If $|G|=12$ with $|Z(G)|=2$, prove that the group $G$ has only one subgroup of order $3$.

I have to solve the following exercise:

If $G$ is a group of order $12$ with the center $Z(G)$ of order $2$, prove that $G$ has only one subgroup of order $3$.

My idea is, since $n_3$, the number of $3$-Sylow can be $1$ or $4$, if $n_3=4$, we have $8$ elements of order $3$ thus an unique subgroup of order $4$, so normal. For each element $x \notin Z(G)$, $|C(x)|=4$ or $|C(x)|=6$, where $|C(x)|=\{g \in G \;|\; gx=xg\}$. From the classes equation $$|G| = |Z(G)| + \sum_{x \notin Z(G)} \frac{|G|}{|C(x)|} \quad \Longrightarrow \quad 10 =3\lambda + 2\mu$$ with solution $(0,5)$ or $(2,2)$. So we have $5$ or $4$ different conjugacy classes. If I'm not wrong, all $3$-Sylow are conjugate thus all $8$ elements of order $3$ are in the same conjugacy class, the other two element out the center are in at most $2$ conjugacy classes. From this can I deduce the absurd?

Solution 1:

Simpler argument: What is the order of normalizer of a sylow 3-group $H$? It contains $H$ and $Z(G),$ so has to be divisible by $6.$ So, it's index is $1$ or $2$. Since we cannot have $2$ Sylow 3-subroups, we must have $1.$