Question

Can there be an element of infinite order in a finite group? Prove or disprove.

Answer 1

No.

Start with a lemma.

Lemma: Let G be a group. If a ∈ G has infinite order, the a^k ≠ aⁿ for k ≠ n where k and n are positive integers.

Proof: Let G be a group and let a ∈ G have infinite order. By way of contradiction, assume a^k = aⁿ for some k ≠ n. Without loss of generality, assume that k < n. By multiplying the equation a^k = aⁿ k times on the left by a⁻¹ , we end up with the equation I = a^n−k where n − k is a positive integer. This is a contradiction since a has infinite order. This concludes the proof of the lemma.

The problem now follows. If a group has an element of infinite order, by the lemma it will have an infinite number of distinct elements (one for each power of a), therefore will be an infinite group.