $2$ in $\mathrm{Aut}(\mathrm{Alt}_6)$ (there are three, including $S_6$ which has an element of order $6$).
Addition (For further context, see the answer below).
Suppose that $p$ is a prime such that:
- $p = 2^i + 2^j + 2^k + 1, i > j > k > 0$,
- $p$ is $4$-super Weiferich: $2^{p-1} \equiv 1 \bmod p^4$.
Note that it is widely expected that there are infinitely many primes satisfying 1. OTOH, it is also widely expected that there are no primes at all satisfying 2 (rough heuristic, the chance of it happening for each random prime is $1/p^3$ which converges, are there are no such primes less than $10^{17}$ so no tiny accidents occur). So we don't expect there to exist primes satisfying both 1or 2 either. However, it seems very hard to rule out their existence using standard techniques from number theory.
Now take the semi-direct product
$$H = (\mathbf{Z}/2)^{4(p-1)} \rtimes \mathbf{Z}/p^4$$
coming from an element of $\mathrm{GL}_{4(p-1)}(\mathbf{F}_2)$ all of whose eigenvalues are primitive $p^4$th roots of unity. Such a matrix exists precisely because of the divisibility (2) above. Suppose that $H$ contained a subgroup of order $4p$. Then it would have to be a semi-direct product of $(\mathbf{Z}/2)^2 \rtimes \mathbf{Z}/p$. Since $p > 3$, this would force the group to be abelian. However, by construction, the action of the element of order $p$ has no fixed points on $(\mathbf{Z}/2)^{4(p-1)}$, and so this is impossible. We now note:
$$(4p)! = 2^{4p - s(4p)} p^4 \cdot M, \ (M,4p) = 1$$
Where $s(n)$ is the number of binary digits of $n$. By assumption (1), $s(4p) = 4$. Hence, letting $G = H \times \mathbf{Z}/M$, we find that $|G| = (4p)!$ but $G$ has no subgroup of order $4p$.
Conclusion I think the result is most likely true (at least for solvable groups) but impossible to prove, since the example above seems very difficult to rule out.
Of course, I may have missed a simple genuine counter-example, but my analysis below suggests to me that the result could well be true, at least for $n = p^{\alpha} q^{\beta}$ and $G$ solvable, which is the main case I have thought about.
End Addition.
Another partial answer.
I'm not sure this is a very natural question, and let me explain why. Let's concentrate on the simplest unknown general case, when $n=2p$ and $p$ is prime. Let's even allow ourself the assumption that $G$ is solvable, so we may reduce to the case (as in YCor's answer) to when $$|G| = 2^m p^2 \ \text{for} \ m = v_2((2p)!).$$ You could now generalize the question slightly and ask: under what conditions on $m$ do all groups of order $2^m p^2$ contain a subgroup of order $2p$? The power of $2$ coming from $n!$ is pretty arbitrary from this perspective. For example, why not ask whether every group of order $2 \cdot n!$ has an element of order $n$. I will show at least that solving the modified problem (if it is true) is beyond the scope of known results in number theory.
Let's see why the answer to which groups of order $2^m p^2$ have a subgroup of order $2p$ depends subtly on $m$. Having established this, note that the particular choice of $m = v_2((2p)!)$ seems a little random. This $m$ can be described more explicitly bu the formula
$$m = v_2((2p)!) = 2p-s(p),$$ where $s(p)$ is the number of digits of $p$ in base binary.
Example: Suppose that $2^m \equiv 1 \bmod p^2$. Then, for any integer $k$, there is a group of order $2^{mk} p^2$ with no subgroup of order $2p$.
Here is the construction. For any $m$, there is an inclusion:
$$\mathrm{GL}_1(\mathbf{F}_{2^m}) = \mathbf{F}^{\times}_{2^m} \subset \mathrm{GL}_m(\mathbf{F}_2)$$
the hypothesis implies that the LHS has an element of order $p^2$, and so its image will be an element $g$ all of whose eigenvalues have order $p^2$. Take the group $G$ to be the semi-direct product of a cyclic group of order $p^2$ acting on $(\mathbf{F}^m_{2})^k$ via $g$ on every factor. Then $G$ will have no subgroup of order $2p$, because $g^p$ has no fixed vectors.
That doesn't contradict anything so far, since the smallest $m$ with this property is usually rather big, for example, $m = 6$ for $p = 3$, whereas $6! = 2^4 \cdot 3^2 \cdot 5$. Similarly, for $p = 5$, $7$, and $11$ we get $m = 20$, $21$, and $110$ respectively, far bigger than $8$, $11$, and $19$. More generally, if $2$ has order $l$ in $(\mathbf{Z}/p)^{\times}$, then $2^{l} \equiv 1 \bmod p$, and $2^{lp} \equiv 1 \bmod p^2$. Usually the order of $2$ modulo $p^2$ will be $m = lp$, which is always bigger than $v((2p)!)=2p-1$. However, there do exist primes for which the order of $2 \bmod p$ is smaller, namely Wieferich primes, for which the order is $l$, not $lp$, e.g.
$$2^{364} \equiv 1 \bmod 1093^2,$$ $$2^{1755} \equiv 1 \bmod 3511^2.$$
To give an indication of the sorry state of our knowledge of Wieferich primes, we currently do not know whether there are finitely many primes which are not Wieferich primes.
This construction doesn't give a counterexample to the problem, because the number $v_2((2p)!)$ is not a multiple of the order of $2 \bmod p^2$ for either $p = 1093$ or $p = 3511$. In fact, it turns out that $v_2((2p)!)$ can never be divisible by the order of $2$ mod $p^2$, Wieferich prime or not. To see this, note that the order of $2$ for Wieferich primes divides $2(p-1)$. So it it also divides $v_2((2p)!)$, it must divide $s(p) - 2$. There are two cases:
$p$ is not a Fermat prime. Here $s(p) > 2$, but certainly $p > 2^{s(p)} - 1$, which makes $p$ too big for even $p$ to divide $2^{s(p)-2}-1$, let alone $p^2$.
$p$ is a Fermat prime. We can't prove that all but finitely many primes are Wieferich, and we can't prove that there aren't infinitely many Fermat primes, but we can prove that Fermat primes are not Wieferich, because
$$2^{2^k} \equiv -1 \bmod p = 2^{2^k} + 1,$$ $$2^{2^{k+1}} \equiv 1 \bmod p = 2^{2^k} + 1,$$ but $$2^{2^{k+1}} - 1 = (p-2)p$$ is not divisible by $p^2$.
So this doesn't lead to a direct counter example. But it says:
If you ask whether all groups of order $2 \cdot n!$ have a subgroup of order $n$, it is beyond curent methods to prove the answer is true even for $n=2p$ with $p$ prime.
The point is one would have to rue out the existence of Wiefrich primes of the form $p = 2^i + 2^j + 1$. Wiefrich primes give groups of order $2^{2(p-1)} p^2$ with no subgroup of order $2p$, and $2(2p)! = 2^{2(p-1)} p^2$ times an integer coprime to $2p$ in this case when $s(p) = 3$.
In the positive direction, it might even be possible to answer the solvable case of order $2p$. To do this, one might try to give a positive answer to the following question:
Question: Suppose that $G$ is a group of order $2^n p^2$ and $2^n \not\equiv 1 \bmod p^2$. Then does there exist a subgroup of order $2p$?
Possibly you might disagree, but this seems like a more natural question.
I will now prove that the smallest possible counter-example to the question has order $2^m$ with $m$ greater than than the order of $2 \bmod p^2$. This proves the $2p$ case of the solvable version of the conjecture for $p < 10^{17}$ except possibly $p = 1093$ and $p = 3511$ (the only Weiferich primes in this range).
Claim: Let $G$ be a group of order $2^n p^2$ for some odd prime $p$. Suppose that $n$ is smaller than the order of $2 \bmod p^2$. Then $G$ has a subgroup of order $2p$.
Possibly one can modify this claim to answer the question above. I started to write up a proof of a stronger (but false!) claim that this was true for all $n$ then stopped when the proof stopped, so it is not written optimally for trying to answer the question above (for example, there is repeated use of induction to groups with smaller $2$-power order.)
Proof: Suppose that $n = 1$. Then the $p$-Sylow $P$ will be normal, and the group will be a semi-direct product of $P$ by an element of order $2$.
If $P$ is cyclic, then the element of order $2$ certainly fixes a subgroup of $P$ of order $p$, and the corresponding extension has order $2p$.
If $P = (\mathbf{Z}/p)^2$, then any order $2$ element acting on $P$ will be diagonalizable and so also fix a subgroup of order $p$.
Now proceed by induction on $n$, so we may assume that $n > 1$. Let $Q$ be a $2$-Sylow, and consider the action of $Q$ on all $p$-Sylows $X$ of $G$. If $P \in X$ has a non-trivial stabilizer, then it is stabilized by an element $x$ of order $2$, and the group generated by $P$ and $x$ has order $2p^2$, and one is reduced to the case $n = 1$ and one is done by induction.
Hence we may assume that every $P$ has no stabilizer. But the action is transitive by the Sylow theorems, so $|X| = |Q|$.
Suppose that $P$ and $P'$ are two distinct $p$-Sylows which have a non-trivial intersection. Let $N$ denote the normalizer of $P \cap P'$. Certainly $N$ contains $P$ and $P'$ (because they have order $p^2$ and so are abelian) and so $N$ has order $2^m p^2$ for some $m > 0$. If $m < n$, then we are done by induction. If $m = n$, then $P \cap P'$ is normal in $G$. But in this case $G/(P \cap P')$ contains an element of order $2$, and the inverse image in $G$ is a subgroup of order $2p$.
Thus we may assume all $p$-Sylow subgroups are distinct and there are $|Q|$ of them, and hence the number of elements of non-trivial $p$-power order is
$$|Q|(p^2 - 1) = |G| - |Q|.$$
But that implies that the $2$-Sylow subgroup $Q$ is normal, because there are only $|Q|$ elements of possible $2$-power order. Hence $G$ is a semi-direct product. Certainly the center $Z$ of $Q$ is characteristic, and so the action of $P$ preserves $Z$, and hence $G$ contains the semi-direct product of $P$ by $Z$. Similarly, we can replace $Z$ by the subgroup of $Z$ of elements of order $2$, and so we may assume that $Q = (\mathbf{Z}/2)^n$. (Thus the semi-direct product corresponds to a map:
$$P \rightarrow \mathrm{GL}_n(\mathbf{F}_2).$$
Suppose that $P$ is not cyclic. Let $V$ denote the representation on the RHS. Clearly $V \otimes \overline{\mathbf{F}}_2$ is a representation of $P$. Any irreducible representation of a non-cyclic group has kernel, so this extension contains invariants under a degree $p$ subgroup $H$ of $P$, and hence:
$$(V \otimes \overline{\mathbf{F}}_2)^{H} \ne 0.$$
But this implies by flatness that $V^{H} \ne 0$. That is, there is an element $x$ of order $2$ which is centralized by $H$, and then $H$ and $x$ generate a group of order $2p$.
Now suppose that $P$ is cyclic. Then we are done unless every representation on the extension is a sum of faithful irreducible representations. But this can only happen if the matrix group has an element of order $p^2$. The eigenalues are defined over fields of degree at most $n$, and thus we are OK unless $n \ge [\mathbf{F}_2(\zeta_{p^2}),\mathbf{F}_2]$ which is equal to the order of $2$ modulo $p^2$.
Bonus! I originally read the question as asking whether any group of order $n!$ necessarily has a subgroup of index $n$. This is false! A pure thought counterexample is
$$A_7 \times \mathrm{Aut}(M_{12})$$
where $M_{12}$ is the Mathieu group. (There are smaller counterexamples but this is the first one I thought of.)
