Finding subfields with Galois group of order 20

I want to prove the following statement. I'd like some feedback on part (a) and maybe a hint for part (b).

Let $$K$$ be a Galois extension of $$F$$ with $$|\text{Gal}(K/F)|=20$$.

(a)Prove that there exists a subfield $$E$$ of $$K$$ containing $$F$$ with $$[E:F]=5$$. Proof.

As $$G=Gal(K/F)$$ is a group of order $$20=2^2\cdot 5$$, by Sylow's theorem, there is a Sylow 2-subgroup $$H$$ of $$G$$, ie $$[G:H]=5$$ and so by the fundamental theorem of Galois Theory there is an intermediate field $$E$$ with degree 5 over $$F$$.

(b) Determine whether there must also exist a subfield $$L$$ of $$K$$ containing $$F$$ with $$[L:F]=10$$.

• Can you find a group of order 20 with no subgroup of order 10? – Gerry Myerson May 28 at 0:39
• @GerryMyerson So I think I've got it. By Cauchy's theorem there is an element of order 2 in $G$, let $H$ be the group generated by this element, thus $[G:H]=10$ so by the fundamental theorem there exists subfield $L$ such that $[L:F]=10$. Is that good? – Jimmy2Goons May 28 at 3:18
• Looks good to me. – Gerry Myerson May 28 at 3:57