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The exponent of a group $G$, denoted $\exp(G)$, is equivalently
the least natural number $n$ such that $g^n = 1$, the identity element, for all $g \in G$,
hence the least common multiple of the orders of the group elements.
Here “least” should, by all rights, refer to the divisibility order on $\mathbb{N}$, where $m \leq n$ means $m$ divides $n$. Notice that with this poset structure, $\mathbb{N}$ is a complete lattice with bottom element $1$ and top element $0$; under this convention, the exponent always exists as the least common multiple of the orders of elements, which is their join (supremum) in the lattice, and is $0$ if (for example) $G$ is a torsion-free group.
If $G$ is a finite group of order $n$, then $\exp(G)$ divides $n$ since $g^n = 1$ for all $g \in G$.
If $G$ is a finite abelian group and $\exp(G) = ord(G)$, then $G$ is cyclic.
Of course $ord(G)$ can’t be $0$, so $e = \exp(G)$ here has a prime factorization $e = p_1^{r_1} p_2^{r_2} \ldots p_k^{r_k}$. Since $e$ is the least common multiple of the orders of elements, the exponent $r_i$ is the maximum multiplicity of $p_i$ occurring in orders of elements; any element realizing this maximum will have order divisible by $p_i^{r_i}$, and some power $y_i$ of that element will have order exactly $p_i^{r_i}$. Then $y = \prod_i y_i$ will have order $e = ord(G)$ by the following lemma and induction, so that powers of $y$ exhaust all $ord(G)$ elements of $G$, i.e., $y$ generates $G$ as desired.
If $m, n$ are relatively prime and $x$ has order $m$ and $y$ has order $n$ in an abelian group, then $x y$ has order $m n$.
Suppose $(x y)^k = x^k y^k = 1$. For some $a, b$ we have $a m - b n = 1$, and so $1 = x^{k a m} y^{k a m} = y^{k a m} = y^k y^{k b n} = y^k$. It follows that $n$ divides $k$. Similarly $m$ divides $k$, so $m n = lcm(m, n)$ divides $k$, as desired.
If $G$ is a finite group, then the prime factors of $\exp(G)$ coincide with the prime factors of $ord(G)$.
Since $\exp(G)$ divides $ord(G)$, every prime factor of $\exp(G)$ is a factor of $ord(G)$. If $p$ is a prime factor of $ord(G)$, then by the Cauchy group theorem, $G$ has an element of order $p$, and thus $p$ divides $\exp(G)$.
If $\exp(G) = 2$, then $G$ is abelian and is a vector space over the field $\mathbb{F}_2$.
Similarly, if $G$ is abelian and its exponent is a prime $p$, then $G$ is a vector space over $\mathbb{F}_p$.
A finitely generated group of exponent $3$, $4$, or $6$ must be finite. On the other hand, it is not known whether a group of exponent $5$ generated by $2$ elements must be finite.
These last facts are part of the lore of the celebrated Burnside problem. The free Burnside group $B(m, n)$ of exponent $n$ with $m$ generators is presented by $\langle x_1, \ldots, x_m\; |\; w^n = 1 \; for\; all\; words\; w\; in \; x_1, \ldots, x_m \rangle$, and one formulation of the original Burnside problem was whether $B(m, n)$ is finite. It is now known that the answer is negative for all $m \gt 1$ and odd $n \gt 665$, and there is a similar result for certain even $n$ as well.
For a fixed prime $p$, a Tarski monster for $p$ is a finitely generated infinite group where every proper nontrivial subgroup is cyclic of order $p$. Such monsters exist for sufficiently large $p$, forming a class of dramatic counterexamples to the Burnside conjecture; they are of course of exponent $p$.
It was shown by Efim Zelmanov that there are only finitely many finite groups with $m$ generators and exponent $n$ (restricted Burnside problem). It was largely on the strength of this work that he was awarded the Fields Medal (1994).
See also
Wikipedia, Burnside problem, web
Last revised on September 2, 2021 at 08:38:07. See the history of this page for a list of all contributions to it.