Clarification on proof that subgroup of cyclic group is cyclic

Asked Jul 06 '21 at 22:12

Active Jul 06 '21 at 22:25

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For the proof that a subgroup of a cyclic group is cyclic, I've seen the use of the Division Theorem to prove that if the group $G$ is generated by $a$, then $a^m$ generates a subgroup $H$, where $m$ is the least positive integer such that $a^m\in H$.

However, I'm confused as to why we use $a^m$. Isn't any subgroup also just generated by $a$?

Sorry if this question seems dumb, I'm pretty new to group theory and proof writing.

edited Jul 06 '21 at 22:25

Shaun

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asked Jul 06 '21 at 22:12

Aweeb3382

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Does this answer your question? – jasnee Jul 06 '21 at 22:17
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When we say a subgroup $H$ is generated by $a$, $a$ must be a member of $H$. The infinite cyclic group $\Bbb{Z}$ is generated by $1$, but its subgroup ${2i \mid i \in \Bbb{Z}}$ is generated by $2$ (or $-2$) and not by $1$. – Rob Arthan Jul 06 '21 at 22:20
I see, thanks for clarifying. – Aweeb3382 Jul 06 '21 at 22:21
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Please use MathJax to typeset mathematics. Here's a tutorial. – Arturo Magidin Jul 06 '21 at 22:26
On the same note, given any element of a group, there is exactly one subgroup that it generates. – Aryaman Maithani Jul 07 '21 at 06:20

Clarification on proof that subgroup of cyclic group is cyclic

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