I am fully aware that this question has been discussed here, but I have thaught a completely different argument without taking any help from google and I would be glad if someone verifies my proof.
Since $|G|=105=3\cdot 5\cdot 7$, and a Sylow $3$-subgroup of $G$ is normal in $G$, a consequence of Sylow's theorem tells us that this Sylow $3$-subgroups is unique, call it $P.$
Therefore this subgroup $P$ is invariant under every automorphism of G. So $P$ is a characteristic subgroup of $G.$
In particular, $P$ is invariant under action by conjugation. One other such subgroup of $G,$ which is invariant under action by conjugation is the center $Z(G)$ of $G.$
But all automorphisms of $G$ can not fix $Z(G).$ For example in $D_4$ for a fixed $g\in D_4,$ $a\mapsto ga, a\in D_4 $ is an automorphism of $D_4$. If we denote $r\colon=$ rotation through $90^\circ$ in the positive sense and $f\colon=$ flip about vertical axis, the center of $D_4$ is $Z(G)=\langle r^2,f \rangle.$ But $ r\cdot Z(G)=\langle r^3,rf \rangle\ne Z(G)$
So, $P$ is contained in $ Z(G)$ and since $P$ is a characteristic subgroup of $G$ we can say that $P$ is a normal subgroup of $Z(G)$. But the smallest normal subgroup containing $P$ is $N_{G}(P),$ the normaliser of $P$ in $G$. Therefore $N_{G}(P)\le Z(G).$
Hence we get, $|G\colon Z(G)|\le|G\colon N_{G}(P)|$
$\implies|G\colon Z(G)|\le|G\colon G|[\because \text{P is normal in G, }N_{G}(P)=G]$
$\implies \dfrac{|G|}{|Z(G)|}\le1$
$\implies |G|\le|Z(G)|$
hence $G=Z(G)$, in other words, $G$ is abelian.
Is my argument correct?