Let $|G|=40=2^3\cdot 5$. Let $n_5$ be the number of Sylow $5$-subgroups. Then

The positive divisors of $8$ are $1,2,4,8$. Among these, only $1$ is congruent to $1$ modulo $5$. Hence

Therefore the Sylow $5$-subgroup is unique and normal. Since its order is $5$, it is nontrivial and proper. Hence $G$ is not simple. $\square$

Let $|G|=56=2^3\cdot 7$. Let $n_7$ be the number of Sylow $7$-subgroups. Then

Thus $n_7=1$ or $n_7=8$. If $n_7=1$, then the Sylow $7$-subgroup is normal and $G$ is not simple. Suppose $n_7=8$. Distinct subgroups of order $7$ intersect only in $\{e\}$. Hence the eight Sylow $7$-subgroups contain

elements altogether. Thus only $7$ nonidentity elements of $G$ lie outside their union. A Sylow $2$-subgroup $P$ has order $8$. None of its nonidentity elements can lie in a subgroup of order $7$. Hence the seven nonidentity elements of $P$ are precisely the seven elements outside the union of the Sylow $7$-subgroups. Every Sylow $2$-subgroup has the same seven nonidentity elements, and therefore all Sylow $2$-subgroups are equal. Thus $n_2=1$, so the Sylow $2$-subgroup is normal and proper. Hence $G$ is not simple. $\square$

Let $|G|=30=2\cdot 3\cdot 5$. Let $n_5$ be the number of Sylow $5$-subgroups. Then

Thus $n_5=1$ or $n_5=6$. If $n_5=1$, then $G$ is not simple. Suppose $n_5=6$. These six subgroups contribute $6(5-1)=24$ nonidentity elements of order $5$. Let $n_3$ be the number of Sylow $3$-subgroups. Then

Thus $n_3=1$ or $n_3=10$. If $n_3=1$, then $G$ is not simple. If $n_3=10$, then the Sylow $3$-subgroups contribute $10(3-1)=20$ nonidentity elements of order $3$. These are distinct from the elements of order $5$. Then $G$ would contain at least

nonidentity elements, impossible since $|G|=30$. Therefore $G$ has a normal Sylow $5$-subgroup or a normal Sylow $3$-subgroup. Hence $G$ is not simple. $\square$

Let $|G|=48=2^4\cdot 3$. Let $P$ be a Sylow $2$-subgroup of $G$. Then

Hence

Let $G$ act on the three left cosets of $P$ by left multiplication. This gives a homomorphism

If $G$ were simple, the kernel would be either $\{e\}$ or $G$. The action is transitive and nontrivial, so the kernel cannot be all of $G$. Thus simplicity would force an embedding of $G$ into $S_3$. But $|S_3|=6$, and a group of order $48$ cannot embed in a group of order $6$. Hence $G$ is not simple. $\square$

1. If $|G|=20=2^2\cdot 5$, then $n_5\equiv 1 \pmod 5$ and $n_5\mid 4$, so $n_5=1$.
2. If $|G|=28=2^2\cdot 7$, then $n_7\equiv 1 \pmod 7$ and $n_7\mid 4$, so $n_7=1$.
3. A Sylow $2$-subgroup has order $32$ and index $3$. The coset action gives a homomorphism into $S_3$. A simple group of order $96$ would embed in $S_3$, impossible.