# Order of Finitely Generated Group of Prime Exponent

Consider the finitely generated group $$G=\langle g_1,g_2,...,g_n \rangle=\{g_1^{r_1}g_2^{r_2}...g_n^{r_n}\mid r_i\in\mathbb{Z}\}$$. The exponent of $$G$$ is $$\exp G=\text{lcm}\left(\left|g_1^{1}g_2^{1}...g_n^{1}\right|,\left|g_1^{1}g_2^{1}...g_n^{2}\right|,...\right),$$ For some prime $$p$$, if $$\exp G=p$$, would $$\left|g_1^{r_1}g_2^{r_2}...g_n^{r_n}\right|=p$$ for all $$r_i\in\mathbb{Z}\backslash\{0\}$$ (excluding the identity $$e$$ with order $$1$$)? Also, what could be concluded about the order of $$G$$ (i.e. must $$G$$ be finite)?

• Since you are defining the exponent as the lcm, clearly the order of any element divides the exponent, in the case $p$. No, $G$ need not be finite: Tarski monsters are $2$-generated infinite groups of exponent $p$. Commented Apr 11, 2022 at 12:50
• PS: your description of $G$ is incorrect. For example, your description excludes $g_2g_1$, which is clearly an element of $G$. Unless you somehow forgot to tell us $G$ is supposed to be abelian. Commented Apr 11, 2022 at 12:51
• @ArturoMagidin Thank you for your comments and reference to the Tarski monsters. I can see now that I have incorrectly assumed $G$ is abelian. What would be a correct definition for a nonabelian $G$? Also, is my definition of the exponent as the lcm of the orders of each element correct or are there alternatives? Commented Apr 11, 2022 at 12:57
• If I remember correctly, Lang uses 'for some $S\subset G$, if $G$ is the intersection of all subgroups $H$ of $G$ containing $S$, then for all $s\in S$, $s$ is a generator of $G$ and $G$ is finitely generated if there exists a finite $S$.' Commented Apr 11, 2022 at 13:04
• See here. You'll need something like $g_{j_1}^{e_1}g_{j_2}^{e_2}\cdots g_{j_k}^{e_k}$, etc. Note also your explicit description of the set in the exponent of $G$ omits $g_1$, $g_2$, etc. Commented Apr 11, 2022 at 13:41

1. Your description of $$\langle S\rangle$$ is incomplete/incorrect. Because we are not assuming $$G$$ is abelian, you can't assume the generators will occur in order. So you want something like the set of all elements of the form $$g_{j_1}^{e_1}g_{j_2}^{e_2}\cdots g_{j_k}^{e_k}$$ where $$k\geq 0$$, $$j_i\in\{1,\ldots,n\}$$ for all $$i$$, $$1\leq i\leq k$$, and $$e_i\in\mathbb{Z}$$.
2. Your description of the set of elements to take the least common multiple is also incorrect, as even in the abelian case it would be missing $$g_1$$, $$g_2$$, $$g_1g_3$$, etc.
3. There are conflicting definitions of "exponent of a group". Note that we say that an element $$g$$ is of "exponent $$n$$" to mean that $$g^n=e$$, not to mean that $$n$$ is the least positive integer for which $$g^n=e$$ (that's called the "order of $$g$$"). But "order of $$G$$" is the size of the underlying set. So some people say that $$G$$ is "of exponent $$n$$" to mean that $$g^n=e$$ for all $$g\in G$$ (so, every element is of exponent $$n$$), and others say that "the exponent of $$G$$ is $$n$$" to mean that $$g^n=e$$ for all $$g\in G$$, and $$n$$ is the least positive integer with this property. This coincides with the least common multiple definition you give, provided such least common multiple exists. In this case, some people also say "$$G$$ is of exponent $$n$$" (but in the first meaning one does not usually say "the exponent of $$G$$ is $$n$$" because the singular definite article would imply that it is unique).
4. If a group $$G$$ is of exponent $$n$$ (in either definition), then for every $$g\in G$$ we necessarily have that $$|g|$$ divides $$n$$. In particular, if $$G$$ is of prime exponent $$p$$, then every non-identity element of $$G$$ must have order $$p$$.
5. Tarski monsters are an extreme example of finitely generated groups of exponent $$p$$ that are not finite. In fact, a Tarski monster is an infinite group that has the property that every nontrivial proper subgroup is of order $$p$$. They exist for every sufficiently large prime $$p$$. For $$p=2$$ and $$p=3$$ we know that a finitely generated group of exponent $$p$$ is necessarily finite. This is related to the Burnside problem.