1. Oct 16, 2014

### Bashyboy

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 $2$?

Last edited: Oct 16, 2014
2. Oct 16, 2014

### HallsofIvy

Staff Emeritus
No, you are missing the point. m was, by hypothesis, the smallest positive integer such that $a^m\in H$. We also have that $a^r\in H$. Once we have that $0\le r< m$, we cannot have r any positive integer because m is the smallest such integer.

Last edited: Oct 16, 2014
3. Oct 16, 2014

### RUber

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 ."