Show that in a group $G$ of order $165=3\cdot 5\cdot 11$, the center $Z(G)$ contains a subgroup of order 11.

Question

I have solved most of the problem, but I still can't figure out the last part.

If $P_{11}\in Syl_{11}(G)$, then $n_{11}\equiv 1 \pmod{11}$ and $n_{11} \mid 3\cdot 5$, i.e. $n_{11}=1$, so that $P_{11}$ is normal in $G$.

Hence an action of $G$ on $P_{11}$ induces an automorphism of $P_{11}$. The stabilizer of this action is $C_G(P_{11}$ (centralizer) and thus $G/C_G(P_{11})$ is isomorphic to a subgroup of $Aut(P_{11})$.

Since $Aut(P_{11})\cong (\mathbb{Z}/11\mathbb{Z})^{\times}\cong \mathbb{Z}/10\mathbb{Z}$, it follows that $|G/C_G(P_{11})|$ divides $10$.

On the other hand, since $P_{11}$ is abelian, $P_{11} \leq C_G(P_{11})$ and thus $|G/C_G(P_{11})|$ divides $165/11=15$.

Therefore, the two possibilities for $|G/C_G(P_{11})|$ are $1$ (in which case we are done) or $5$.

However, I can't figure out how $|G/C_G(P_{11})|=5$ would lead to $P_{11}\leq Z(G)$. Any help would be really appreciated.

Answer 1

As Sean Eberhard pointed out in the commentary, it is not true that the centre of such a group has elements of order $11$. Indeed, by applying the command $GAP$

AllSmallGroups(165);

we really make sure that there is exactly one non-Abelian group whose centre consists of three elements.

However, there is a simple way to prove this without using $GAP$.

Indeed, let $G$ be a group of order $165$. Let us prove that the centre of the group $G$ contains a subgroup of order 3. Hence it follows easily that there exists exactly one non-Abelian group of order $165$ and its centre consists of exactly three elements.

1. As the author of the question has already noticed, the Sylow $11$-subgroup $Q$ is normal in $G$.
2. The quotient group $G/Q$ is of order $15$ and it follows from Sylow's theorem that the group $G/Q$ is abelian.
3. It follows from (2) that $G$ has a normal subgroup $H$ of order $33$. Again by Sylow's theorem $H$ is abelian.
4. Let $P$ be a Sylow $3$-subgroup of group $H$. Further we can reason differently. A simpler one is. Since $H\leq N_G(P)$ and hence the number $n_3$ of Sylow $3$-subgroups в $G$ is $n_3=|G:N_G(P)|\leq5$ it follows that $n_3=1$.
5. It follows that $G$ contains a subgroup $F$ of order $15$, which as we already know is abelian. Now the order of the centraliser $|C_G(P)|$ is divisible by $11$, $5$, and $3$, so $G=C_G(P)$.