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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 $G$.

citadel
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Ryan
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  • 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
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    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

2 Answers2

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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.

mesel
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(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 $2$).

$^\dagger$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 $C_3\times S_3$.

citadel
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    A simpler non-Sylow way to show the existence of a subgroup of order $9$ is to consider the Cayley representation of $G$. There exists an element of order two, and that acts as a product of nine 2-cycles, an odd permutation. So the subgroup acting by even permutations has index two, and thus order $9$. See this for a generalization. – Jyrki Lahtonen Oct 14 '22 at 15:04