In our research, we need to know that whether every group $G$ of order $2520 = 2^3 \cdot 3^2 \cdot 5 \cdot 7=\frac{7!}{2}$ or $5040 = 2^4 \cdot 3^2 \cdot 5 \cdot 7=7!$ has a proper subgroup non-isomorphic to the following groups \begin{gather*} C_1,\; C_2,\; C_2\times C_2,\; C_2\times C_2\times C_2,\; C_2\times C_2\times C_2\times C_2, \\ C_3,\; C_3\times C_3,\; C_4,\; C_5,\; S_3,\; C_7,\; C_4\times C_2 ,\; D_8,\; Q_8 \end{gather*} (i.e., all groups of orders $\leq 9$ except $C_6$, $C_8$, $C_9$, together with $C_2\times C_2\times C_2\times C_2$).
Note that
the above list of groups contains at least a $p$-group, for all existing primes $p$ (here), up to their powers.
if $G$ is nilpotent, then it is true (because $G$ would contain the subgroup $C_{p_1p_2}$ for all distinct prime divisors $p_1$ and $p_2$ of $|G|$).
from Groups of order $2520$ it seems the answer is positive for $\lvert G\rvert=2520$.