Show that in any group of order $23 \cdot 24$, the $23$-Sylow subgroup is normal. 
Show that in any group of order $23 \cdot 24$, the $23$-Sylow subgroup is normal. 

Let $P_k$ denote the $k$-Sylow subgroup and let $n_3$ denote the number of conjugates of $P_k$. 
$n_2 \equiv 1 \mod 2$ and $n_2 | 69 \implies n_2= 1, 3, 23, 69$
$n_3 \equiv 1 \mod 3$ and $n_3 | 184 \implies n_3 = 1, 4, 184$
$n_{23} \equiv 1 \mod 23$ and $n_{23} | 24 \implies n_{23} = 1, 24$
Suppose for contradiction that $n_{23} = 24$. Then $N(P_{23})=552/24=23$. So the normalizer of $P_3$ only contains the identity and elements of order $23$.
It is impossible for $n_2$ to equal $23$ or $69$, or for $n_3$ to equal $184$, since we would then have more than $552$ elements in G. 
Case 1: 
Let $n_2 = 1$. Then we have $P_2 \triangleleft G$, and so we have a subgroup $H=P_2P_{23}$ in G. We know that $|H|=\frac{|P_2||P_{23}|}{|P_2 \cap P_{23}} = 184$. We also know that the $23$-Sylow subgroup of $H$ is normal, so it's normalizer is of order $184$. Since a $p$-sylow subgroup is the largest subgroup of order $p^k$ for some $k$; and since the order of the $23$-Sylow subgroup of H and G both equal $23$, they must coincide. But that means that elements of orders not equal to $23$ normalize $P_{23}$, which is a contradiction.
Case 2:
Suppose that $n_3=1$. Then we have $P_3 \triangleleft G$, and so we have a subgroup $K=P_2P_{23}$ in $G$. Since $|K|=\frac{|P_3||P_{23}|}{|P_2 \cap P_{23}}= 69$, the $23$-sylow subgroup of $K$ is normal. Again, as in case 1, we have an element of order not equal to $23$ that normalizes $P_{23}$. 
Case 3:
Let $n_3=4$ and $n_2=3$. But then $N(P_2)=184$. So this is the same as case 1, since we have a subgroup of G of order $184$.
Do you think my answer is correct?
Thanks in advance 
 A: As commented back in the day, the OP's solution is correct. Promoting a modified (IMHO simplified) form of the idea from Thomas Browning's comment to an alternative answer.
The group $G$ contains $23\cdot24-24\cdot22=24$ elements outside the Sylow $23$-subgroups. Call the set of those elements $X$. Clearly $X$ consists of full conjugacy classes. If $s\in G$ is an element of order $23$, then consider the conjugation action of $P=\langle s\rangle$ on $X$. The identity element is obviously in an orbit by itself. By Orbit-Stabilizer either $X\setminus\{1_G\}$ is a single orbit, or the action of $P$ has another non-identity fixed point in $X$.

*

*In the former case all the elements of $X\setminus\{1_G\}$ must share
the same order, being conjugates, in violation of Cauchy's theorem
implying that there must exist elements of orders $2$ and $3$ in $X$
at least.

*In the latter case $P$ is centralized, hence normalized, by a non-identity element $\in X$. But this contradicts the fact that $P$ is known to have $24=[G:P]$ conjugate subgroups, and hence must be self-normalizing, $N_G(P)=P$.

