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I was reading this Wikipedia article and came across this

For groups of order $n<60$, the simple groups are precisely the cyclic groups $\mathbb{Z}_n$ for prime $n$.

I was wondering, why is this true for $n<60$? What is so special about this value?

Trogdor
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    $A_5$ is the smallest non-abelian simple group. Smaller groups aren't complicated enough to be simple, except for the groups of prime order. – Daniel Fischer Nov 08 '15 at 13:17
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    Outside maths, would we ever say: "not complicated enough to be simple"? – badjohn Dec 13 '20 at 07:48
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    @badjohn Yes, we would say this in art and music, too. Mozart's music sounds so simple, although it is complicated. Or about Mozart himself. For example, Rolando Villazón says about on Mozart: "I think he was so complicated because he was so simple." – Dietrich Burde Dec 01 '21 at 11:42

3 Answers3

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Well if the order of a group is divisible by at most two primes then it is either a cyclic group of prime order or has a proper normal subgroup (or the trivial group of order $1$). So any non-abelian finite simple group is divisible by at least three primes (see Burnside's Theorem)

The first possibility with at least three primes is $30=2\times 3 \times 5$ - but if this were simple, Sylow's theorems would tell us that there would be six subgroups of order $5$ containing $24$ elements of order $5$ and ten subgroups of order $3$ containing $20$ elements of order $3$. So that doesn't work.

Then [added because my arithmetic is poor, in response top a comment] there is $42=2\times 3\times 7$ which Sylow tells us has a normal subgroup of order $7$. The next possibility is $60$ which does work.

The Sylow Theorems can be awkward to use in general (for example they don't provide any easy proof of Burnside's Theorem), but they can tell us a lot about groups with simple structures. For example, if we have a factor $3$ in the order of the group we also need one of $4,7,10,13,16 \dots$. A factor of five comes with a factor from $6, 11, 16, 21 \dots$, and seven comes with $8, 15, 22 \dots $

Now suppose, for example, we have $4$ Sylow $3$ subgroups in our simple group $G$. Then $G$ acts transitively by conjugation on those subgroups, and this action gives a homomorphism from $G$ onto a non-trivial subgroup of $S_4$. Since $G$ is simple, this must have trivial Kernel and $G$ is isomorphic to a subgroup of $S_4$. So four subgroups won't do.

If we have a factor seven, we need at least $8$ as another factor, plus another prime. If we take $3$ we get $168$ which is the order of the next non-abelian simple group after order $60$.

Daniel Fisher's comment that smaller groups "aren't complicated enough to be simple" is spot on.

Mark Bennet
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All groups of order smaller than $60$ are solvable. And the simple groups that are solvable are precisely those of prime order.

Nicky Hekster
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Note that 60 is the least non square free product of 3 powers of distinct primes(= $2^2\bullet3\bullet5$), all non-negative integers less than 60 is one of the following cases and any group of this order is either abelian or not simple or both:

p-group : Either cyclic (abelian) or not-simple (By the 1st Sylow)

$p^a q^b$ : Burnside's $p^a q^b$ theorem states they are not simple.

pqr : Assume some group of order pqr is simple and p<q<r then calculate the smallest numbers can be $n_p, n_q, n_r$, then count the smallest number of elements of Sylow p,q,r-subgroups reveals that the sum is exceed pqr which is a contradiction.

0 : $<e>$, which is abelian and not simple since everything is equal to e, and has no proper normal subgroup.

MathGuest
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    This doesn't really add anything that's not present in other answers, and it could use some clarification (60 is not the least square-free product of three distinct primes). – KReiser Dec 13 '20 at 07:45