Step 1: Recheck (A) using the partition-counting formula directly.
The number of Abelian groups of order $n = p_1^{a_1}p_2^{a_2}\cdots$ is $\prod_i p(a_i)$, the product of partition counts. For $n = 45 = 3^2\cdot 5^1$, this is $p(2)\cdot p(1) = 2 \cdot 1 = 2$: namely $\mathbb{Z}_{45}$ and $\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_5$. That is 2 groups, so claiming "3" in (A) is wrong.
Step 2: Recheck (B) with a different counterexample.
The quaternion group $Q_8 = \{\pm1,\pm i,\pm j,\pm k\}$ has order $8 = 2^3$, so it is a $2$-group, yet $ij = k$ while $ji = -k$, so $Q_8$ is non-abelian. This second, independent counterexample confirms (B) fails.
Step 3: Recheck (C) by identifying the group explicitly.
By CRT, $\mathbb{Z}_2\times\mathbb{Z}_3 \cong \mathbb{Z}_6$, cyclic of order $6$. Since $6=2\times3$ involves two distinct primes, this group is not a $p$-group for any single prime $p$. (C) fails.
Step 4: Recheck (D) via the normal Sylow-7 subgroup and Cauchy's theorem.
For $|G|=42$, $n_7 \mid 6$ and $n_7 \equiv 1 \pmod 7$ gives $n_7=1$, so the Sylow $7$-subgroup $P \trianglelefteq G$ is normal and cyclic of order $7$, giving an element of order $7$. Since $\gcd(|P|, |G/P|) = \gcd(7,6) = 1$, the Schur-Zassenhaus theorem gives a complement subgroup $H \le G$ with $|H| = 6$ and $G = PH$, exhibiting the order-6 and order-7 structure of (D). Having independently eliminated (A), (B), (C), (D) is the correct option.
Step 5: Conclusion.\[\boxed{\text{Option (D)}}\]