Would the following brief argument suffice in showing that there are (up to isomorphism) only two groups of order $99$, being $Z_{99}$ or $Z_{3} \oplus Z_{33}$? I have seen rather "fuller" proofs (e.g., here), but I believe that certain arguments therein can be omitted.
- Suppose $G$ is a group of order $99 = 3^2 \times 11$.
- Let $H$ be a Sylow $3$-subgroup of $G$, let $K$ be a Sylow $11$-subgroup of $G$.
- By Sylow's Third Theorem, the number $n_3$ of such Sylow $3$-subgroups must be congruent to $1 \text{mod} 3$ and must divide $11$, i.e., $n_3 = 1$. Hence, as $H$ is the unique Sylow $3$-subgroup, $H \triangleleft G$.
- Since $H \triangleleft G, K \leq G$, $HK \leq G$. Moreover, since $\gcd(9, 11) = 1$, $H \cap K = \{e\}$. Thus, $$|HK| = \frac{|H||K|}{|H \cap K|} = 9 \times 11 = 99 = |G|,$$ at once implying that $HK = G$.
- Moreover, since $K$ is of prime order, $K$ is cyclic and thus Abelian. Since $H$ is of prime-squared order, $H$ is Abelian. Thus, $G = HK$ is Abelian.
- One can use the Fundamental Theorem of Finite Abelian Groups to conclude that $G \approx Z_{99}$ or $G \approx Z_3 \oplus Z_3 \oplus Z_{11} \approx Z_3 \oplus Z_{11}.$
In particular, I do not believe that it is required to show that $K \triangleleft G$, or that the elements of $H$ and $K$ themselves commute. (Or are these two actually needed to conclude that $G$ is Abelian?)
This step is missing an argument. You don't know that an arbitrary element of $H$ commutes with an arbitrary element of $K$ (i.e., you know that $G$ is the product of these two subgroups, but it might not be a direct product).