Algebra Solutions: In a group, the set of powers of a fixed element is a subgroup

Let $G$ be a group and let $x \in G$ . Prove that $A = \{ x^n \ |\ n \in \mathbb{Z} \}$ is a subgroup of $G$ (called the cyclic subgroup of $G$ generated by $x$ ).

By a previous exercise, it suffices to show that $A$ is nonempty and that $A$ is closed under multiplication and inverses.

We have $x^0 = 1 \in A$ , so that $A$ is not empty.
Given some $x^a, x^b \in A$ , we have $x^a x^b = x^{a+b} \in A$ by a previous example.
Given some $x^a \in A$ , we have $(x^a)^{-1} = x^{-a} \in A$ by a previous exercise.

Thus $A$ is indeed a subgroup of $G$ .

Algebra Solutions

In a group, the set of powers of a fixed element is a subgroup

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