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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 $x$).


Solution: By Exercise 26, it suffices to show that $A$ is nonempty and that $A $is closed under multiplication and inverses.

(1) We have $x^0 = 1 \in A$, so that $A$ is not empty.

(2) Given some $x^a$, $x^b \in A$, we have $x^a x^b = x^{a+b} \in A$ by Exercise 19.

(3) Given some $x^a \in A$, we have $(x^a)^{-1} = x^{-a} \in A$ by Exercise 19.

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

Linearity

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