SO I know that I'm suppose to prove it by contradiction and assume that the element has a positive power. I'm not really sure how to answer it though.
2026-03-26 22:16:14.1774563374
Let G be an infinite cyclic group. Prove that G cannot have any non-identity elements of finite order.
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Suppose $g$ is the generator for the group, and suppose $g^a$ has finite order $b$. Then $e=(g^{a})^b=g^{ab}$. And so the generator has finite order, which implies the group has finite order. Since if $n>ab$ then $n=cab+r$ and then $g^n=g^{cab}+g^r=eg^r$. So your group has at most $ab$ elements.