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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)?

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    $\begingroup$ 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 $p$. $\endgroup$ Commented Apr 11, 2022 at 12:50
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    $\begingroup$ 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. $\endgroup$ Commented Apr 11, 2022 at 12:51
  • $\begingroup$ @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? $\endgroup$
    – UNOwen
    Commented Apr 11, 2022 at 12:57
  • $\begingroup$ 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 $S$.' $\endgroup$
    – UNOwen
    Commented Apr 11, 2022 at 13:04
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    $\begingroup$ 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. $\endgroup$ Commented Apr 11, 2022 at 13:41

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As noted in comments:

  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.

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