# Does every subgroup of an abelian group have to be abelian?

My original problem is to show that E/L is an abelian extension over L and L/F is an abelian extension over F, given that E/F is an abelian extension over F and that L is a normal extension of F such that $$F\subseteq L \subseteq E$$.

So far I have proved that E is a normal extension of F, E is a normal extension of L, and L is a normal extension of F. I know that to prove abelian extension I must also prove that Gal(E/L) is an abelian group. I have shown that Gal(E/L) $$\subseteq$$ Gal (E/F). In my mind it makes sense that I cannot lose commutativity therefore my subgroup must be Abelian too. How do I show this in a proof? Is it enough to show two elements in the subgroup must also exist in the larger group and that they must be commutative in the larger group? I feel like I know what needs to be done, just not how to phrase it.

Showing this is pretty easy. Take an abelian group $$G$$ with subgroup $$H$$. Then we know that, for all $$a,b\in H$$, $$ab=ba$$ since it must also hold in $$G$$ (as $$a,b \in G \ge H$$ and $$G$$ is given to be abelian).
If $$G$$ is an abelian group and $$H$$ is a subgroup, suppose $$x, y\in H$$. Then in particular $$x, y\in G$$, so $$xy=yx$$. Since $$x, y$$ were arbitrary, $$H$$ is abelian.