Here is something I plan on doing over the winter of 2019 - show groups of order \(n\) are (not) simple for as many values of \(n\) as I can.
We shall constantly be appealing to the Sylow Theorems and be using the relevant notations.
Notations: Let \(G\) be a group with order \(p^nm\) where \(p\) is a prime, \(n\) is a positive integer, and \(p\not\mid m\).
Syl\(_p(G) = \{H \le G : |H| = p^n\} =\) set of subgroups of \(G\) that have order \(p^n\).
An element of Syl\(_p(G)\) is called a Sylow-\(p\) subgroup of \(G\).
\(n_p := |\operatorname{Syl}_p(G)|\), the number of Sylow-\(p\) subgroups of \(G\).
Now, we list the Sylow theorems:
(1) \(n_p \ge 1\).
(2) (i) All Sylow-\(p\) subgroups are conjugates.
(ii) If \(H \le G\) and \(|H| = p^m\) for some \(1 \le m \le n\), then \(H\) is contained in a Sylow-\(p\) subgroup.
(3) \(n_p \mid m\) and \(n_p \equiv 1 \mod p\).
Keeping the same notations as before, suppose \(m > 1\) and \(n_p = 1\), then \(G\) is not simple.
To see this, let \(H\) be the unique Sylow-\(p\) subgroup of \(G\). (This is unique because \(n_p = 1\).)
This is clearly a nontrivial proper subgroup of \(G\). Moreover, let \(g \in G\). Then, \(gHg^{-1}\) is also a Sylow-\(p\) subgroup of \(G\). By uniqueness, it follows that \(gHg^{-1} = H\). Thus, \(H\) is normal and we are done.
This corollary will be used frequently in concluding that a group is not simple.
The letters \(p,\) \(q,\) and \(r\) are used for primes. Moreover, they are assumed to be distinct. \(n\) is a positive integer.
Here is the list of all order \(\le 200\) that are taken care of, by the above classifications.