Need help in understanding a certain step of a certain proof in finite group theory and group actions

Question

A proof is from Aluffi's textbook "Algebra: Chapter 0".

A statement:

There are no simple groups of order $24$.

The proof from the book:

Let $G$ be a group or order $24 = 2^33$, and consider its $2$-Sylow subgroups; by the third Sylow theorem, there are either $1$ or $3$ such subgroups. If there is $1$, the $2$-Sylow subgroup is normal, and $G$is not simple. Otherwise, $G$ acts (nontrivially) by conjugation on this set of three $2$-Sylow subgroups; this action gives a nontrivial homomorphism $G \to S_3$, whose kernel is a proper, nontrivial normal subgroup of $G$-thus again $G$ is not simple.

What I don't understand is why the action of $G$ by conjugation on the set of $3$ $2$-Sylow subgroups of $G$ is not trivial and why kernel of $G \to S_3$ is proper and nontrivial.

Answer 1

The action is non-trivial because for a fixed $p$, all Sylow-$p$ subgroups of a group $G$ are conjugate to one another. This is the second Sylow theorem.

So the kernel of $G\to S^3$ is proper, since there are elements of $G$ that does a non-trivial permutation of the three Sylow-$2$ subgroups. The kernel is also non-trivial, since it's a homomorphism from a group of $24$ elements to a group of $6$ elements, so it cannot be injective.

Answer 2

The second Sylow theorem says that all of the Sylow $2$-subgroups are conjugate to one another. If they are $H_1,H_2$, and $H_3$, there are $g_2,g_3\in G$ such that $H_2=H_1^{g_2}$ and $H_3=H_1^{g_3}$, and of course then $H_3=H_2^{g_2^{-1}g_3}$. Thus, the action is clearly not trivial.

The kernel is proper, since the action doesn’t collapse the group, and it’s non-trivial, since $|S_3|=6<24=|G|$.