---
tag: TE19BGFFF1
kind: object
name: Tarski monster group
msc: ["20F50", "20F06"]
instance_of: [JZNQTTMG58, 0YMYMNQ0QZ]
sources:
  - url: https://en.wikipedia.org/wiki/Tarski_monster_group
same_as:
  wikidata: Q7686763
  wikipedia: Tarski monster group
---

## Description

Let $p$ be a prime.
A *Tarski monster group* for $p$ is an infinite [[group|G5S1M9]] $G$ in which every [[subgroup|0X1JXT47GA]] other than $\set{1}$ and $G$ is cyclic of order $p$.
Ol'shanskii proved, with [[small cancellation|3A1D37TVF6]] methods, that a Tarski monster group exists for every prime $p > 10^{75}$.

## Remarks

Let $g \ne 1$ in $G$.
If $\langle g \rangle = G$, then $G \cong \Z$, which has the subgroup $2\Z$ of infinite order; this is impossible.
So $\langle g \rangle$ is a proper subgroup other than $\set{1}$, and $g$ has order $p$.
Two elements $x, y$ that do not lie in one subgroup of order $p$ generate $G$, since $\langle x, y \rangle$ is neither trivial nor cyclic of order $p$.
So $G$ is generated by $2$ elements and has [[exponent|0ZDW0H0VX4]] $p$, and it answers the bounded [[Burnside problem|XWEMJ7T2XV]] in the negative for this $p$.

$G$ is [[simple|0YMYMNQ0QZ]].
Let $N \ne \set{1}$ be a [[normal subgroup|0CB0DX5BVV]] with $N \ne G$; then $N$ has order $p$.
Take $x \notin N$; then $\langle x \rangle N$ is a subgroup of order at most $p^2$, so it is proper since $G$ is infinite.
It contains $N$ and $x$, so it is not cyclic of order $p$, a contradiction.

## Explanations

### L1 human

In a Tarski monster, every subgroup except the trivial one and the whole group has exactly $p$ elements.
So there are no medium-sized subgroups at all: any two elements that are not in the same small subgroup generate the whole infinite group.
Groups of this kind were long expected not to exist, which is why they are called monsters.