Is this type of proof by induction correct in Sylow's Theorem?

Question

The following is the first part of the Sylow's Thm:

My question is: if order of $G$ was $p^a$ (and not $p^am$) then we could start with $|G|=1$ which means $a=0$. Then supposed that the theorem holds for all $a$'s up to $p-1$ then proved for $p$. But the problem here is that statement by induction involved more than one parameter, i.e. not only $p^a$ but also $m$. Why $m$ is absolutely irrelevant? Or otherwise why we don't go another induction on $m$, since especially when we say $|G|=1$ we don't mean arbitrary $m$ and we consider $m=1$ the smallest number in induction?

Added: in the 2nd paragraph of the proof it says "If we let P be a subgroup of G containing N such that ..."; How we know that such P definitely exists?

Answer 1

The induction is only on the size of the group $G$. So the induction hypothesis is that for any group $H$ of order less than $|G|$, if $|H|=p^\alpha m$ for any prime $p$, any $\alpha$, and any $m$ not divisible by $p$ (so long as $p^\alpha m<|G|$), $\text{Syl}_p(G) \neq \emptyset$.

Thus the base case is $|G|=1$, in which case there are no primes dividing $|G|$ and there is nothing to show.

Answer 2

Note that the author uses induction on $|G|$ and not induction on $a$. In addition if we assume that $G$ is of order equal to $p^a$ for some prime number $p$ then there is nothing to prove because the theorem holds by assumption($G$ is the $p$-Sylow subgroup of $G$.)

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