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If the group and an element have the same finite order then the group is cyclic

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.3 Exercise 2.3.2

Let $G$ be a finite group and let $x \in G$. Prove that if $|x| = |G|$ then $G = \langle x \rangle$. Give an explicit example to show that if $G$ is infinite this need not be true.

Solution: Recall that $|x|$ is short for $|\langle x \rangle|$. So in fact we have $|G| = |\langle x \rangle|$. Since $\langle x \rangle \subseteq G$ and $G$ is finite, $G = \langle x \rangle$.

Now consider $r \in D_\infty$; $r$ has infinite order, so $|r| = |D_\infty|$. But $D_\infty \neq \langle r \rangle$ since $s \notin \langle r \rangle$.


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