# Tarski Monster group with prime $3$ or $5$

Is there any Tarski Monster Groups for the prime $3$ or $5$? I know that there is no Tarski Monster Groups with prime $2$, but I don't know for the prime $3$ and $5$.

• Are the Tarski Monsters not only known to exist for very large primes? – Tobias Kildetoft Jul 23 '13 at 8:38
• What i know is that for primes >1003 such group always exist. But this does not mean that for prime 3 or 5, such group does not exist. – Dei Jul 23 '13 at 8:45
• Could you give a reference for that existence? I only remember having seen it for much larger primes than that. – Tobias Kildetoft Jul 23 '13 at 8:50
• S I Adyan and I G Lysënok, ON GROUPS ALL OF WHOSE PROPER SUBGROUPS ARE FINITE CYCLIC, Mathematics of the USSR-Izvestiya Volume 39 Number 2 – Dei Jul 23 '13 at 8:54
• Finitely generated groups of exponents 2,3,4 and 6 have been proved finite. Exponent 5 (together with lots of other small exponents) is unknown. – Derek Holt Jul 23 '13 at 10:17

There does not exist a Tarski monster for $p=3$, while I am unsure about the $p=5$ case. However, I understand that it is an open problem to prove that every two-generated group of exponent $5$ is finite (if such groups were finite then there would be no Tarski monster for $p=5$).
For $p=3$, this is because any finitely-generated group of exponent three is finite (this is a "left to the reader" exercise at the start of Chapter 6, using the start of Chapter 4, of Ol'shanskii's book Geometry of Defining Relations in Groups). Indeed, the free Burnside group of rank $m$ and exponent three has order $3^{m+{m \choose 2} + {m\choose 3}}$ (see MathOverflow).
To see that every finitely generated group $G$ of exponent three is finite, begin by using $a^3=1$, $b^3=1$, $(ab)^3=1$ and $(a^{-1}b)^3=1$ to get that $a(bab^{-1})=(bab^{-1})a$ (this is Figure 25, p113, of Ol'shanskii's book). Thus, $a$ commutes with its conjugates, and so $a$ lies in some abelian normal subgroup $N$ of $G$, $N=\langle a^b: b\in G\rangle$. As $a$ is arbitrary, every element lies in an abelian normal subgroup of $G$. This implies that $G$ is finite, as take $a$ to be a generator and look at $G_1=G/N$. Then $G_1$ is generated by fewer elements than $G=G_0$, and $G_2=G_1/N_1$ is generated by fewer elements than $G_1$, etc. If $G_1$ is infinite then this process will obtain after finitely many steps a finite group, $G_n$ say, such that $G_{n-1}$ is infinite. We then have $G_0\rightarrow G_1\rightarrow G_2\rightarrow \cdots\rightarrow G_n$. We shall prove that $G_{n-1}$ is finite, a contradiction, and so $G=G_1$ is finite. As $G_n$ is finite, $N_{n-1}$ has finite index in $G_{n-1}$. As $G_{n-1}$ is finitely generated, and finite index subgroups of finitely generated groups are finitely generated, we have that $N_{n-1}$ is finitely generated. It is also abelian of exponent three, and so $N_{n-1}$ is finite. As $N_{n-1}$ is a finite finite index subgroup of $G_{n-1}$, we have that $G_{n-1}$ is finite, and the proof is complete. (This ending is sloppy. Any ideas for a neater way?)
I have given up trying to find decent references for the $p=5$ case. Perhaps this would be a good MathOverflow question. Mark Sapir frequents there, and he should know this (my "looking for a reference" consisted of trawling through his papers...).