Sylow subgroups and homomorphism of $S_4$ onto $S_3$.

Question

"It's known that $S_3$ has exactly three 2-Sylows. Show that there exists a surjective homomorphism from $S_4$ onto $S_3$".

Well, I don't know how to start. Can someone help me with this exercise?

Answer 1

Hint: $S_3$ permutes three objects.There is only one set of three objects (the Sylow subgroups) in view. Consider the action of $S_4$ on the three Sylow subgroups by conjugation.

Answer 2

Let $\pi=\{P1, P2, P3\}$ be the set containing the $2$-Sylow subgroups of $S_4$. Let $\phi:S_4 \rightarrow S_\pi$ be the homomorphism induced by the action of $S_4$ on $\pi$ by conjugation. Since $S_\pi \approx S_3$ it's enough to show that $\phi$ is surjective. We will show that $|ran(\phi)|=6$ by showing that $|ker(\phi)|=4$.

We know that $g \in ker(\phi) \Leftrightarrow gP_1g^{-1}=P_1 \wedge gP_2g^{-1}=P_2 \wedge gP_3g^{-1}=P_3 \Leftrightarrow g \in \bigcap_{n=1}^3N(P_n)$. Notice that for each $i$, $3=\frac{24}{|N(P_i)|}$, therefore $|N(P_i)|=8$, so $N(P_i)=P_i$. So all that remains to show is that $|P_1 \cap P_2 \cap P_3|=4$.

Let $n=|P_1 \cap P_2 \cap P_3|$. Since $P_1 \cap P_2 \cap P_3\leq P_1$, $n$ is $1$, $2$, $4$ or $8$. $n$ cannot be 8 since we would have $P_1=P_2=P_3$. Since $|ran(\phi)|\leq6$, $n=|\ker(\phi)|\geq4$, therefore $n=4$.