What am I missing?What is the Proof for Cyclic Groups Being Subgroups?

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SUMMARY

The discussion centers on the proof that subgroups of cyclic groups are also cyclic, specifically addressing the conclusion that if \( a^m \in H \), then \( r \) must equal \( 0 \). The participant clarifies that \( m \) is defined as the smallest positive integer such that \( a^m \in H \), which directly implies \( r \) cannot be any positive integer less than \( m \). The critical point is the closure property of groups, which ensures that if \( a^m \in H \), then all powers of \( a^m \) and its inverse are also in \( H \).

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Bashyboy
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Hello everyone,

I am trying to understand the proof given in this link:

https://proofwiki.org/wiki/Subgroup_of_Cyclic_Group_is_Cyclic

I understand everything up until the part where they conclude that ##r## must be ##0##. Their justification for this is, that ##m## is the smallest integer, and so this forces ##r=0##. Is it not possible that ##m=3## could be the smallest integer? Couldn't ##r## then be ##2##?
 
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No, you are missing the point. m was, by hypothesis, the smallest positive integer such that [itex]a^m\in H[/itex]. We also have that [itex]a^r\in H[/itex]. Once we have that [itex]0\le r< m[/itex], we cannot have r any positive integer because m is the smallest such integer.
 
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It seems like the critical information is in this line "Since ## a^m∈H## so is ##(a^m)^{−1}## and all powers of the inverse by closure ."
 

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