If $|G| = n < 60$, and $n$ is composite, then $G$ is not a simple group. Why? [duplicate]

I'm given a theorem (call it Theorem 1) which states that $G$ is solvable iff there exists a chain of normal subgroups $\{e\} = G_0 \unlhd \ldots \unlhd G_n = G$ such that $G_i / G_{i-1}$ is abelian for $1 \le i \le n$. I am also given (call it Theorem 2) that $G$ is solvable iff it contains a normal subgroup $N$ such that $N$ and $G/N$ are also solvable. With these information, particularly Theorem 2, how do I go to show that a group of order less than $60$ is solvable?

I tried working with a group $G_i$, which is somewhere in the chain of normal subgroups. Since $G_i \unlhd G$, then $|G_i| < 60$. Also, $G_i$ contains a normal subgroup $G_{i-1}$, and $G_i / G_{i-1}$ is abelian, which means $G_i / G_{i-1}$ is also solvable. Is this sufficient to show that $G_i$ is solvable, which is what I want?

Solution 1:

Step 1: If $\# G =2^n 3^m<60$ then $G$ is solvable. (Proof by induction)

Step 2: General case: Consider $p$ to be the maximum prime factor of $\# G$. If $p=3$ then this case is covered in Step 1. If $p\geq7$ then write $\# G = pk$ with $k<p$ then the number of Sylow p-subgroups of $G$ is 1, and we can go to the quotient and proceed by induction on the order of $G$. If p = 7. The only case that we have more than one 7-Sylow subgroup is when $\# G=56$. In this case the number of 7-subgroups are 8 and so there are 48 elements of order 8 and hence there is one 2-Sylow subgroup, which is normal. If p = 5. The only case that we have more than one 5-Sylow subgroup is when $\# G=30$. In this case there are six 5-Sylow subgroups so we have 24 elements of order 5. The remaining elements form a 3-Sylow subgroup, which is normal.

Source: Check a complete proof here.