Question
Let $G$ be a group of order $24$ having no normal subgroups of order $3$. Show that $G$ has four subgroups of order $6$.
Attempt
It is $24=3\cdot 2^3$ and $n_3\mid 2^3,\ n_3\equiv 1\pmod{3}\Rightarrow n_3=1,4$, $n_3\not=1$ from the hypothesis. Hence $n_3=4$. It is $n_2\mid 3$ and $n_2\equiv 1\pmod{2}\Rightarrow n_2=1,3$. If $n_2=1$ then there is a unique normal Sylow $2$-subgroup $P_2$ [and by taking the product with the $4$ Sylow $3$-subgroups we have the four subgroups of order $6$ needed.] {that's wrong}
Problem : If $n_2=3$ how could I proceed? EDIT: if $n_2=1 ?$
Best Answer
We know that $n_3=4$. Consider the conjugation action of $G$ on the four Sylow 3-subgroups of $G$, and let $I$ be the image of this action. So $I$ is a subgroup of $S_4$.
Since the Sylow subgroups are all conjugate in $G$, the action is transitive. So $|I|$ is divisible by 4. Also, since no Sylow subgroup can normalize another, the action of a Sylow 3-subgroup $P$ is a fixed point (i.e. $P$ itself), together with a 3-cycle. So $|I|$ is divisible by 3.
So $I$ is a subgroup of $S_4$ of order divisible by 12, and it must be $A_4$ or $S_4$.
Case 1. $I=A_4$. So the kernel $K$ of the action has order 2. Since $A_4$ has no subgroups of order $6$, the subgroups of order 6 in $G$ must contain $K$, and so they are the inverse images of the four subgroups of order $3$ in $I$. So there are four such subgroups in total, which are cyclic. (This is the case $n_2=1$.)
Case 2. $I=S_4$, so $I \cong G$, and as HallaSurvivor pointed out, there are also exactly four subgroups of order 6 (isomorphic to $S_3$)in this case. (This is the case $n_2=3$.)