I was wondering if somebody could explain why '$G$ is clearly non-Abelian' just from knowing that $\operatorname{Aut}(G)$ is abelian and of order $p^5$, with $p$ prime (not 2) in this proof.
If $\operatorname{Aut}(G)$ is abelian order $p^5$, is $G$ non-abelian
automorphism-groupfinite-groupsgroup-theory
Related Solutions
There is only one non-abelian group of order $pq$ with primes $p<q$ and $p\mid q-1$, which is proved in Theorem $3.8$ of Keith Conrad's notes here. It is indeed given by a semidirect product; and as a subgroup of the affine group ${\rm Aff}(\mathbb{Z}/(q))$.
Proposition: Let $G$ be a group of order $pq$ with primes $p<q$. If $q\not\equiv 1 \bmod p$, then $G \cong C_{pq}$. If $q\equiv 1 \bmod p$, then $G$ is isomorphic to either $C_{pq}$, or to the non-abelian group $$ \left\{ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \mid a\in (\Bbb{Z}/(q))^{\times}, b\in \Bbb{Z}/(q), a^p\equiv 1 \bmod q \right\}. $$
Remark: When $q \equiv 1 \bmod p$, so ${\rm Aut}(\mathbb Z/q\mathbb Z) = (\mathbb Z/q\mathbb Z)^\times$ is cyclic of order $q-1$ and thus contains a unique subgroup of order $p$, the matrix $(\begin{smallmatrix}a&b\\0&1\end{smallmatrix})$ can be written as $(\begin{smallmatrix}1&b\\0&1\end{smallmatrix})(\begin{smallmatrix}a&0\\0&1\end{smallmatrix})$. Therefore the matrix group in the proposition can be described as a semidirect product $\cong C_q\rtimes C_p$ where $C_p = \{a \in (\mathbb Z/q\mathbb Z)^\times : a^p \equiv 1 \bmod q\}$ acts as automorphisms of $C_q = \mathbb Z/q\mathbb Z$ by multiplication
Proof: Let $P$ be a Sylow $p$-subgroup of $G$, and $Q$ be a Sylow $q$-subgroup of $G$. We have $P\cong C_p$, $Q\cong C_q$ and $(G:Q)=p$, which is the smallest prime dividing $(G:1)$. So $Q$ is normal. Because $P$ maps bijectively onto $G/Q$, we have that $G=Q\rtimes P$. Since ${\rm Aut}(Q)\cong C_{q-1}$ we obtain $G=Q\times P\cong C_q\times C_p\cong C_{pq}$, unless $p\mid (q-1)$, i.e., $q\equiv 1 \bmod p$. In that case the cyclic group ${\rm Aut}(Q)$ has a unique subgroup $A$ of order $p$. In fact, $A$ consists of the automorphisms $x\mapsto x^i$ for $i\in \Bbb{Z}/q\Bbb{Z}$ with $i^p=1$. Let $a$ and $b$ be generators of $P$ and $Q$ respectively, and let the action of $a$ on $Q$ by conjugation be $x\mapsto x^j$ with $j\neq 1$ in $\Bbb{Z}/q\Bbb{Z}$. Then $$ G=\langle a,b\mid a^p=b^q=1,\; aba^{-1}=b^j\rangle, $$ which is the semidirect product $Q\rtimes P$ with this action of $P$ on $Q$ by conjugation. Choosing a different $j$ amounts to choosing a different generator $a$ for $P$, and so gives a group isomorphic to $G$. By definition, this group is non-abelian. In fact it is isomorphic to the subgroup of ${\rm Aff}(\Bbb{Z}/(q))$ given above.
Note: The proof above uses the Sylow theorems and the terminilogy of semidirect products, but the proof in the link does not.
Take $G=\mathbb{Z}^2$ its automorphism group $Gl(2,\mathbb{Z})$ is not commutative.
Best Answer
If $G$ is abelian and some $g\in G$ has $g^2\not = 1$, then $\operatorname{Aut}(G)$ contains the element $x \to x^{-1}$ of order $2\!\!\not|\,p^5$. The remaining case $G = \mathbb{Z}_2^{\oplus n}$ and $\operatorname{Aut}(G) = GL_n(\mathbb{Z}_2)$ can be handled directly. (That's assuming $G$ is finite, given the tags and the proof of the theorem.)