Final answer:
In a cyclic group with an element of infinite order, there are no elements of finite order. Every non-identity element in such a group also has infinite order, as having an element of finite order would contradict the group's definition of infinite order.
Step-by-step explanation:
In the context of group theory in mathematics, a cyclic group is characterized by the presence of a single generating element, which means every other element of the group can be derived from it through the group operation, usually multiplication or addition. Now, if a cyclic group contains an element of infinite order, that single element can be repeatedly operated on itself to produce an infinite sequence of distinct elements. This implies that the group is infinite as well.
Coming to the crux of the question, in a cyclic group with an element of infinite order, there are no elements of finite order. Why is that? Because if there were an element of finite order, then there would also be a finite number of operations to return to the identity element of the group, which contradicts the definition of infinite order.
Thus, in a cyclic group with an element of infinite order, every non-identity element also has infinite order. This scenario is consistent with the infinite cyclic group, which is isomorphic to the group of integers under addition. This group essentially has one element of infinite order (which can be thought of as 1 or -1 in the integers), and every other element can be written as some integer multiple of this generator.