If $|G|=24$ or $48,36,56,72,80,90$ or $96$ prove that $G$ is an abelian or simple group . This is problem from book Dumit, Foote

If $G$ is non-abelian simple group of order $<100,$ prove that $G\cong A_5.$

I eliminate abelian groups (its orders are prime or prime squares).
For several groups I use Sylow's theorem to show that those groups are not simple.
Also, I know groups of order $p^n$, ($p$ is prime) is not simple.
I can't prove that groups of following order are abelian or have a normal subgroup: $$24=2^3\cdot 3$$
$$48=2^4\cdot 3$$
$$36=2^2\cdot 3^2$$
$$56=2^3\cdot 7$$
$$72=2^3\cdot 3^2$$
$$80=2^4\cdot 5$$
$$90=2\cdot 3^2 \cdot 5$$
$$96=2^5\cdot 3$$
 A: 1. Let $H$ be a subgroup of index $k$ in $G$. So, there are exactly $k$ cosets of $H$ in $G$. Define 
$$\varphi\colon G\rightarrow S_k,\hskip1cm
g\mapsto 
\begin{pmatrix} 
H & x_2H & x_3H & \cdots & x_kH\\
gH & gx_2H & gx_3H & \cdots & gx_kH
\end{pmatrix}
$$
Then (easy to show that) $\varphi$ is non-trivial homomorphism.
2. Remember: let $P$ be a Sylow-$p$ subgroup, and $N(P)$ be the largest subgroup of $G$ in which $P$ is normal. Then (the number of Sylow-$p$ subgroups is $m$) means (index of $N(P)$ in $G$ is $m$)
3. Now consider groups of order $2^3.3, 2^4.3, 2^5.3$. If Sylow-$2$ subgroup is not unique, then they will be three in number and hence we obtain a non-trivial homomorphism from $G$ to $S_3$. Since $|G|$ exceeds order of $S_3$, this homomorphism can not be $1-1$. Also, the homomorphism is non-trivial (noted above). Thus, $\ker\phi$ is non-trivial normal subgroup of $G$, $G$ is not simple.
4. For $2^2.3^2$, consider here number of Sylow-3 subgroups. If it is not $1$, then it should be $4$, and we obtain again a non-trivial homomorphism from $G$ to $S_4$. You can check that this homomorphism can not be 1-1 (compare orders) and also non-trrivial. Hence, $G$ contains proper normal subgroup. 
5. Same method works for order $80=2^4.5$. Here number of Sylow-$2$ subgroups is $1$ or $5$. If it is $5$, we get a homomorphism from $G$ to $S_5$. This homomorphism can not be 1-1 since $|G|$ do not divides $|S_5|$. Kernel of this homomorphism is now a proper normal subgroup of $G$.
6. You can see now where this method is applicable. For order $2^3.3^2$, the number of Sylow-3 subgroups is $1$ or $4$. If it is $4$, proceed as in 4.
7. For order $56$, the number of Sylow-7 subgroups is $1$ or $8$. If it is eight, let $K_1, K_2, \cdots, K_8$ be the wight Sylow-7 subgroups. Then $K_i\cap K_j=\{1\}$ for $i\neq j$ (can you prove this?) Thus, from each $K_i$ pick up non-identity elements; totally there will be $48$. Remaining elements will be from Sylow-2 subgroup, and so there is unique Sylow-2 subgroup; it is normal.
8. For order $90$, it is of the form $2.m=2.odd$. Then $G$ necessarily contains normal subgroup of order $m$ (See this.)
