# Simple Group of Order Less than 60 is Prime

## Theorem

Let $G$ be a simple group.

Let $\order G < 60$, where $\order G$ denotes the order of $G$.

Then $G$ is a prime group.

## Proof

First it is noted that Prime Group is Simple.

We also note from Alternating Group is Simple that the alternating group $A_5$, which is of order $60$, is simple.

Hence the motivation for the result.

It remains to be shown that all groups of composite order such that $\order G < 60$ are not simple.

Let $S$ be the set:

- $S = \set {n \in \Z: 0 < n < 60: \text { there exists a simple group of order $n$ such that $n$ is composite} }$

The aim is to show that $S$ is empty.

From Abelian Group is Simple iff Prime, all abelian groups of composite order are not simple.

Thus any simple group must be non-abelian.

Let $p$ and $q$ be prime.

From Prime Power Group has Non-Trivial Proper Normal Subgroup, no group of order $p^n$ is simple.

Thus:

- $\forall n \in \Z_{>0}: p^n \notin S$

Thus, with the primes and prime powers eliminated, we have:

- $S \subseteq {6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36, 38, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58}$

From Group of Order $p q$ has Normal Sylow $p$-Subgroup:

- $p q \notin S$

The set of non-square semiprimes less than $60$ is:

- $\set {6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58}$

Thus we have so far:

- $S \subseteq {12, 18, 20, 24, 28, 30, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56}$

From Group of Order $p^2 q$ is not Simple:

- $p^2 q \notin S$

This eliminates:

\(\ds 12\) | \(=\) | \(\ds 2^2 \times 3\) | ||||||||||||

\(\ds 18\) | \(=\) | \(\ds 3^2 \times 2\) | ||||||||||||

\(\ds 20\) | \(=\) | \(\ds 2^2 \times 5\) | ||||||||||||

\(\ds 28\) | \(=\) | \(\ds 2^2 \times 7\) | ||||||||||||

\(\ds 44\) | \(=\) | \(\ds 2^2 \times 11\) | ||||||||||||

\(\ds 45\) | \(=\) | \(\ds 3^2 \times 5\) | ||||||||||||

\(\ds 50\) | \(=\) | \(\ds 5^2 \times 2\) | ||||||||||||

\(\ds 52\) | \(=\) | \(\ds 2^2 \times 13\) |

Thus we are left with:

- $S \subseteq {24, 30, 36, 40, 42, 48, 54, 56}$

From Normal Subgroup of Group of Order 24, a group of order $24$ has a normal subgroup either of order $4$ or order $8$.

Hence $24 \notin S$.

From Group of Order 56 has Unique Sylow 2-Subgroup or Unique Sylow 7-Subgroup, a group of order $56$ has at least one normal subgroup.

Hence $56 \notin S$.

From Group of Order 30 is not Simple:

- $30 \notin S$

Thus we are left with:

- $S \subseteq {36, 40, 42, 48, 54}$

$40$ is eliminated by Group of Order 40 has Normal Subgroup of Order 5.

$42$ is eliminated by Group of Order 42 has Normal Subgroup of Order 7.

$54$ is eliminated by Group of Order 54 has Normal Subgroup of Order 27.

Thus remains:

- $S \subseteq {36, 48}$

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 59 \eta$