# Prove that there is no simple group of order $48$.

I wrote a proof of the aforementioned statement. I am aware that there is another post similar to this one, but my proof differs significantly from all of the answers in that post, and I'd like to know if my proof is valid.

By the third Sylow theorem, the number of Sylow 2-subgroups $n_2$ divides $3$, and $n_2 \equiv 1 \mod 2$, so $n_2 \in \{ 1, 3 \}$. Similarly, $n_3$ divides $16$, and $n_3 \equiv 1 \mod 3$. So, $n_3 \in \{ 1,4,16 \}$. If $n_2 = 1$, we are done, as the Sylow 2-subgroup is the desired normal subgroup. If $n_2 = 3$, take any two Sylow 2-subgroups $P$ and $Q$. Their intersection $P \cap Q$ has order either equal to $1, 2, 4$ or $8$. Using the formula $|PQ| = \frac{|P||Q|}{|P \cap Q|}$, we can eliminate $1$, $2$ and $4$ (since in these cases, $|PQ|$ is greater than the order of the group itself). So, it must be the case that $|P \cap Q| = 8$. Then, since $P \cap Q$ has index 2 in both $P$ and $Q$, it must be the case that $P \cap Q$ is normal in $P$ and $Q$. So, the normalizer of $P \cap Q$, which I denote $N(P \cap Q)$, has order at least $|P \cup Q| = 24$, and the order divides $48$. So, $N(P \cap Q)$ is either $24$ or $48$. If it is $24$, then $N(P \cap Q)$ is normal in $G$ (since it has index $2$). If it is $48$, then $P \cap Q \triangleleft G$. This completes the proof.

• Looks good. Note that you don't really need to "number" the Sylow theorems: they are a collection of theorems, and people in different parts of the world number them differently (or don't number them at all), so it is slightly useless to do so. – Pedro Tamaroff Nov 8 '15 at 17:16
• Why not add your proof as a new answer on the other Question? You don't seem to be clearly asking a new math question. – hardmath Nov 8 '15 at 17:16
• The proofs given here are much shorter and better to understand. – Dietrich Burde May 29 at 8:48