# Prove that there is no nonabelian simple group of order less than 60 [duplicate]

Any tips for this question? I don't want the answer itself, just figure out how must I proceed.

## marked as duplicate by Randall, YuiTo Cheng, Xander Henderson, Theo Bendit, TianlaluMay 24 at 3:59

• what tools do you know? – Rylee Lyman May 24 at 2:34

A nice trick is Burnside's $$p^aq^b$$ theorem: it says that if $$p,q$$ are distinct primes and $$G$$ is a group of order $$p^aq^b$$, $$a,b\geq 1$$, then $$G$$ must be solvable.
But a nonabelian simple group cannot be solvable: its commutator subgroup $$G'$$ is normal, so $$G=G'$$, which means the lower central series cannot terminate.
In this way you can rule out a lot of numbers $$<60$$. (Plus: you also can't have a group of order $$p^a$$, because all $$p$$-groups are solvable.)
For example, the smallest number with at least three primes in its factorization is $$2\cdot 3\cdot 5 = 30$$. Another is $$2\cdot 3\cdot 7 = 42$$. In fact I'm pretty sure those are the only two such numbers $$<60$$ :) Can you rule them out?
Basically you need Sylow's Theorem. Using the notation from the Wikipedia article: if $$n_p=1$$, then in fact the Sylow $$p$$-subgroup must be normal. For $$|G|=30$$ you can show $$n_3=1$$, and for $$|G|=42$$ you can show $$n_7=1$$. So no such $$G$$'s can be simple.