Let G be a group of order 24 that is not isomorphic to S4. Then
one of its Sylow subgroups is normal.
This is the proof from my textbook.
Proof
Suppose that the 3-Sylow subgroups are not normal. The number of 3-Sylow
subgroups is 1 mod 3 and divides 8. Thus, if there is more than one 3-Sylow subgroup,
there must be four of them.
Let X be the set of 3-Sylow subgroups of G. Then G acts on X by conjugation, so we
get a homomorphism $f : G → S(X) \cong S_4$. As we’ve seen in the discussion on G-sets, the
kernel of f is the intersection of the isotropy subgroups of the elements of X. Moreover,
since the action is that given by conjugation, the isotropy subgroup of H ∈ X is $N_G(H)$ (the normalizer of H in G). Thus,
$$ker f = \cap_{H \in X} N_G(H).$$
For H ∈ X, the index of $N_G(H)$ is 4, the number of conjugates of H. Thus, the order
of $N_G(H)$ is 6. Suppose that K is a different element of X. We claim that the order of
$N_G(H) \cap N_G(K)$ divides 2.
To see this, note that the order of $N_G(H) \cap N_G(K)$ cannot be divisible by 3. This is
because any p-group contained in the normalizer of a p-Sylow subgroup must be contained in the p-Sylow subgroup itself (Corollary 5.3.5). Since the 3-Sylow subgroups have prime
order here, they cannot intersect unless they are equal. But if the order of $N_G(H) \cap N_G(K)$ divides 6 and is not divisible by 3, it must divide 2.
In consequence, we see that the order of the kernel of f divides 2. If the kernel has
order 1, then f is an isomorphism, since G and $S_4$ have the same number of elements.
Thus, we shall assume that ker f has order 2. In this case, the image of f has order
12. But by Problem 2 of Exercises 4.2.18, $A_4$ is the only subgroup of $S_4$ of order 12, so we must have im f = $A_4$.
By Problem 1 of Exercises 4.2.18, the 2-Sylow subgroup, $P_2$, of $A_4$ is normal. But
since ker f has order 2, $f^{−1}P_2$ has order 8, and must be a 2-Sylow subgroup of G. As
the pre-image of a normal subgroup, it must be normal, and we’re done.
My Question
I'm just confused about the last part. I kind of got lost when it was explaining how/why $f^{-1}P_2$ has order 8. I'm not really sure how that's related to the kernel of f.
Thank you in advance
Best Answer
It has to do with the fact that every (non-empty) fiber of a homomorphism is a coset of the kernel. That is, if $\varphi:G\to H$ is a homomorphism, and $h\in\operatorname{im}\varphi,$ then the fiber of $h$ under $\varphi$ is the set $$\{g\in G:\varphi(g)=h\},$$ and is a coset of $\ker\varphi$ in $G$. I outline the proof of this fact (from a linear algebra standpoint) in my answer here, and not much changes in the more general case.
Since $\ker f$ has order two, then for any $\sigma\in S_4,$ we have $f^{-1}(\sigma)$ has cardinality either $2$ or $0$. Since we're assuming that $A_4=\operatorname{im}f,$ then for each $\sigma\in A_4$ (and in particular for each $\sigma\in P_2$) we have $f^{-1}(\sigma)$ has cardinality $2$. Since $P_2$ has $4$ elements by the referenced exercise, then $f^{-1}(P_2)$ is a union of $4$ pairwise disjoint sets of cardinality $2$, meaning that $f^{-1}(P_2)$ has order $8$.
Does that help?