Suppose I wanted to show that, for a specific odd order, any finite group of that order is solvable. What tools are available to solve such a question?
I'm asking since I was thinking about Feit-Thompson and wondered whether or not people conjectured it to be true before its proof came, and so I thought "how could one gain heuristic evidence to be partially convinced?"
I have read about all groups of finite order below 60 are solvable, which are proven a lot through various Lemmas, including:
-$p$-groups are solvable.
-If $|G| = p^a m, p \nmid m, a \geq 1$, the number of Sylow $p$-subgroups of a group $G$, denoted $s_p (G)$, satisfies $s_p(G) \mid m$ and $s_p \equiv 1 \mod p$. If $a = 1$, then $G$ has $(p-1)s_p(G)$ elements of order $p$. If $s_p(G) = 1$, then the unique Sylow $p$-subgroup is normal to $G$.
-If $G$ has a subgroup of index $m>1$ and $m! < |G|$, then $G$ has a proper normal subgroup.
And then proceeding by strong induction: if we find a proper normal subgroup $K$ then $K$ and $[G : K]$ are both solvable, hence $G$ solvable.
What else is there? I can't imagine these work for too much longer, especially since the proofs of some groups (such as orders 56, 30, and 45) seem to rely more on "luck of the numbers" more than anything.