# 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 $G$.

• Yes, all fine. A straightforward application of Sylow Theorems. Just observe that a Sylow $\;3$ - subgroup of $\;G\;$ has order $\;9\;$ not only because "nine apperas in the group order" (this can be highly confusing), but rather because two is the highest power of three that divides the order of the group. – Timbuc Oct 30 '14 at 11:30
• Is there anything else I need to do to show that the $3$-Sylow subgroup I found is the only subgroup of order $9$ in $G$. This was our midterm problem today, and someone else told me that I didn't do enough to show that this is the only group of order $9$ in $G$. – Ryan Oct 30 '14 at 11:37
• If ...you are allowed to rely on Sylow Theorems then you don't have to do anything else at all, imo. – Timbuc Oct 30 '14 at 11:42
• Any $3$-group $\le G$ is contained in some $3$-Sylow group, isn't it? – Hagen von Eitzen Oct 30 '14 at 12:25

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.