**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.35**

Let $G$ be a group, and let $x \in G$ be an element of finite order; say $|x| = n$. Use the Division Algorithm to show that any integral power of $x$ is equal to one of the elements in the set $A = \{ 1, x, x^2, \ldots, x^{n-1} \}$. Conclude that $A$ is precisely the set of distinct elements of the cyclic subgroup of $G$ generated by $x$.

Solution: Let $k \in \mathbb{Z}$. By the Division Algorithm, there exist unique integers $(q,r)$ such that $k = qn + r$ and $0 \leq r < n$. Thus $$x^k = x^{qn + r} = (x^n)^q x^r = x^r,$$ where $x^r \in A$. Hence the cyclic subgroup of $G$ generated by $x$ is $A$; moreover, by Exercise 1.1.34, the elements of $A$ are all distinct.