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, Tianlalu May 24 at 3:59

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  • $\begingroup$ what tools do you know? $\endgroup$ – 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?

Here's the answer:

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.


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