Problem 5.2.6 (Properties of the exponent)

Proof. Let $G$ be a finite group, and let $n$ be the least common multiple of the orders of the elements of $G$. If $x\in G$, then the order of $x$ divides $n$, thus $x^n=1$. Since $x^n=1$ for all $x$, the exponent of $G$ is finite.

By Exercise 5.1.18(a), every element $x$ of

satisfies $x^2=1$, so the exponent of $G\ne \{1\}$ is $2$.

Since $S_3$ contains some elements of order $2$ and of order $3$, the exponent of $S_3$ is at least $6$. Moreover ${\sigma }^6=1$ for all $\sigma \in S_3$, thus the exponent of $S_3$ if $6$.

Nevertheless, there is no element of order $6$ in $S_3$ ( $S_3$ is not cyclic). A finite group of exponent $m$ does not always contain an element of order $m$. □