Every infinite cyclic group has non-trivial proper subgroups

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Every infinite cyclic group contains non-trivial proper subgroups, which differ from finite cyclic groups primarily in that they do not have a finite order. In infinite cyclic groups, there is no integer n such that a^n equals the identity element e for any element a other than e itself. The structure of an infinite cyclic group can be represented as C = ⟨a^n | n ∈ ℤ⟩, encompassing all integer powers of a. Non-trivial subgroups must include the identity and additional elements, leading to the conclusion that proper subgroups exist. Thus, infinite cyclic groups inherently possess non-trivial proper subgroups.
Mr Davis 97
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Homework Statement


Every infinite cyclic group has non-trivial proper subgroups

Homework Equations

The Attempt at a Solution


I know that if we have a finite cyclic group, it only has non-trivial proper subgroups if the order of the group is not prime. But I'm not sure how to make this argument with infinite groups
 
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What is a infinite cyclic group and how does it differ from a finite one?
 
fresh_42 said:
What is a infinite cyclic group and how does it differ from a finite one?
Doesn't it differ by the fact that there exists no integer ##n## for which ##a^n=e##, where ##a## is an element of the group besides e, the neutral element?
 
Mr Davis 97 said:
Doesn't it differ by the fact that there exists no integer ##n## for which ##a^n=e##, where ##a## is an element of the group besides e, the neutral element?
Yes. So we have ##C = \langle a^n\,\vert \,n \in \mathbb{Z} \rangle = \{\ldots , a^{-2},a^{-1},e,a,a^2,\ldots\}##.
Non-trivial subgroup ##U## means ##\{e\}\subsetneq U \subsetneq C##.
So what can we conclude from ##\{e\}\subsetneq U##?
What could be a subgroup?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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