Talk:Existence of Unique Subgroup Generated by Subset/Singleton Generator

I'm puzzled as to the need for this proof by induction. The initial proof was all that was needed: to prove that any subgroup containing $a$ has $H$ as a subgroup. As $K$ is a group, it is closed and therefore $a^n \in K$. Job done. --prime mover (talk) 01:35, 23 May 2017 (EDT)

The initial proof started with $K \le G: a \in K$, and then immediately concluded that $\forall n \in \Z: a^n \in K$ with no explanation at all. I understand that this is a simple result, but there has to be some explanation as to how one gets that result. Even if you were to mention the closure axiom, that alone simply wouldn't be enough. I also shortened and simplified my proof a little bit to make it more readable. --HumblePi (talk) 17:42, 23 May 2017 (EDT)

But if $a \in K$ then $a^n \in K$ by dint of $K$ being a subgroup of $G$ and so a group. If we need to prove that $a \in K \implies a^n \in K$ we link the appropriate result.

Otherwise why don't we derive the fundamental theorem of calculus in longhand every time we calculate a primitive? --prime mover (talk) 17:46, 23 May 2017 (EDT)

Would you like for me to make my work a second proof? Perhaps the initial proof can be presented as an informal one, and the one that I made can be presented as formal, like with Cartesian Product of Countable Sets is Countable? --HumblePi (talk) 17:58, 23 May 2017 (EDT)

The first proof is completely formal -- or it would be if it is made explicit that the power of a group element is itself an element of a group. It's such a fundamental result that including it in full detail in this proof seems the wrong place for it.

If you think about it, the very expression $H = \left\langle {a}\right\rangle = \left\{{a^n: n \in \Z}\right\}$ already takes for granted that fact, otherwise you wouldn't be able to express this proof in the first place. So if it's taken for granted for $H$, then it can be taken for granted for $K$ as well. --prime mover (talk) 18:24, 23 May 2017 (EDT)

Very well, I'll undo my edits.

Also, I'm sorry about the mistake; it was a huge oversight on my part. --HumblePi (talk) 20:48, 23 May 2017 (EDT)

:-) --prime mover (talk) 00:54, 24 May 2017 (EDT)