$|\mathrm{Gal}(L/K)| = 105$ and subgroups $A,B \leq \mathrm{Gal}(L/K)$ with $|A| = 21$ and $|B| = 35$. Prove that $L^{A}\cap L^{B} = K$.

Question

Let $L/K$ be an Galois extension such that $|\mathrm{Gal}(L/K)| = 105$ and subgroups $A,B \leq \mathrm{Gal}(L/K)$ with $|A| = 21$ and $|B| = 35$. Prove that $L^{A}\cap L^{B} = K$.

We have

$$L^{A} = \lbrace \alpha \in L \mid \sigma(\alpha)=\alpha,\quad \forall \sigma \in A \rbrace,$$ $$L^{B} = \lbrace \beta \in L \mid \tau(\beta)=\beta,\quad \forall \tau \in B \rbrace$$ and $$L^{A}\cap L^{B} = \lbrace \rho \in L \mid \sigma(\rho)=\rho=\tau(\rho)\rbrace.$$

Since for every $h \in \mathrm{Gal}(L/K)$, $h$ is a $K$-automorphism so, $K \subset L^{A}\cap L^{B}$.

I couldn't show that $K \supset L^{A}\cap L^{B}$. Can someone help me? I tried to use the Fundamental Theorem of Galois Theory, but I couldn't development anything.

Answer 1

By Fundamental Theorem of Galois Theory, we have $$[L^{A}:K] = 5$$ and $$[L^{B}:K] = 3.$$

Note that

$$[L^{A}:K] = [L^{A}:L^{A}\cap L^{B}][L^{A}\cap L^{B}:K] = 5$$ and $$[L^{B}:K] = [L^{B}:L^{A}\cap L^{B}][L^{A}\cap L^{B}:K] = 3.$$

Therefore, $[L^{A}\cap L^{B}:K]$ divides $3$ and $5$, then $[L^{A} \cap L^{B} : K] = 1$. Thus, $L^{A} \cap L^{B} = K$

Answer 2

You can do a slightly more group theoretic proof. We have that $L^A \cap L^B$ is fixed by both $A$ and $B$ and so it must be fixed by the subgroup generated by $A$ and $B$. Now we have that:

$$|AB| = \frac{|A||B|}{|A \cap B|} = \frac{735}{|A \cap B|}$$

Now note that $|A \cap B| = 1$ or $7$, as $|A \cap B|$ divides both $21$ and $ 35$. This means that we must have $|AB| \ge 105$ from above. But also $|AB| \subseteq \text{Gal}(L/K)$, so $|AB| \le 105$. This yields that $|AB| = 105$ and in fact $AB = \text{Gal}(L/K)$. So $L^A \cap L^B$ is fixed by the whole Galois group, so it must be $K$.

Hence the proof.