Group of order $18$ contains exactly one subgroup of order $9$ 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 $G$.
 A: 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.
A: (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, but not the existence of a subgroup of order $9$. As for this latter, a Sylow-free argument is the following.

*

*If $G$ is abelian, then it has a subgroup of order any divisor of $|G|$ (see e.g. here).

*If $|Z(G)|=9$, then we  are done.

*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 $6$.

*If $|Z(G)|=3$, then the noncentral conjugacy classes have size any among $6$, $3$ and $2$. There must be conjugacy classes of size $3$, because otherwise the class equation would yield:
$$18=3+6k+2l$$
and $2\nmid 15$. Now, since $|G/Z(G)|=6$ and $G/Z(G)\ncong C_6$ (see point 3), we get $\operatorname{Inn}(G)\cong$ $G/Z(G)\cong$ $S_3$, which has elements of order $2$. Therefore, there must be conjugacy classes of even size as well (either $2$ or $6$). If there are conjugacy classes of size $2$, then we are done because the centralizers have then order $9$. If there are conjugacy classes of size $6$ (whose elements have centralizers of order $3$), then there are at least $6$ elements of order $3$, and hence at least $3$ subgroups of order $3$. So, said $H$ some subgroup of order $3$ distinct from $Z(G)$, $Z(G)H$ is a subgroup of $G$ of order $9$$^\dagger$.

*If $|Z(G)|=2$, then elements' centralizers can have order $2$ or $6$, only. Therefore, the class equation yields:
$$18=2+9k+3l$$
contradiction, because $3\nmid 16$. So, there's no group of order $18$ with center of order $2$.

*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 $2$).


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