Let G be a group of order 24 that is not isomorphic to S4. Then one of its Sylow subgroups is normal.

Question

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

Answer 1

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?