Theorem 7, Section 2.5 of Hungerford’s Abstract Algebra

Question

Theorem 5.7. (First Sylow Theorem) Let $G$ be a group of order $p^nm$, with $n\geq 1$, $p$ prime, and $(p,m)=1$. Then $G$ contains a subgroup of order $p^i$ for each $1\leq i\leq n$ and every subgroup of $G$ of order $p^i$ ($i\lt n$) is normal in some subgroup of order $p^{i+1}$.

Proof: Since $p\mid |G|$, $G$ contains an element $a$, and therefore, a subgroup $\langle a\rangle$ of order $p$ by Cauchy's Theorem. Proceeding by induction assume $H$ is a subgroup of $G$ of order $p^i(1\leq i\lt n)$. Then $p\mid [G:H]$ and by Lemma 5.5 and Corollary 5.6 $H$ is normal in $N_G(H)$, $H\neq N_G(H)$ and $1\lt |N_G(H)/H|= [N_G(H):H]\equiv [G:H]\equiv 0 \pmod p$. Hence $p\mid |N_G(H)/H|$ and $N_G(H)/H$ contains a subgroup of order $p$ as above. By Corollary 1.5.12 this group is of the form $H_1/H$ where $H_1$ is a subgroup of $N_G(H)$ containing $H$. Since $H$ is normal in $N_G(H)$, $H$ is necessarily normal in $H_1$. Finally $|H_1| = |H||H_1/H| = p^ip = p^{i+1}$.

Question: I don’t see where Author used $(p,m)=1$ hypothesis in the proof. At least it is not been used explicitly.

Answer 1

There are two important parts to Sylow's First Theorem:

1. If $p$ is a prime, $i\geq 1$, and $p^i$ divides $|G|$, then $G$ has a subgroup of order $p^i$.
2. If $H$ is a subgroup of order $p^i$, and $p^{i+1}$ divides $|G|$, then $H$ is contained in a subgroup of order $p^{i+1}$ (and is in fact normal in that subgroup).

The fact that $\gcd(p,m)=1$ is not used in establishing these results. It is used in "limiting" their reach. In a sense, item 1 can be read as saying

Let $G$ be a finite group, $p$ a prime, and $i\geq 1$. Then $G$ has a subgroup of order $p^i$ if and only if $p^i$ divides $|G|$.

It's just that the "only if" clause follows from Lagrange's Theorem, so the important part of Sylow's First Theorem in this item is the "if" clause.

Likewise, item $2$ is trading on the fact that $p^{i+1}$ must also divide $|G|$ (otherwise there is no subgroup of order $p^{i+1}$ at all, let alone one that contains $H$). You could likewise rephrase it as an "if and only if":

Let $G$ be a finite group, $p$ a prime, $i\geq 0$, and assume that $H$ is a subgroup of order $p^i$. Then $H$ is contained in a subgroup of order $p^{i+1}$ (and is normal in that subgroup) if and only if $p^{i+1}$ divides $|G|$.

The "only if" clause again holds by Lagrange's Theorem; the $i=0$ case holds by Cauchy's Theorem and that the trivial subgroup is always normal. So the meat of this result is the "if" clause, which requires $p^{i+1}$ to be a divisor of $|G|$.

So the fact that $p^n$ is the largest powers of $p$ that divides $p^nm$ is being used in the way the important part of the statement, the part that was not already known through Lagrange's Theorem, is made: restricting $i$ to $i\leq n$ in the first part, and to $i\lt n$ in the second part.