Every group of order $45$ is abelian

Given: The hypotheses and objects in the Statement.

Let $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$. Then the number of Sylow $p$-subgroups satisfies $n_p(G)\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p),n_p(G)\mid m\mathrm{.}$. (Sylow III: $n_p\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}p)$ and $n_p\mid m$ when $\mathrm{\mid }G\mathrm{\mid }=p^{a}m$ with $p\nmid m$).

A Sylow $p$-subgroup of a finite group is normal if and only if it is the unique Sylow $p$-subgroup. (A Sylow $p$-subgroup is normal if and only if it is unique).

Normal Sylow subgroups for distinct primes centralize one another. (Distinct normal Sylow subgroups centralize one another).

If $p$ is prime and $G$ is a group of order $p^2$, then $G$ is abelian. (Every group of order $p^2$, for prime $p$, is abelian).

Let $N_0,\dots ,N_{r-1}⊴G$. The following are equivalent: the $N_i$ form an internal direct product of $G$; every $g\in G$ has a unique expression $g=n_0\cdots n_{r-1}$ with $n_i\in N_i$; and the multiplication map $\mu :{\prod }_{i<r}{N}_i\to G$ is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).