Group of order $18$ contains exactly one subgroup of order $9$

Question

I'm trying to prove the following:

Proposition: A finite group $G$ of order $18$ has a unique subgroup of order $9$.

Here is my attempt:

Observe that $18 = 3^2 \times 2$. Let's count the number of $3$-Sylow subgroups in $G$ (which have order $9$ since $3^2$ appears in the group order). Let $n_3$ denote the number of $3$-Sylow subgroups in $G$. By Sylow's Theorem, $n_3 \equiv 1 \ (\text{mod 3})$ and $n_3 \ | \ 2$. This implies that $n_3 = 1$. Thus, there is a unique $3$-Sylow subgroup of order $9$. By a theorem, this means that there is a single subgroup of order $9$ in $G$.

Answer 1

Without using Sylow's theorem;

Assume we have two $H,K$ with order $9$ then $|HK|=\dfrac{|H||K|}{|H\cap K|}\geq 27> 18$ as $|H\cap K|$ at most $3$. Thus, it is impossible we are done.

Answer 2

(By Sylow I, there is a subgroup of order $9$. By Sylow III, it is unique.)

Without Sylow. The other answer nicely shows the uniqueness of a subgroup of order $9$, but not its existence. As for this latter, a Sylow-free argument is the following.

1. If $G$ is abelian, then it has a subgroup of order any divisor of $|G|$ (see e.g. here).
2. If $|Z(G)|=9$, then we are done.
3. If $|Z(G)|=6$, then $G/Z(G)\cong C_3$, and hence $G$ is abelian: contradiction. So, there's no group of order $18$ with center of order $6$.
4. If $|Z(G)|=3$, then necessarily $\operatorname{Inn}(G)\cong$ $G/Z(G)\cong$ $S_3$, which has (class-preserving) elements of order $2$, and hence there are conjugacy classes of size $2$$^\dagger$, and finally centralizers of order $9$$^{\dagger\dagger}$.
5. If $|Z(G)|=2$, then noncentral elements' centralizers can have order $6$, only. Therefore, the class equation yields: $$18=2+3l$$ contradiction, because $3\nmid 16$. So, there's no group of order $18$ with center of order $2$.
6. If $|Z(G)|=1$, then elements' centralizers can have order any divisor of $18$, and hence the class equation yields: $$18=1+9k+6l+3m+2n \tag 1$$ If $n\ne 0$ we are done, because then there are centralizers of order $9$. If $n=0$, then $(1)$ leads to a contradiction, as $3\nmid 17$ (so, there are no centerless groups of order $18$ without any conjugacy class of size $2$).

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$^\dagger$Otherwise, $G\setminus Z(G)$ would be split into $5$ classes of size $3$; but then an inner automorphism of order $2$ would fix one element in each class, and hence there would be some $a\in G$ such that $|C_G(a)|=8$, contradiction because $8\nmid 18$.

$^{\dagger\dagger}$Actually such a $G$ does exist: it is $C_3\times S_3$.