Let $p_1^{\alpha_1}p_2^{\alpha_2}...p_r^{\alpha_r}$ be the prime decomposition of $n$. Prove: $(|G|=n ⇒ G~\text{is abelian})⇔∀i,j∈\{1,...,r\}(\alpha_i∈\{1,2\} \land p_i \nmid p_j^{\alpha_j}-1)$.
Proof: Recall that a direct product is abelian if and only if all of its factors are abelian (5.1.1).
($⇒$) (1) Assume there is a prime factor $p_k^{\alpha_k}$ in the prime decomposition of $n$ such that $\alpha_k \geq 3$. Obtain a nonabelian group $P$ of order $p_j^3$ as described in the material and we have $P \times Z_{n/p_j^3}$ is a nonabelian group of order $n$, contrary to assumption.
(2) Assume $p_i \mid p_j^{\alpha_j}-1$ for some $i,j$. If $\alpha_j = 1$, then obtain the nonabelian group $P$ of order $p_ip_j$ also as presented in the material and we have $P \times Z_{n/(p_ip_j)}$ is a nonabelian group of order $n$, also contrary to assumption. If $\alpha_j=2$, then we have $p_i \mid p_j^2-1=(p_j-1)(p_j+1)$. Now, $|Aut(Z_{p_ j} \times Z_{p_j})|=p_j(p_j-1)^2(p_j+1)$, so it contains a cyclic group of order $p_i$ by Cauchy, to which $φ$ shall nontrivially map $Z_{p_i}$. We obtain $H = (Z_{p_j} \times Z_{p_j}) \rtimes_φ Z_{p_i}$ is a nonabelian group of order $p_ip_j^2$, so that $H \times Z_{n/(p_ip_j^2)}$ is a nonabelian group of order $n$, a contradiction.
($\Leftarrow$) Assume $G$ is a nonabelian group of order $n$. Inductively, every proper subgroup of $G$ is abelian (since any proper subgroup divides $G$'s order, and any of the factors whose exponents are reduced from 2 to 1 in the process are insured since $p \nmid q^2 - 1 = (q-1)(q+1) ⇒ p \nmid q-1$), so that by 4.5.56 $G$ is solvable. Let $N$ be the normal subgroup such that $G/N$ is abelian, and in particular $N$ is not trivial due to the nonabelianness of $G$. If $N$ has a prime decompositional width greater than 1 (i.e. is divisible by two distinct primes), then two distinct Sylow subgroups $A$ and $B$ of $N$ are normal in $N$ due to $N$'s proper-subgroup abelianness, and are thus characteristic in $N$, leading to them being normal in $G$. Since $G/A$ and $G/B$ are of smaller order than $G$ and holding to the a priori qualities of $G$, and thus abelian, we have by 5.4.15 that $G/(A∩B)=G/1$ is abelian, so $N$ is of width 1, and of order $p_i$ or $p_i^2$.
If $G$ is of width 1, then by the abelianness of groups of order $p$ or $p^2$ we know that $G$ is abelian, so assume $G$ is of width greater than 1. By Cauchy, there is a subgroup $\bar{H} \trianglelefteq \bar{G}$ (normal since the latter is abelian) with an order of $p_k$ - a prime distinct from $p_i$ - so that $H \trianglelefteq G$ is of order $p_ip_k$ or $p_i^2p_k$. If $G = H$ of order $p_ip_k$, then $G$ is abelian by previous material. If $G=H$ of order $p_i^2p_k$, then since $p_i \nmid p_k - 1$ we have $p_k \not ≡ 1 \mod{p_i}$, leading to the Sylow $p_i$-subgroup $C$ being normal so that $G = C \rtimes D$ for a $p_k$-subgroup $D$, but $Aut(C)$ has order either $p_i(p_i-1)^2(p_i+1)$ or $p_i(p_i-1)$ and $p_k \nmid p_i^2-1 = (p_i-1)(p_i+1)$, so we have it is the trivial homomorphism in the semidirect product and $G$ is abelian. Therefore $H$ is a proper subgroup and thus abelian so that its Sylow $p_k$-subgroup $P_k$ is characteristic in $H$ and thus normal in $G$, so that $G/P_k$ is also abelian and now $G/(N∩P_k)=G/1≅G$ is abelian. $\square$
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