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

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.

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 $8$.
• when you say that "every (non-empty) fiber of a homomorphism is a coset of the kernel", you mean the first isomorphism theorem, right? Because we know that $G/kerf \cong im f$ So the cosets of the kernel must have the same order as the corresponding elements in the image. Also when you say that "$f^{-1}(P_2)$ is a union of 4 pairwise disjoint sets of cardinality 2, meaning $f^{-1}(P_2)$ has order 8", you are just using the formula $|HK| = \frac{|H||K|}{|H \cap K|}$, right? – user58289 Jun 2 '13 at 0:11
• @Artus, yes it is the 1st iso thm. What does "cosets of the kernel must have the same order as the corresponding elements in the image" mean? The elements in the image are not sets, and their order in the group-element sense is irrelevant. For the last part, it is not even that complicated: for any map $f:X\to Y$, the preimage $f^{-1}[A]$ has size $\sum_{a\in A} |f^{-1}(a)|$. In this case, $2+2+2+2=8$. – anon Jun 2 '13 at 0:19
• @Artus: Yes, and no. In the first case, that fact is fundamentally tied up in the proof of the first isomorphism theorem. In the second case, I was acting purely set-theoretically: given any function $f:X\to Y$ and any subset $A$ of the range of $f$, we know that $$f^{-1}(A)=\bigcup_{a\in A}\{x\in X:f(x)=a\},$$ and that this union is a pairwise disjoint union of non-empty fibers. – Cameron Buie Jun 2 '13 at 0:21
• @Artus: An application of that set theoretic bit from my previous comment is that, if $f:X\to Y$ is an $n$-to-$1$ function--that is, if all the non-empty fibers under $f$ are of cardinality $n$--then for any subset $B$ of $Y$, we have $$|f^{-1}(B)|=n\cdot|B\cap\operatorname{range}(f)|.$$ This is pretty much exactly what we used here. This homomorphism was assumed to be a $2$-to-$1$ function, with $P_2$ a subset of the image, and we knew that $|P_2|=4,$ so concluded that $$|f^{-1}(P_2)|=2\cdot|P_2|=2\cdot4=8.$$ – Cameron Buie Jun 2 '13 at 0:50